Codeforce 373 D Counting Rectangles is Fun (统计全0子矩阵)

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题目链接：http://codeforces.com/contest/373/problem/D

题意:求一个矩阵中有多少个全0子矩阵。

思路：动态规划,dp[a][b][c][d]为待求量，矩阵大小只有40*40，询问数量可以达到3*10^5，提示应该进行预处理。

状态转移方程:dp[a][b][c][d]= dp[a][b][c-1][d] + dp[a][b][c][d-1] - dp[a][b][c-1][d-1]+val

val代表在矩阵(a,b)(c,d)中以(c,d)为右下角的全0矩阵的个数

预处理数组top[i][j]表示点(i,j)向上能延伸到最远的0的位置，O(n)向左扫出矩阵的个数。

总体复杂度O(n^5)

#include <cstdio>#define min(a,b) ((a)<(b)?(a):(b))const int N=42;int dp[N][N][N][N];char g[N][N];int top[N][N];int main(){    int n,m,q,i,j;    scanf("%d%d%d",&n,&m,&q);    for (i=1;i<=n;i++)    {        scanf("%s",g[i]+1);        for (j=1;j<=m;j++)            if (g[i][j]=='0')                top[i][j]=top[i-1][j]+1;            else                top[i][j]=0;    }    int a,b,c,d;    for (a=1;a<=n;a++) for (b=1;b<=m;b++)        for (c=a;c<=n;c++) for (d=b;d<=m;d++)        {            dp[a][b][c][d]= dp[a][b][c-1][d] + dp[a][b][c][d-1] - dp[a][b][c-1][d-1];            int minH = min(top[c][d],c-a+1);            for (i=d;i>=b;i--)            {                minH = min(top[c][i],minH);                if (minH==0) break;                else dp[a][b][c][d] += minH;            }        }    while (q--)    {        scanf("%d%d%d%d",&a,&b,&c,&d);        printf("%d\n",dp[a][b][c][d]);    }    return 0;}

下面一份代码参考了：http://mochavic.blog.163.com/blog/static/2196651172013111402949522/

利用了部分和的思想，复杂度O(n^4)，但因为循环次数多，所以实际速度没有上面的快

#include <cstdio>const int N=42;int s[N][N];char str[N];int Cal (int x0, int y0, int x1, int y1){    x0--; y0--;    return s[x1][y1] - s[x0][y1] - s[x1][y0] + s[x0][y0];}int dp[N][N][N][N];int val[N][N][N][N];int main (){    int n,m,q,i,j;    scanf("%d%d%d",&n,&m,&q);for (i=1;i<=n;i++)    {        scanf("%s",str);for (j=0;j<m;j++)            s[i][j + 1] = s[i][j] + str[j] - '0';    }for (i=1;i<=n;i++)        for (j=0;j<m;j++)            s[i][j] += s[i - 1][j];//s[i][j]预处理出了(0,0,i,j)范围内所有1的个数    int x0, x1, y0, y1;    for (x0 = 1; x0 <= n; x0++)        for (x1 = x0; x1 <= n; x1++)            for (y0 = 1; y0 <= m; y0++)                for (y1 = y0; y1 <= m; y1++)                    if (Cal(x0, y0, x1, y1) == 0)                        val[x0][y0][x1][y1]++;  //(x0,y0,x1,y1)范围内全0for (x1 = 1; x1 <= n; x1++)for (y1 = 1; y1 <= m; y1++){for (i = n; i >= 1; i--)for (j = m; j >= 1; j--)val[i][j - 1][x1][y1] += val[i][j][x1][y1];for (i = n; i >= 1; i--)for (j = m; j >= 1; j--)                    val[i - 1][j][x1][y1] += val[i][j][x1][y1];}//val[x0][y0][x1][y1]为以x1,y1为右下角的左上角不超过x0,y0的全0矩阵个数    for (x0 = 1; x0 <= n; x0++) for (x1 = x0; x1 <= n; x1++)for (y0 = 1; y0 <= m; y0++) for (y1 = y0; y1 <= m; y1++)dp[x0][x1][y0][y1] = dp[x0][x1 - 1][y0][y1] + dp[x0][x1][y0][y1 - 1] - dp[x0][x1 - 1][y0][y1 - 1] + val[x0][y0][x1][y1];    while (q--){        scanf("%d%d%d%d", &x0, &y0, &x1, &y1);        printf("%d\n", dp[x0][x1][y0][y1]);    }    return 0;}

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