ZOJ2067 White Rectangles

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贪心, 数矩形的个数.

可以先把(x,y)为终点往左的矩形数记录下来, 这里可以应用DP的方法

然后O(n^3)循环, 说不清楚, 看代码吧.

/******************************************************************************* # Author : Neo Fung # Email : neosfung@gmail.com # Last modified: 2012-02-29 20:24 # Filename: ZOJ2067 White Rectangles.cpp # Description :  ******************************************************************************/#ifdef _MSC_VER#define DEBUG#define _CRT_SECURE_NO_DEPRECATE#endif#include <fstream>#include <stdio.h>#include <iostream>#include <string.h>#include <string>#include <limits.h>#include <algorithm>#include <math.h>#include <numeric>#include <functional>#include <ctype.h>#define MAX 110using namespace std;char board[MAX][MAX];int sum[MAX][MAX];int main(void){#ifdef DEBUG    freopen("../stdin.txt","r",stdin);  freopen("../stdout.txt","w",stdout); #endif    int n;  while(~scanf("%d",&n))  {    memset(sum,0,sizeof(sum));    memset(board,'#',sizeof(board));    getchar();    for(int i=1;i<=n;++i)      gets(board[i]+1);int ans=0;for(int i=1;i<=n;++i)for(int j=1;j<=n;++j)if(board[i][j]=='.'){if(board[i][j-1]!='.')sum[i][j]=1;else            sum[i][j]=sum[i][j-1]+1;}for(int i=1;i<=n;++i)for(int j=1;j<=n;++j){int len=INT_MAX;for(int k=i;k<=n && sum[k][j];++k){len=min(len,sum[k][j]);ans+=len;}}printf("%d\n",ans);  }  return 0;}