HDU 2119 Matrix(二分图最小边覆盖)

题意：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.

思路：二分图简单题,前面好像有几道都与本题类似.

建图:把行号放在二分图左边点集,列号放在右边点集.如果(i,j)格子是1,那么就连左i与右j的无向边.

本题的问题就是求该二分图的最小边覆盖点集.使得每条边都被至少1个点覆盖. 最小覆盖集大小==最大匹配数.

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int maxn=100+5;

struct Max_Match
{
int n,m;
vector<int> g[maxn];
bool vis[maxn];
int left[maxn];

void init(int n,int m)
{
this->n=n;
this->m=m;
for(int i=1;i<=n;i++) g[i].clear();
memset(left,-1,sizeof(left));
}

bool match(int u)
{
for(int i=0;i<g[u].size();i++)
{
int v=g[u][i];
if(!vis[v])
{
vis[v]=true;
if(left[v]==-1 || match(left[v]))
{
left[v]=u;
return true;
}
}
}
return false;
}

int solve()
{
int ans=0;
for(int i=1;i<=n;i++)
{
memset(vis,0,sizeof(vis));
if(match(i)) ++ans;
}
return ans;
}
}MM;

int main()
{
int n,m;
while(scanf("%d",&n)==1 && n)
{
scanf("%d",&m);
MM.init(n,m);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
int v;
scanf("%d",&v);
if(v) MM.g[i].push_back(j);
}
printf("%d\n",MM.solve());
}
return 0;
}

Description

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