《后缀数组练习》

SP694 DISUBSTR - Distinct Substrings

求本质不同的子串数量。

考虑容斥：对于rk[i] 与 rk[i -1]他们的lcp就是会重复出现的子串数量。就是这段会被重复计算在两个人的前缀中，所以减去。

对于rk[i] 和 rk[i - 2]会被减掉在rk[i - 1]

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 1e3 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

char s[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
void solve() {
    int ca;ca = read();
    while(ca--) {
        scanf("%s",s + 1);
        n = strlen(s + 1);
        m = 122;
        SA();
        LCP();
        LL ans = 1LL * n * (n + 1) / 2;
        for(int i = 1;i <= n;++i) ans -= height[i];
        printf("%lld\n",ans);
    }
}
int main() {
    solve();
    //system("pause");
    return 0;
}

View Code

P2870 [USACO07DEC]Best Cow Line G

显然这题有一个思路就是从两端贪心取，如果一样就比较里面，一直比到不一样。

比较的操作暴力复杂度过高，我们可以发现，这个过程其实就是比较原串的后缀和原串反转之后的后缀的大小。

我们可以把原串反转接入原串的结尾，注意在中间插入一个无穷小字符，这样不影响后缀排序。然后直接比较rk就行了。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 1e6 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

char s[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
vector<char> ans;
void solve() {
    n = read();
    string t;
    for(int i = 1;i <= n;++i) {
        char c;scanf("%c",&c);
        getchar();
        t += c;
    }
    string ss = t;
    ss += 'A' - 1;
    reverse(t.begin(),t.end());
    ss += t;
    n = ss.size();
    m = 122;
    for(int i = 1;i <= n;++i) s[i] = ss[i - 1];
    SA();
    LCP();
    int L = 1,r = (n - 1) / 2,tot = 0;
    while(L <= r) {
        //printf("L is %d r is %d rk[L] is %d rk[r] is %d\n",L,r,rk[L],rk[n - r + 1]);
        if(rk[L] <= rk[n - r + 1]) printf("%c",s[L++]);
        else printf("%c",s[r--]);
        ++tot;
        if(tot == 80) printf("\n"),tot = 0;
    }
}
int main() {
    solve();
    system("pause");
    return 0;
}

View Code

P4051 [JSOI2007]字符加密

这题其实不难：就是对环形字符串进行一下后缀排序就可以了。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 2e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

char s[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
vector<pii> vec;
bool cmp(pii a,pii b) {
    return a.first < b.first;
}
void solve() {
    string t;cin >> t;
    t += t;
    n = t.size();
    for(int i = 1;i <= n;++i) s[i] = t[i - 1];
    m = 122;
    SA();
    LCP();
    int x = n / 2;
    for(int i = 1;i <= x;++i) {
        vec.push_back(pii{rk[i],i});
    }
    sort(vec.begin(),vec.end(),cmp);
    for(auto v : vec) {
        printf("%c",s[v.second + x - 1]);
    }
}
int main() {
    solve();
    system("pause");
    return 0;
}

View Code

P2852 [USACO06DEC]Milk Patterns G

求可重叠的出现至少k次的子串。

对于连续的k - 1个rk，他们的min(height)就是他们出现的最小重叠子串。

我们求一下所有的连续k - 1的min(height)的max即可。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 2e4 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

int s[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
int a[N],b[N];
deque<int> Q;
void solve() {
    int k;n = read(),k = read();
    for(int i = 1;i <= n;++i) a[i] = read(),b[i] = a[i];
    sort(b + 1,b + n + 1);
    int len = unique(b + 1,b + n + 1) - b - 1;
    for(int i = 1;i <= n;++i) {
        a[i] = lower_bound(b + 1,b + len + 1,a[i]) - b;    
    }
    for(int i = 1;i <= n;++i) s[i] = a[i];
    m = 122;
    SA();
    LCP();
    k--;
    int ans = 0;
    for(int i = 1;i <= n;++i) {
        while(!Q.empty() && i - Q.front() >= k) Q.pop_front();
        while(!Q.empty() && height[Q.back()] >= height[i]) Q.pop_back();
        Q.push_back(i);
        if(i >= k) ans = max(ans,height[Q.front()]);
    }
    printf("%d\n",ans);

}
int main() {
    solve();
    system("pause");
    return 0;
}

View Code

P4248 [AHOI2013]差异

这里我们前面的两个的和还是很好处理的，主要是求后面的那个lcp。

考虑后缀排序之后的所有后缀：因为对于两个后缀lcp(x,y) = min{height[i]} (i = [x,y])

那么我们可以考虑每个height的贡献，很显然height的贡献就是一段区间满足里面的都height都>=他，那么这个区间内的任意横跨i的子区间都满足条件。

这是个很经典的维护操作，我们考虑单调栈维护出L[i]表示左边<=他的最小的位置，r[i]表示右边<他的最小的位置，这里我们右边不取到等号，是因为防止端点重复计算。

然后横跨的区间数量就是左半边任选一点 * 右半边任选一点的方案数。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 5e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

char s[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
stack<int> S;
LL dp[N];
void solve() {
    scanf("%s",s + 1);
    m = 122;
    n = strlen(s + 1);
    SA();
    LCP();
    LL ans = 1LL * n * (n + 1) * (n - 1) / 2;
    for(int i = 1;i <= n;++i) {
        int t = 0;
        while(!S.empty() && height[S.top()] > height[i]) S.pop();
        if(!S.empty()) t = S.top();
        dp[i] = dp[t] + 1LL * (i - t) * height[i];
        ans -= dp[i] * 2LL;
        S.push(i);
    }
    printf("%lld\n",ans);
}
int main() {
    solve();
    system("pause");
    return 0;
}

View Code

P3181 [HAOI2016]找相同字符

这里其实和上面一题很像：首先很显然考虑容斥。

字符串拼接之后产生的相同子串数量 - 没拼接前产生的相同子串数量。

那么主要问题就变成了求相同子串数量。因为这里没有规定长度，显然对于任意的height都满足条件。

那么问题也就变成了求$\sum_{i = 1}^{n}\sum_{j = i + 1}^{n}lcp(i,j)$

那么就是上题是一样的东西了。注意中间特殊字符拼接。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 5e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

char s[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
stack<int> S;
int L[N],r[N];
LL query() {
    n = strlen(s + 1);
    m = 122;
    SA();
    LCP();
    while(!S.empty()) S.pop();
    LL ans = 0;
    for(int i = 1;i <= n;++i) L[i] = i - 1,r[i] = i + 1;
    for(int i = 1;i <= n;++i) {
        while(!S.empty() && height[S.top()] > height[i]) {
            r[S.top()] = i;
            S.pop();
        }
        if(!S.empty()) L[i] = S.top();
        S.push(i);
    }
    while(!S.empty()) r[S.top()] = n + 1,S.pop();
    for(int i = 2;i <= n;++i) ans += 1LL * (i - L[i]) * (r[i] - i) * height[i];
    //dbg(ans);
    return ans;
}   
void solve() {
    string s1,s2;cin >> s1 >> s2;
    for(int i = 1;i <= s1.size();++i) s[i] = s1[i - 1];
    LL ma1 = query();
    for(int i = 1;i <= s2.size();++i) s[i] = s2[i - 1];
    s[s2.size() + 1] = '\0';
    LL ma2 = query();
    s1 += 'A' - 1;
    s1 += s2;
    for(int i = 1;i <= s1.size();++i) s[i] = s1[i - 1];
    s[s1.size() + 1] = '\0';
    LL ans = query() - ma1 - ma2;
    printf("%lld\n",ans);
}
int main() {
    solve();
    system("pause");
    return 0;
}

View Code

P4070 [SDOI2016]生成魔咒

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 5e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e9
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

int s[N],b[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
struct Node{int L,r,mi;}node[N << 2];
void Pushup(int idx) {node[idx].mi = min(node[idx << 1].mi,node[idx << 1 | 1].mi);}
void build(int L,int r,int idx) {
    node[idx].L = L,node[idx].r = r,node[idx].mi = INF;
    if(L == r) {
        node[idx].mi = height[L];
        return;
    }
    int mid = (L + r) >> 1;
    build(L,mid,idx << 1);
    build(mid + 1,r,idx << 1 | 1);
    Pushup(idx);
}
int query(int L,int r,int idx) {
    if(node[idx].L >= L && node[idx].r <= r) return node[idx].mi;
    int mid = (node[idx].L + node[idx].r) >> 1,ans = INF;
    if(mid >= L) ans = min(ans,query(L,r,idx << 1));
    if(mid < r) ans = min(ans,query(L,r,idx << 1 | 1));
    return ans;
} 
set<int> S;
void solve() {
    n = read();
    for(int i = n;i >= 1;--i) s[i] = read(),b[i] = s[i];
    sort(b + 1,b + n + 1);
    int len = unique(b + 1,b + n + 1) - b - 1;
    for(int i = n;i >= 1;--i) {
        int pos = lower_bound(b + 1,b + len + 1,s[i]) - b;
        s[i] = pos;
    }
    m = 122;
    SA();
    LCP();
    build(1,n,1);
    LL ans = 0;
    for(int i = n;i >= 1;--i) {
        S.insert(rk[i]);
        int pre = -1,nxt = -1;
        auto it = S.lower_bound(rk[i]);
        if(it != S.begin()) pre = *(--it);
        it = S.upper_bound(rk[i]);
        if(it != S.end()) nxt = *it;
        ans += n - i + 1;
        if(pre != -1) ans -= query(pre + 1,rk[i],1);
        if(nxt != -1) ans -= query(rk[i] + 1,nxt,1);
        if(pre != -1 && nxt != -1) ans += query(pre + 1,nxt,1);
        printf("%lld\n",ans);
    }

}
int main() {
    solve();
    system("pause");
    return 0;
}

View Code

ST表:

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 5e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e9
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

int s[N],b[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
int f[N][20],lg[N];
set<int> S;
void init() {
    lg[1] = 0;for(int i = 2;i <= n;++i) lg[i] = lg[i >> 1] + 1;
    for(int i = 1;i <= n;++i) f[i][0] = height[i];
    for(int j = 1;j <= 19;++j) {
        for(int i = 1;i <= n && i + (1 << j) - 1 <= n;++i) {
            f[i][j] = min(f[i][j - 1],f[i + (1 << (j - 1))][j - 1]);
        }
    }
}
int query(int x,int y) {
    int k = lg[y - x + 1];
    return min(f[x][k],f[y - (1 << k) + 1][k]);
}
void solve() {
    n = read();
    for(int i = n;i >= 1;--i) s[i] = read(),b[i] = s[i];
    sort(b + 1,b + n + 1);
    int len = unique(b + 1,b + n + 1) - b - 1;
    for(int i = n;i >= 1;--i) {
        int pos = lower_bound(b + 1,b + len + 1,s[i]) - b;
        s[i] = pos;
    }
    m = 122;
    SA();
    LCP();
    init();
    LL ans = 0;
    for(int i = n;i >= 1;--i) {
        S.insert(rk[i]);
        int pre = -1,nxt = -1;
        auto it = S.lower_bound(rk[i]);
        if(it != S.begin()) pre = *(--it);
        it = S.upper_bound(rk[i]);
        if(it != S.end()) nxt = *it;
        ans += n - i + 1;
        if(pre != -1) ans -= query(pre + 1,rk[i]);
        if(nxt != -1) ans -= query(rk[i] + 1,nxt);
        if(pre != -1 && nxt != -1) ans += query(pre + 1,nxt);
        printf("%lld\n",ans);
    }

}
int main() {
    solve();
    system("pause");
    return 0;
}

View Code

SP1811 LCS - Longest Common Substring

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 5e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e9
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

char s[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
stack<int> S;
int f[N][20];
int lg[N],st,r[N],L[N];
void init() {
    lg[1] = 0;for(int i = 2;i <= n;++i) lg[i] = lg[i >> 1] + 1;
    for(int i = 1;i <= n;++i) f[i][0] = height[i];
    for(int j = 1;j <= 19;++j) {
        for(int i = 1;i <= n && i + (1 << j) - 1 <= n;++j) {
            f[i][j] = min(f[i][j - 1],f[i + (1 << (j - 1))][j - 1]);
        }
    }
}
void solve() {
    string s1,s2;cin >> s1 >> s2;
    st = s1.size();
    s1 += 'A' - 1;
    s1 += s2;
    n = s1.size();
    for(int i = 1;i <= n;++i) s[i] = s1[i - 1];
    m = 122;
    SA();
    LCP();
    int ans = 0;
    for(int i = 2;i <= n;++i) {
        if(sa[i] <= st && sa[i - 1] > st || sa[i] > st && sa[i - 1] <= st) ans = max(ans,height[i]);
    }
    printf("%d\n",ans);
}
int main() {
    solve();
    system("pause");
    return 0;
}

View Code

SP705 SUBST1 - New Distinct Substrings

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 5e4 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

char s[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
void solve() {
    int ca;ca = read();
    while(ca--) {
        scanf("%s",s + 1);
        n = strlen(s + 1);
        m = 122;
        SA();
        LCP();
        LL ans = 1LL * n * (n + 1) / 2;
        for(int i = 1;i <= n;++i) ans -= height[i];
        printf("%lld\n",ans);
    }
}
int main() {
    solve();
    //system("pause");
    return 0;
}

View Code

P5341 [TJOI2019]甲苯先生和大中锋的字符串

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 1e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-10;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e9
#define dbg(ax) cout << "now this num is " << ax << endl;
inline int read() {
    int f = 1;int x = 0;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
    return x*f;
}
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

char s[N];
int x[N],y[N],c[N],sa[N],rk[N],height[N],n,m;
void SA() {
    memset(c,0,sizeof(c));
    for(int i = 1;i <= n;++i) x[i] = s[i],c[x[i]]++;
    for(int i = 2;i <= m;++i) c[i] += c[i - 1];
    for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
    for(int k = 1;k <= n;k = (k << 1)) {
        int num = 0;
        for(int i = n - k + 1;i <= n;++i) y[++num] = i;
        for(int i = 1;i <= n;++i) {
            if(sa[i] > k) {
                y[++num] = sa[i] - k;
            }
        }
        for(int i = 1;i <= m;++i) c[i] = 0;
        for(int i = 1;i <= n;++i) c[x[i]]++;
        for(int i = 2;i <= m;++i) c[i] += c[i - 1];
        for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i],y[i] = 0;
        swap(x,y);
        num = 1,x[sa[1]] = 1;
        for(int i = 2;i <= n;++i) {
            if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
            else x[sa[i]] = ++num;
        }
        if(num == n) break;
        m = num;
    }
}
void LCP() {
    int k = 0;
    for(int i = 1;i <= n;++i) rk[sa[i]] = i;
    for(int i = 1;i <= n;++i) {
        if(rk[i] == 1) {
            height[1] = 0;
            continue;
        }
        if(k) k--;
        int j = sa[rk[i] - 1];
        while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
    }
}
int f[N][20],lg[N];
void init() {
    lg[1] = 0;for(int i = 2;i <= n;++i) lg[i] = lg[i >> 1] + 1;
    for(int i = 1;i <= n;++i) f[i][0] = height[i];
    for(int j = 1;j <= 19;++j) {
        for(int i = 1;i <= n && i + (1 << j) - 1 <= n;++i) {
            f[i][j] = min(f[i][j - 1],f[i + (1 << (j - 1))][j - 1]);
        }
    }
}
int query(int x,int y) {
    if(x == y) return height[x];
    int k = lg[y - x + 1];
    return min(f[x][k],f[y - (1 << k) + 1][k]);
}
int cnt[N];
void solve() {
    int ca;ca = read();
    while(ca--) {
        string s1;cin >> s1;
        n = s1.size();
        for(int i = 1;i <= n;++i) s[i] = s1[i - 1];
        int k;k = read();
        m = 122;
        SA();
        LCP();
        height[n + 1] = 0;
        init();
        memset(cnt,0,sizeof(cnt));
        for(int i = 1;i <= n && i + k - 1 <= n;++i) {
            int L = i,r = i + k - 1,lx,rx;
            if(L + 1 > r) rx = n - sa[r] + 1;
            else rx = query(L + 1,r); 
            lx = max(height[L],height[r + 1]) + 1;
           // printf("lx is %d rx is %d\n",lx,rx);
            if(lx <= rx) cnt[lx]++,cnt[rx + 1]--;
        }
        for(int i = 1;i <= n;++i) cnt[i] += cnt[i - 1];
        int mx = 1,ans = -1;
        for(int i = 1;i <= n;++i) {
            mx = max(mx,cnt[i]);
            if(cnt[i] >= mx) {
                ans = i;
            }
        }
        printf("%d\n",ans);
    }
}
int main() {
    solve();
    system("pause");
    return 0;
}

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posted @ 2021-09-23 18:32 levill 阅读(37) 评论(0) 收藏举报

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