(最小点覆盖) hdu 1498

50 years, 50 colors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1817    Accepted Submission(s): 996

Problem Description

On Octorber 21st, HDU 50-year-celebration, 50-color balloons floating around the campus, it's so nice, isn't it? To celebrate this meaningful day, the ACM team of HDU hold some fuuny games. Especially, there will be a game named "crashing color balloons".

There will be a n*n matrix board on the ground, and each grid will have a color balloon in it.And the color of the ballon will be in the range of [1, 50].After the referee shouts "go!",you can begin to crash the balloons.Every time you can only choose one kind of balloon to crash, we define that the two balloons with the same color belong to the same kind.What's more, each time you can only choose a single row or column of balloon, and crash the balloons that with the color you had chosen. Of course, a lot of students are waiting to play this game, so we just give every student k times to crash the balloons.

Here comes the problem: which kind of balloon is impossible to be all crashed by a student in k times.

 

 

Input

There will be multiple input cases.Each test case begins with two integers n, k. n is the number of rows and columns of the balloons (1 <= n <= 100), and k is the times that ginving to each student(0 < k <= n).Follow a matrix A of n*n, where Aij denote the color of the ballon in the i row, j column.Input ends with n = k = 0.

 

 

Output

For each test case, print in ascending order all the colors of which are impossible to be crashed by a student in k times. If there is no choice, print "-1".

 

 

Sample Input

1 1 1 2 1 1 1 1 2 2 1 1 2 2 2 5 4 1 2 3 4 5 2 3 4 5 1 3 4 5 1 2 4 5 1 2 3 5 1 2 3 4 3 3 50 50 50 50 50 50 50 50 50 0 0

 

 

Sample Output

-1 1 2 1 2 3 4 5 -1

 

 

Author

8600

 

题意：

题意：给定一张图，图中的每个数代表一种颜色的气球

求：哪种颜色的气球不能在K次中被消灭（ 每次只能消灭一行或者一列 ）

 

分析：就是要寻找一种消灭方法 使得所有同种气球被消灭，显然最小点覆盖（最大匹配），直接匈牙利就好了。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<stack>
#include<set>
using namespace std;
int n,k,mp[105][105],cnt[55],ans[55],link[105],mark[105];
bool dfs(int x,int col)
{
    for(int i=1;i<=n;i++)
    {
        if(mp[x][i]==col&&mark[i]==-1)
        {
            mark[i]=1;
            if(link[i]==-1||dfs(link[i],col))
            {
                link[i]=x;
                return true;
            }
        }
    }
    return false;
}
int km(int x)
{
    memset(link,-1,sizeof(link));
    int ret=0;
    for(int i=1;i<=n;i++)
    {
        memset(mark,-1,sizeof(mark));
        if(dfs(i,x))
            ret++;
    }
    return ret;
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n==0&&k==0)
            break;
        memset(cnt,0,sizeof(cnt));
        memset(mp,0,sizeof(mp));
        memset(ans,0,sizeof(ans));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&mp[i][j]);
                cnt[mp[i][j]]++;
            }
        }
        int tmp=0;
        for(int i=1;i<=50;i++)
        {
            if(cnt[i]&&km(i)>k)
            {
                ans[++tmp]=i;
            }
        }
        if(tmp==0)
        {
            printf("-1\n");
            continue;
        }
        printf("%d",ans[1]);
        for(int i=2;i<=tmp;i++)
            printf(" %d",ans[i]);
        printf("\n");
    }
    return 0;
}