Oil Skimming HDU - 4185 （二分图最大匹配）

Oil Skimming

 HDU- 4185 

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 suchoil 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 coveredin 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 representsa 10m square of water, and each cell is marked as either being covered in oil or pure water.InputThe 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 thecells of a row in the grid. A character of '#' represents an oily cell, and a character of '.' represents a pure water cell.OutputFor 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

code：

/**题目让求有多少个像##这样的东西可能横着可能竖着，实际上可以转化成求二分图最大匹配**/#include <iostream>#include <cstdio>#include <cstring>using namespace std;char mp[660][660];int g[660][660];int relate[660][660];int match[660];int vis[660];int k,n;int tmp;int dfs(int u){int i;for(i = 0; i < tmp; i++){if(g[u][i] && !vis[i]){vis[i] = 1;if(!match[i] || dfs(match[i])){match[i] = u;return 1;}}}return 0;}int hungry(){int i;int cnt = 0;memset(match,0,sizeof(match));for(i = 0; i < tmp; i++){memset(vis,0,sizeof(vis));if(dfs(i))cnt++;}return cnt;}int main(){cin >> k;int cas = 0;while(k--){cin >> n;getchar();memset(relate,0,sizeof(relate));memset(g,0,sizeof(g));int i,j;tmp = 0;for(i = 0; i < n; i++){cin >> mp[i];getchar();for(j = 0; j < n; j++){if(mp[i][j] == '#')relate[i][j] = tmp++;}}for(i = 0; i < n; i++){for(j = 0; j < n; j++){if(mp[i][j] != '#') continue;if(mp[i-1][j] == '#') g[relate[i][j]][relate[i-1][j]] = 1;if(mp[i+1][j] == '#') g[relate[i][j]][relate[i+1][j]] = 1;if(mp[i][j-1] == '#') g[relate[i][j]][relate[i][j-1]] = 1;if(mp[i][j+1] == '#') g[relate[i][j]][relate[i][j+1]] = 1;}}int ans = hungry();printf("Case %d: %d\n",++cas,ans/2);//除以2是因为每个点都作为起点看了一遍，即正反都看了一遍所以是真正答案的两倍}return 0;}