Hdu 5093 Battle Ship

每个海面要么放要么不放，因此可以用二分图匹配， 考虑把同一行内的能互相看到的点放到一个行块里，同一列内能看到的点放到一个列块里，然后每一个行块都可以和该行块里所有海面的列块连边，选了这个行块，就必须选且只选择一个该行块里的一个海面对应的列块。

#include <iostream>	
#include <cmath>	
#include <cstdio>	
#include <algorithm>
#include <cstring>	
#define N 10100
using namespace std;
struct edg {
	int to, nex;
}e[N];
char ma[101][101];				
int n, m, T, n1, n2, cnt, cntx, cnty;
int flag[N], lin[N], x[101][101], y[101][101], vis[N];
void add(int f, int t)			
{
	e[++cnt].to = t;
	e[cnt].nex = lin[f];
	lin[f] = cnt;
}
bool dfs(int now)
{
	for (int i = lin[now]; i; i = e[i].nex)
	{
		int to = e[i].to;
		if (vis[to]) continue;
		vis[to] = 1;
		if (flag[to] == -1 || dfs(flag[to]))
		{
			flag[to] = now;
			return 1;
		}
	}
	return 0;
}
void clea()
{
	memset(flag, -1, sizeof(flag));
	memset(e, 0, sizeof(e));
	memset(lin, 0, sizeof(lin));
	memset(vis, 0, sizeof(vis));
	cnt = 0;
	cntx = cnty = 1;
}
int main()
{
 	scanf("%d", &T);
 	while (T--)
 	{ 
	 	clea();
	 	scanf("%d%d", &n, &m);
	 	char ca  = getchar();;
	 	for (int i = 1; i <= n; i++, ca = getchar())
	 		for (int j = 1; j <= m; j++)
	 			scanf("%c", &ma[i][j]);
	 	for (int i = 1; i <= n; i++, cntx++)
	 		for (int j = 1; j <= m; j++)
	 		{
	 			if (ma[i][j] == '*')//把属于同一行的数并到一起 
	 				x[i][j] = cntx;
	 			if (ma[i][j] == '#')
	 				++cntx;
	 		}
	 	for (int i = 1; i <= m; i++, cnty++)
	 		for (int j = 1; j <= n; j++)
	 		{
	 			if (ma[j][i] == '*')
	 				y[j][i] = cnty;
	 			if (ma[j][i] == '#')
	 				++cnty;
	 		}
	 	for (int i = 1; i <= n; i++)
	 		for (int j = 1; j <= m; j++)
	 			if (ma[i][j] == '*')
	 				add(x[i][j], y[i][j]);
	 	n1 = cntx, n2 = cnty;
	 	int ans = 0;
	 	for (int i = 1; i <= n1; i++)
	 	{
	 		memset(vis, 0, sizeof(vis));
	 		if (dfs(i))
	 			ans++;
	 	}
	 	for (int i = 1; i <= n2; i++)
			if (flag[i]!=-1)
				printf("%d %d
", flag[i], i); 
	 	printf("%d
", ans);
	}
	return 0;
}//
/*
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*/