BZOJ1397 Ural 1486 Equal squares

首先二分答案ans，然后只要进行判断答案ans是否可行即可。

验证方法：首先对每一个位置，求出它开始长度为ans的横行的hash值

然后求出每一个hash值的长度为ans的竖列的Hash值

查看是否有两个Hash值相同即可（比如我们可以基数排序。。。做什么大死！）

/**************************************************************
    Problem: 1397
    User: rausen
    Language: C++
    Result: Accepted
    Time:2084 ms
    Memory:5052 kb
****************************************************************/
 
#include <cstdio>
#include <algorithm>
 
using namespace std;
typedef long long ll;
 
const ll B1 = 103;
const ll B2 = 211;
const int N = 505;
 
int n, m;
char a[N][N];
ll p1[N], p2[N];
ll h[N][N], hash[N * N];
 
bool check(int x) {
  int i, j, t;
  ll tmp;
  for (i = 1; i <= n; ++i) {
    for (tmp = 0, j = 1; j < x; ++j)
      (tmp *= B1) += a[i][j], h[i][j] = 0;
    for (j = x; j <= m; ++j)
      h[i][j] = tmp = tmp * B1 - p1[x] * a[i][j - x] + a[i][j];
  }
  for (t = 0, i = x; i <= m; ++i) {
    for (tmp = 0, j = 1; j < x; ++j)
      (tmp *= B2) += h[j][i];
    for (j = x; j <= n; ++j)
      hash[t++] = tmp = tmp * B2 - p2[x] * h[j - x][i] + h[j][i];
  }
  sort(hash, hash + t);
  for (i = 1; i < t; ++i)
    if (hash[i] == hash[i - 1]) return 1;
  return 0;
}
 
int main() {
  int i, j, l, r, mid;
  scanf("%d%d
", &n, &m);
  for (i = 1; i <= n; ++i) {
    gets(a[i] + 1);
    for (j = 1; j <= m; ++j)
      a[i][j] -= 'a' - 1;
  }
  l = 1, r = min(n, m) + 1, p1[0] = p2[0] = 1;
  for (i = 1; i <= r; ++i)
    p1[i] = p1[i - 1] * B1, p2[i] = p2[i - 1] * B2;
  while (l + 1 < r) {
    mid = (l + r) >> 1;
    if (check(mid)) l = mid;
    else r = mid;
  }
  printf("%d
", l);
  return 0;
}