Oil Skimming
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
Problem Description
Thanks to a certain “green” resources company, there is a new profitable industry of oil skimming. There are large slicks of crude oil floating in the Gulf of Mexico just waiting to be scooped up by enterprising oil barons. One such oil baron has a special plane that can skim the surface of the water collecting oil on the water’s surface. However, each scoop covers a 10m by 20m rectangle (going either east/west or north/south). It also requires that the rectangle be completely covered in oil, otherwise the product is contaminated by pure ocean water and thus unprofitable! Given a map of an oil slick, the oil baron would like you to compute the maximum number of scoops that may be extracted. The map is an NxN grid where each cell represents a 10m square of water, and each cell is marked as either being covered in oil or pure water.
Input
The input starts with an integer K (1 <= K <= 100) indicating the number of cases. Each case starts with an integer N (1 <= N <= 600) indicating the size of the square grid. Each of the following N lines contains N characters that represent the cells of a row in the grid. A character of ‘#’ represents an oily cell, and a character of ‘.’ represents a pure water cell.
Output
For each case, one line should be produced, formatted exactly as follows: “Case X: M” where X is the case number (starting from 1) and M is the maximum number of scoops of oil that may be extracted.
Sample Input
1 6
……
.##…
.##…
….#.
….##
……
Sample Output
Case 1: 3
解题心得：
- 题意很简单，就是要你在图中去截取1x2的方块，问最多能截取多少个。
- 就是一个二分匹配问题，还是很简单的，有两种建图的方法
- 第一种是给每一个坐标编号，查找每一个‘#’的四周图形，然后建一个双向的图，得到的答案除2就可以了。
- 第二种就更高级了，仔细观察可以观看一个点的行列之和与他四周的四个格子的行列和奇偶性一定是不同的，这样就可以参考国际象棋的棋盘，黑色的方格匹配白色方格，二分匹配。
弱智建图写代码：
#include<bits/stdc++.h>
using namespace std;
const int maxn = 610;
char maps[maxn][maxn];
bool vis[maxn*maxn];
vector <int> ve[maxn*maxn];
int n,dir[4][2] = {0,1,0,-1,-1,0,1,0},match[maxn*maxn];void init()
{scanf("%d",&n);for(int i=0;i<maxn;i++)ve[i].clear();memset(vis,0,sizeof(vis));memset(maps,0,sizeof(maps));memset(match,-1,sizeof(match));for(int i=1;i<=n;i++)scanf("%s",maps[i]+1);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(maps[i][j] == '#')for(int k=0;k<4;k++){int x = i + dir[k][0];int y = j + dir[k][1];if(maps[x][y] == '#')ve[i*n+j].push_back(x*n+y);//如果四周是‘#’就建立联系}}bool dfs(int x)//match
{for(int i=0;i<ve[x].size();i++){int v = ve[x][i];if(!vis[v]){vis[v] = true;if(match[v] == -1 || dfs(match[v])){match[v] = x;return true;}}}return false;
}int solve()
{int ans = 0;for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(maps[i][j] == '#'){memset(vis,0,sizeof(vis));if(dfs(i*n+j))ans++;}return ans;
}int main()
{int t;scanf("%d",&t);int cas = 1;while(t--){init();int ans = solve();printf("Case %d: %d\n",cas++,ans/2);//因为建立的是双向图，所以一定要除2}return 0;
}
)
