[luoguP2601] [ZJOI2009]对称的正方形（二维Hash + 二分 || Manacher）

  1 #include <cstdio>
  2 #include <iostream>
  3 #define UI unsigned int
  4 
  5 const int MAXN = 2010, bs1 = 19260817, bs2 = 20011001;
  6 int n, m, ans;
  7 UI sum[4][MAXN][MAXN], base1[MAXN], base2[MAXN];
  8 
  9 inline int read()
 10 {
 11     int x = 0, f = 1;
 12     char ch = getchar();
 13     for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
 14     for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
 15     return x * f;
 16 }
 17 
 18 inline int min(int x, int y)
 19 {
 20     return x < y ? x : y;
 21 }
 22 
 23 inline bool pd(int x, int y, int l)
 24 {
 25     UI t, h;
 26     h = sum[0][x + l - 1][y + l - 1]
 27       -    sum[0][x - l][y + l - 1] * base1[l + l - 1]
 28       - sum[0][x + l - 1][y - l] * base2[l + l - 1]
 29       + sum[0][x - l][y - l] * base1[l + l - 1] * base2[l + l - 1];
 30     t = sum[1][x + l - 1][y - l + 1]
 31       - sum[1][x - l][y - l + 1] * base1[l + l - 1]
 32       - sum[1][x + l - 1][y + l] * base2[l + l - 1]
 33       + sum[1][x - l][y + l] * base1[l + l - 1] * base2[l + l - 1];
 34     if(h ^ t) return 0;
 35     t = sum[2][x - l + 1][y + l - 1]
 36       - sum[2][x + l][y + l - 1] * base1[l + l - 1]
 37       - sum[2][x - l + 1][y - l] * base2[l + l - 1]
 38       + sum[2][x + l][y - l] * base1[l + l - 1] * base2[l + l - 1];
 39     if(h ^ t) return 0;
 40     t = sum[3][x - l + 1][y - l + 1]
 41       - sum[3][x + l][y - l + 1] * base1[l + l - 1]
 42       - sum[3][x - l + 1][y + l] * base2[l + l - 1]
 43       + sum[3][x + l][y + l] * base1[l + l - 1] * base2[l + l - 1];
 44     if(h ^ t) return 0;
 45     return 1;
 46 }
 47 
 48 inline int work(int i, int j)
 49 {
 50     int mid, s = 0, x = 1, y = min(min(i, n - i + 1), min(j, m - j + 1));//二分半径 
 51     while(x <= y)
 52     {
 53         mid = (x + y) >> 1;
 54         if(pd(i, j, mid)) s = mid, x = mid + 1;
 55         else y = mid - 1;
 56     }
 57     return s;
 58 }
 59 
 60 int main()
 61 {
 62     int i, j, k, x;
 63     n = read();
 64     m = read();
 65     n = n << 1 | 1;
 66     m = m << 1 | 1;
 67     for(i = 2; i <= n; i += 2)
 68         for(j = 2; j <= m; j += 2)
 69         {
 70             x = read();
 71             for(k = 0; k < 4; k++) sum[k][i][j] = x;
 72         }
 73     base1[0] = base2[0] = 1;
 74     for(i = 1; i <= n; i++) base1[i] = base1[i - 1] * bs1;
 75     for(i = 1; i <= m; i++) base2[i] = base2[i - 1] * bs2;
 76     for(i = 1; i <= n; i++)
 77         for(j = 1; j <= m; j++)
 78             sum[0][i][j] += sum[0][i - 1][j] * bs1;
 79     for(i = 1; i <= n; i++)
 80         for(j = 1; j <= m; j++)
 81             sum[0][i][j] += sum[0][i][j - 1] * bs2;
 82     for(i = 1; i <= n; i++)
 83         for(j = m; j; j--)
 84             sum[1][i][j] += sum[1][i - 1][j] * bs1;
 85     for(i = 1; i <= n; i++)
 86         for(j = m; j; j--)
 87             sum[1][i][j] += sum[1][i][j + 1] * bs2;
 88     for(i = n; i; i--)
 89         for(j = 1; j <= m; j++)
 90             sum[2][i][j] += sum[2][i + 1][j] * bs1;
 91     for(i = n; i; i--)
 92         for(j = 1; j <= m; j++)
 93             sum[2][i][j] += sum[2][i][j - 1] * bs2;
 94     for(i = n; i; i--)
 95         for(j = m; j; j--)
 96             sum[3][i][j] += sum[3][i + 1][j] * bs1;
 97     for(i = n; i; i--)
 98         for(j = m; j; j--)
 99             sum[3][i][j] += sum[3][i][j + 1] * bs2;
100     for(i = 1; i <= n; i++)
101         for(j = 1; j <= m; j++)
102             if((i ^ j ^ 1) & 1)
103                 ans += work(i, j) >> 1;
104     printf("%d\n", ans);
105     return 0;
106 }