貪心, 數矩形的個數.
可以先把(x,y)為終點往左的矩形數記錄下來, 這裡可以應用DP的方法
然後O(n^3)迴圈, 說不清楚, 看代碼吧.
/******************************************************************************* # 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;}