poj2029 2010.4.14

poj2029 2010.4.14

 

跟分割矩阵的存储方式相同

dp[i][j]：以(1,1)为左上角，以(i,j)为右下角的矩阵柿子树的数目。

#include <cstdio>#include <cstring>#define LEN 102int dp[LEN][LEN];int w,h;int s,t;int n;void init(){int i;for(i=1;i<=n;i++){int x,y;scanf("%d %d",&x,&y);dp[y][x]=1;}}int calit(int x,int y){int xx=x-t,yy=y-s;int ans=dp[x][y]-dp[xx][y]-dp[x][yy]+dp[xx][yy];return ans;}int max(int x,int y){if (x>y) return x;else return y;}int main(){int i,j;while (scanf("%d",&n),n){scanf("%d %d",&w,&h);memset(dp,0,sizeof(dp));init();for(i=1;i<=h;i++)for(j=1;j<=w;j++)dp[i][j]+=(dp[i-1][j]-dp[i-1][j-1]+dp[i][j-1]);int ans=0;scanf("%d %d",&s,&t);for(i=t;i<=h;i++)for(j=s;j<=w;j++)ans=max(ans,calit(i,j));printf("%d\n",ans);}return 0;}