hdu 5769 Substring 后缀数组 + KMP

http://acm.hdu.edu.cn/showproblem.php?pid=5769

题意：在S串中找出X串出现的不同子串的数目? 其中1 <= |S| < $10^5$

官方题解： 处理出后缀数组中的sa[]数组和height[]数组。在不考虑包含字符X的情况下，不同子串的个数为

如果要求字符X，只需要记录距离sa[i]最近的字符X的位置(用nxt[sa[i]]表示)即可，个数

理解：后缀数组height[i]就是sa[i]与sa[i-1]的LCP,在后缀数组中求解全部的不同子串(之前只写过SAM处理所有不同子串..)还是比较好理解的，在一定要含有子串x时，需要先kmp求出所有匹配的位置，在处理到第i个后缀时，取max即可表示一定含有X,并且是不同的子串;

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define pb push_back
#define MK make_pair
#define A first
#define B second
#define eps 1e-8
#define zero(x) (((x)>0?(x):-(x))<eps)
#define bitnum(a) __builtin_popcount(a)
#define lowbit(x) (x&(-x))
#define clear0 (0xFFFFFFFE)
#define mod 1000000007
#define K(x) ((x)*(x))
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
template<typename T>
void read1(T &m)
{
    T x = 0,f = 1;char ch = getchar();
    while(ch <'0' || ch >'9'){ if(ch == '-') f = -1;ch=getchar(); }
    while(ch >= '0' && ch <= '9'){ x = x*10 + ch - '0';ch = getchar(); }
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>9) out(a/10);
    putchar(a%10+'0');
}
inline ll gcd(ll a,ll b){ return b == 0? a: gcd(b,a%b); }
template<class T1, class T2> inline void gmax(T1& a, T2 b){ if(a < b) a = b;}
template<class T1, class T2> inline void gmin(T1& a, T2 b){ if(a > b) a = b;}
const int maxn = 100001;
int sa[maxn], t[maxn],t2[maxn],c[maxn],w[maxn];
int cmp(int *r,int a,int b,int l)
{
    return r[a] == r[b] && r[a+l] == r[b+l];
}
void build_sa(char *r,int n,int m)
{
    int i, j, p, *x = t, *y = t2;
    for(i = 0; i < m;i++) c[i] = 0;
    for(i = 0; i < n;i++) c[x[i] = r[i]]++;
    for(i = 1; i < m;i++) c[i] += c[i-1];
    for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;

    for(j = 1, p = 1; p < n; j <<= 1, m = p){
        for(p = 0, i = n - j; i < n;i++) y[p++] = i;
        for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
        for(i = 0; i < n; i++) w[i] = x[y[i]];
        for(i = 0; i < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[w[i]]++;
        for(i = 0; i < m; i++) c[i] += c[i-1];
        for(i = n-1; i >= 0; i--) sa[--c[w[i]]] = y[i];
        swap(x,y);
        for(p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i-1], sa[i], j)? p-1: p++;
        if(p >= n) break;
        m = p;
    }
}
char S[maxn], X[maxn];
int  height[maxn], rk[maxn];
void getHeight(int n)
{
    for(int i = 1; i<= n; i++) rk[sa[i]] = i;
    for(int i = 0, j, k = 0; i < n; height[rk[i++]] = k)
        for(k? k--:0, j = sa[rk[i] - 1]; S[i+k] == S[j+k]; k++);
}

int f[maxn];
void getfail(char *p)
{
    f[0] = f[1] = 0;
    int n = strlen(p);
    for(int i = 1;i < n;i++){
        int j = f[i];
        if(j && p[i] != p[j]) j = f[j];
        f[i+1] = (p[i] == p[j] ?j+1:0);// ｉ＋１会递推到第ｎ位
    }
}
vector<int> vec;
void Find(char *T, char *p)
{
    vec.clear();
    ll j = 0,n = strlen(T),m = strlen(p);
    for(int i = 0;i < n;i++){
        while(j && T[i] != p[j]) j = f[j];
        if(T[i] == p[j]) j++;
        if(j == m){
            vec.pb(i);
            j = 0;
            i -= m-1;
        }
    }
    sort(vec.begin(), vec.end());
}

int main()
{
    //freopen("data.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T, kase = 1;
    scanf("%d",&T);
    while(T--){
        scanf("%s%s", X, S);
        int len = strlen(S), m = strlen(X);
        S[len] = '#';S[len+1] = 0;
        build_sa(S,len+1,'z'+1);
        getHeight(len);

        getfail(X);
        Find(S,X);
        vec.pb(len);
        ll ans = 0;
        rep1(i,1,len){
            if(sa[i]+m-1 > len) continue;
            int nxt = lower_bound(vec.begin(), vec.end(), sa[i]+m-1) - vec.begin();
            ans += len - max(vec[nxt],sa[i] + height[i]);
        }
        printf("Case #%d: %lld\n", kase++, ans);
    }
    return 0;
}