UOJ Round #15 [构造 | 计数 | 异或哈希 kmp]

UOJ Round #15

大部分题目没有AC,我只是水一下部分分的题解...


225【UR #15】奥林匹克五子棋

  • 题意:在n*m的棋盘上构造k子棋的平局

  • 题解:

    玩一下发现k=1, k=2无解,然后间隔着,上下两行相同:

    010101
    010101
    101010
    101010

    这样构造下来就行了。

    然后要特判n=1 或 m=1,这时候k=2可以有解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 505;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m, k, a[N][N], t[N][N];
void paint(int *a, int f) {for(int j=1; j<=m; j++) a[j] = (j&1) ^ f;}
inline void _walk(int &x, int &y) {
    y++;
    if(y == m+1) x++, y=1;
}
void walk(int &x, int &y, int f) {
    while(a[x][y] != f) _walk(x, y);
}

void solve1() {
    if(k == 1) {puts("-1"); return;}
    if(n == 1) for(int i=1; i<=m; i++) printf("%d %d\n", 1, i);
    else for(int i=1; i<=n; i++) printf("%d %d\n", i, 1);
}   
int main() {
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
    n = read(); m = read(); k = read();
    if(min(n, m) == 1) {solve1(); return 0;}
    if(k == 1 || k == 2) {puts("-1"); return 0;}

    int flag = 0;
    if((m&1) && (~n&1)) swap(n, m), flag = 1;
    int lim = (n>>1) + 1;
    for(int i=1; i <= lim; i++) paint(a[(i<<1)-1], i&1), paint(a[i<<1], i&1);
    if(flag) {
        for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) t[j][i] = a[i][j];
        swap(n, m);
        for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) a[i][j] = t[i][j];
    }
    
    int b = 0, w = 0;
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++) {
            if(a[i][j] == 1) b++;
            else w++;
        }
    if(b < w) b = 0, w = 1; else b = 1, w = 0;
    
    int bx = 1, by = 1, wx = 1, wy = 1;
    
    lim = n * m;
    for(int i=1; i <= lim; i++) {
        if(i & 1) walk(bx, by, b), printf("%d %d\n", bx, by), _walk(bx, by);
        else walk(wx, wy, w), printf("%d %d\n", wx, wy), _walk(wx, wy);
    }
    
    //puts("\ntest");
    //for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) printf("%d%c", a[i][j], j==m ? '\n' : ' ');
}

226【UR #15】奥林匹克环城马拉松

  • 题意:有向 树 / 环 / 基环树 的欧拉回路计数
  • 题解:题解太神了看不懂啊,我只能看懂树的...
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
const int N = 1e5+5, P = 998244353;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m, u, v, w, de[N], t[N];
struct edge {int v, ne;} e[N<<1];
int cnt, h[N];
inline void ins(int u, int v, int w) {
    e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;
    e[++cnt] = (edge) {u, h[v]}; h[v] = cnt;
}
ll inv[N*10], fac[N*10], facInv[N*10];
inline ll C(int n, int m) {return fac[n] * facInv[m] %P * facInv[n-m] %P;}
int main() {
    freopen("in", "r", stdin);
    n = read(); m = read();
    int flag = 1, lim = 0;
    for(int i=1; i<=m; i++) {
        u = read(), v = read(), w = t[i] = read(), de[u] += w, de[v] += w;
        lim = max(lim, w);
        if(w & 1) flag = 0;
    }

    for(int i=1; i<=n; i++) {
        if(de[i] & 1) {puts("0"); return 0;}
        de[i] >>= 1;
        lim = max(lim, de[i]);
    }
    
    inv[1] = fac[0] = facInv[0] = 1;
    for(int i=1; i <= lim; i++) {
        if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;
        fac[i] = fac[i-1] * i %P;
        facInv[i] = facInv[i-1] * inv[i] %P;
    }

    if(m == n-1) {
        if(!flag) {puts("0"); return 0;}    
        ll ans = 1;
        for(int i=1; i<=n; i++) ans = ans * fac[de[i] - 1] %P; //printf("ans %lld\n", ans);
        for(int i=1; i<=m; i++) ans = ans * C(t[i], t[i] >> 1) %P * (t[i] >> 1) %P;
        
        printf("%lld\n", ans);
    }
}

227【UR #15】奥林匹克数据结构

  • 题意:长为n的序列a,Q次询问,给出一个长m的序列b,求a有多少长m的子序列和b的相对大小关系相同

  • 题解:

    太神了不会写啊,看了题解之后写了一下午一晚上才有50部分分

    一开始想的是把a的子序列离散化然后和b判断相等

    这样对于\(m_i \le 25\)的点可以哈希一下,预处理\(O(nm^2)\),结果T了

    发现题解里的这些点的预处理是\(O(nm\log n)\)的!

    想了一下,发现不需要离散化,可以定义每个点的名次为前面比它小的个数+1,照样可行!

    然后用树状数组+异或哈希就可以拿这20分啦


    试着写了一下AC自动机+主席树的离线做法,还是有细节不会处理啊,又去写了kmp的做法。

    其实就是匹配b的名次序列在a中出现次数,只不过a中元素的名次依赖于序列的起始位置。

    kmp的border是单调递减的,所以用bit维护当前的border,移动now时暴力修改,求名次用bit求

    好喵啊!

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <map>
    #include <ctime>
    double tim() {return (double) clock() / CLOCKS_PER_SEC;}
    using namespace std;
    typedef long long ll;
    const int N = 1e5+5;
    inline int read() {
        char c=getchar(); int x=0,f=1;
        while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    
    int type, n, Q, a[N], b[N * 5];
    
    void decode(int *b, int m, int ans) {
      static int c[N];
      for(int i=1; i<=m; i++) c[(i+ans-1) % m + 1] = b[i];
      for(int i=1; i<=m; i++) b[i] = c[i];
    }
    
    
      namespace bit {
          int c[N], t[N], T;
          inline void add(int p, int v) {
              for(; p <= n; p += p&-p) {
                  if(t[p] != T) t[p] = T, c[p] = v;
                  else c[p] += v;
              }
          }
          inline int sum(int p) {
              int ans = 0;
              for(; p; p -= p&-p) 
                  if(t[p] == T) ans += c[p];
              return ans;
          }
      }
    
    namespace m_25 {
      int s[26][N];
      int h[N];
    
      using bit::add; using bit::sum;
    
      void ini_hash() {
          srand(317);
          for(int i=1; i<=25; i++) h[i] = rand() << 15 ^ rand();
    
          for(int i=1; i<=n; i++) {
              int ans = 0; bit::T++;
              for(int m = 1; m <= 25 && i+m-1 <= n; m++) 
                  ans = ans << 1 ^ h[sum(a[i+m-1])], s[m][++s[m][0]] = ans, add(a[i+m-1], 1);
          }
          for(int m = 1; m <= 25; m++) sort(s[m] + 1, s[m] + s[m][0] + 1);
      }
      int get_hash(int *b, int m) {
          int ans = 0; bit::T++;
          for(int i=1; i<=m; i++) ans = ans << 1 ^ h[sum(b[i])], add(b[i], 1);
          return ans;
      }
      int Count(int *a, int x) {
          return upper_bound(a + 1, a + a[0] + 1, x) - lower_bound(a + 1, a + a[0] + 1, x);
      }
      int ans;
      void solve() {
          ini_hash();
          for(int i=1; i<=Q; i++) {
              int m = read();
              for(int i=1; i<=m; i++) b[i] = read();
              if(type) decode(b, m, ans);
              int x = get_hash(b, m);
              ans = Count(s[m], x);
              printf("%d\n", ans); 
          }
      }
    }
    
    namespace kmp {
      int fail[N], rnk[N], _a[N];
      void build(int *s, int n) {
          bit::T++;
          for(int i=1; i<=n; i++) rnk[i] = bit::sum(s[i]) + 1, bit::add(s[i], 1);
          bit::T++;
          fail[1] = 0;
          for(int i=2; i<=n; i++) {
              int now = fail[i-1];
              //printf("iiii %d  %d\n", i, now);
              while(now && bit::sum(s[i]) + 1 != rnk[now+1]) {
                  for(int j = i-now; j <= i-fail[now]-1; j++) bit::add(s[j], -1);
                  now = fail[now];
              }
              if(rnk[now+1] == bit::sum(s[i]) + 1) fail[i] = now+1, bit::add(s[i], 1);
              else fail[i] = 0;
          }
          //for(int i=1; i<=n; i++) printf("%d ", rnk[i]); puts("  rnk");
          //for(int i=1; i<=n; i++) printf("%d ", fail[i]); puts("  fail");
      }
    
      void walk(int *a, int n, int *b, int m) {
          bit::T++;
          int now = 0, ans = 0;
          for(int i=1; i<=n; i++) { //printf("-----i %d   %d\n", i, now);
              while(now && bit::sum(a[i]) + 1 != rnk[now+1]) {
                  //printf("lose [%d, %d]\n", i-now, i-fail[now]-1);
                  for(int j = i-now; j <= i-fail[now]-1; j++) bit::add(a[j], -1);
                  now = fail[now]; 
              }
              if(bit::sum(a[i]) + 1 == rnk[now+1]) now++, bit::add(a[i], 1);
              if(now == m) ans++; 
          }
          printf("%d\n", ans);
      }
    }
    
    namespace type_0 {
      void solve() {
          for(int i=1; i<=Q; i++) {
              int m = read();
              for(int j=1; j<=m; j++) b[j] = read();
              kmp::build(b, m);
              kmp::walk(a, n, b, m);
              for(int i=1; i<=m; i++) kmp::fail[i] = kmp::rnk[i] = 0;
          }
      }
    }
    
    int main() {
      freopen("in", "r", stdin);
      freopen("out", "w", stdout);
      type = read(); 
      n = read(); Q = read();
      for(int i=1; i<=n; i++) a[i] = read();
      if(type == 0) type_0::solve();
      else m_25::solve();
      return 0;
    }
    

    下面这个不会写的代码只是留作纪念

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <map>
    #include <ctime>
    double tim() {return (double) clock() / CLOCKS_PER_SEC;}
    using namespace std;
    typedef long long ll;
    const int N = 1e5+5;
    inline int read() {
        char c=getchar(); int x=0,f=1;
        while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    
    int type, n, Q, a[N], b[N * 5];
    
    void decode(int *b, int m, int ans) {
      static int c[N];
      for(int i=1; i<=m; i++) c[(i+ans-1) % m + 1] = b[i];
      for(int i=1; i<=m; i++) b[i] = c[i];
    }
    
    
      namespace bit {
          int c[N], t[N], T;
          inline void add(int p, int v) {
              for(; p <= n; p += p&-p) {
                  if(t[p] != T) t[p] = T, c[p] = v;
                  else c[p] += v;
              }
          }
          inline int sum(int p) {
              int ans = 0;
              for(; p; p -= p&-p) 
                  if(t[p] == T) ans += c[p];
              return ans;
          }
      }
    
    namespace m_25 {
      int s[26][N];
      int h[N];
    
      using bit::add; using bit::sum;
    
      void ini_hash() {
          srand(317);
          for(int i=1; i<=25; i++) h[i] = rand() << 15 ^ rand();
    
          for(int i=1; i<=n; i++) {
              int ans = 0; bit::T++;
              for(int m = 1; m <= 25 && i+m-1 <= n; m++) 
                  ans = ans << 1 ^ h[sum(a[i+m-1])], s[m][++s[m][0]] = ans, add(a[i+m-1], 1);
          }
          for(int m = 1; m <= 25; m++) sort(s[m] + 1, s[m] + s[m][0] + 1);
      }
      int get_hash(int *b, int m) {
          int ans = 0; bit::T++;
          for(int i=1; i<=m; i++) ans = ans << 1 ^ h[sum(b[i])], add(b[i], 1);
          return ans;
      }
      int Count(int *a, int x) {
          return upper_bound(a + 1, a + a[0] + 1, x) - lower_bound(a + 1, a + a[0] + 1, x);
      }
      int ans;
      void solve() {
          ini_hash();
          for(int i=1; i<=Q; i++) {
              int m = read();
              for(int i=1; i<=m; i++) b[i] = read();
              if(type) decode(b, m, ans);
              int x = get_hash(b, m);
              ans = Count(s[m], x);
              printf("%d\n", ans); 
          }
      }
    }
    
    namespace tr {
      #define lc(x) t[x].l
      #define rc(x) t[x].r
      struct meow {int l, r, size;} t[N];
      int sz, root[N];
      void insert(int &x, int l, int r, int p) {
          t[++sz] = t[x]; x = sz;
          t[x].size ++;
          if(l == r) return;
          int mid = (l+r) >> 1;
          if(p <= mid) insert(lc(x), l, mid, p);
          else insert(rc(x), mid+1, r, p);
      }
      int sum(int x, int y, int l, int r, int ql, int qr) {
          //if(ql > qr) return 0;
          if(ql <= l && r <= qr) return t[y].size - t[x].size;
          else {
              int mid = (l+r) >> 1, ans = 0;
              if(ql <= mid) ans += sum(lc(x), lc(y), l, mid, ql, qr);
              if(mid < qr)  ans += sum(rc(x), rc(y), mid+1, r, ql, qr);
              return ans;
          }
      }
    }
    using tr::root; 
    inline int rank(int x, int l, int r) {return tr::sum(root[l-1], root[r], 1, n, 1, a[x]);}
    
    namespace ac {
      #define fir first
      #define sec second
      int pos[N];
      struct meow {
          map<int, int> ch; 
          map<int, int> mp;
          int fail, cnt, rnk;
          int count(int c) {return ch.count(c);}
          int has(int x) {return mp.count(x);}
      } t[N * 5];
      map<int, int>::iterator it;
      int sz, deep[N];
      void insert(int *s, int m, int id) { printf("\ninsert %d\n", id);
          int u = 0;
          bit::T++;
          for(int i=1; i<=m; i++) {
              int c = s[i];
              if(!t[u].ch[c]) {
                  t[u].ch[c] = ++sz;
                  deep[sz] = deep[u] + 1;
                  t[sz].rnk = bit::sum(s[i]) + 1;
                  t[u].mp[t[sz].rnk] = sz;
              }
              bit::add(s[i], 1);
              u = t[u].ch[c];
              printf("u %d  %d %d\n", u, deep[u], t[u].rnk);
          }
          pos[id] = u;
      }
      void build() {
          static int q[N], head = 1, tail = 1;
          for(it = t[0].ch.begin(); it != t[0].ch.end(); it++) q[tail++] = it -> sec;
          while(head != tail) {
              int u = q[head++];
              for(it = t[u].ch.begin(); it != t[u].ch.end(); it++) {
                  int v = it -> sec, rnk = t[v].rnk, now = t[u].fail;
                  while(now && !t[now].has(rnk)) now = t[now].fail;
                  if(t[now].has(rnk)) t[v].fail = t[now].mp[rnk];
                  q[tail++] = v;
              }
          }
      }
    
      void walk(int *a, int n) { puts("\nwalk");
          int u = 0;
          for(int i=1; i<=n; i++) {
              int l = i - deep[u], r = i, rnk = rank(i, l, r);
              while(u && !t[u].has(rnk)) u = t[u].fail;
              if(t[u].has(rnk)) u = t[u].mp[rnk];
              t[u].cnt ++; printf("u %d\n", u);
          }
      }
    
      int f[N * 5];
      void dfs(int u) { //printf("dfs %d\n", u);
          f[u] = t[u].cnt;
          for(map<int, int>::iterator it = t[u].ch.begin(); it != t[u].ch.end(); it++) 
              dfs(it -> sec), f[u] += f[it -> sec];
      }
    }
    
    namespace type_0 {
      void solve() {
          for(int i=1; i<=n; i++) root[i] = root[i-1], tr::insert(root[i], 1, n, a[i]);
          //puts("rank");
          //for(int l=1; l<=n; l++) for(int r=l; r<=n; r++) printf("rank [%d, %d]    %d\n", l, r, rank(r, l, r));
          for(int i=1; i<=Q; i++) {
              int m = read();
              for(int j=1; j<=m; j++) b[j] = read();
              ac::insert(b, m, i);
          }
          ac::build();
          ac::walk(a, n); 
          ac::dfs(0); //puts("hi");
          using ac::f; using ac::pos;
          for(int i=1; i<=Q; i++) printf("%d\n", f[pos[i]]);
      }
    }
    
    int main() {
      freopen("in", "r", stdin);
      //freopen("out", "w", stdout);
      type = read(); 
      n = read(); Q = read();
      for(int i=1; i<=n; i++) a[i] = read();
      if(type == 0) type_0::solve();
      else m_25::solve();
      return 0;
    }

posted @ 2017-05-13 21:51 Candy? 阅读(...) 评论(...) 编辑 收藏