hdu 2119 Matrix（二分图最大匹配）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2119

给你一个N*M的0/1矩阵,你每次可以选特定的某行或某列,然后删除该行/列的所有1,问你最少需要几次操作能删除矩阵的所有1.

Give you a matrix(only contains 0 or 1),every time you can select a row or a column and delete all the '1' in this row or this column . Your task is to give out the minimum times of deleting all the '1' in the matrix.     Input There are several test cases. The first line contains two integers n,m(1<=n,m<=100), n is the number of rows of the given matrix and m is the number of columns of the given matrix. The next n lines describe the matrix:each line contains m integer, which may be either ‘1’ or ‘0’. n=0 indicate the end of input.     Output For each of the test cases, in the order given in the input, print one line containing the minimum times of deleting all the '1' in the matrix.     Sample Input   3 3 0 0 0 1 0 1 0 1 0 0     Sample Output   2

 

前面有好几道类似的题，也没什么好说的了。是1的点，就i->j连边。

#pragma GCC optimize(2)
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn = 1010;
const int inf = 0x3f3f3f3f;
typedef long long ll;
int n, m;
vectorvec[maxn];
int linker[maxn];
bool vis[maxn];
void init()
{
	for (int i = 0; i <= n; i++)
	{
		vec[i].clear();
	}
	return;
}
bool dfs(int x)
{
	for (int i = 0; i < vec[x].size(); i++)
	{
		int v = vec[x][i];
		if (!vis[v])
		{
			vis[v] = true;
			if (linker[v] == -1 || dfs(linker[v]))
			{
				linker[v] = x;
				return 1;
			}
		}
	}
	return 0;
}
int mx()
{
	int ans = 0;
	memset(linker, -1, sizeof(linker));
	for (int i = 1; i <= n; i++)
	{
		memset(vis, false, sizeof(vis));
		ans += dfs(i);
	}
	return ans;
}
int main()
{
	//freopen("C://input.txt", "r", stdin);
	while (scanf("%d", &n)&&n)
	{
		scanf("%d", &m);
		init();
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= m; j++)
			{
				int t;
				scanf("%d", &t);
				if (t)
				{
					vec[i].push_back(j);
				}
			}
		}
		int _max = mx();
		printf("%d\n", _max);
	}
	return 0;
}