集训队专题（6）1003 50 years, 50 colors

50 years, 50 colors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2072    Accepted Submission(s): 1155

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

 

此题又是进行整行整列的处理，就像小编在上一题(1002)的题解中说到的一样，对于整行整列处理的题目，都可以一横纵坐标来作为两大集合进行二分图建图处理，题目中说到一共有50种颜色，把出现过的颜色进行一次二分图建图就好了。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100+10;
int n;
int map[maxn][maxn],g[maxn][maxn],linker[maxn],result[55];
bool used[maxn],color[55];
bool dfs(int u)
{
	int v;
	for(v=1; v<=n; v++)
		if(g[u][v] && !used[v])
		{
			used[v] = true;
			if(linker[v]==-1 || dfs(linker[v]))
			{
				linker[v] = u;
				return true;
			}
		}
	return false;
}
int hungary()
{
	int res = 0;
	int u;
	memset(linker,-1,sizeof(linker));
	for(u=1; u<=n; u++)
	{
		memset(used,0,sizeof(used));
		if(dfs(u)) res++;
	}
	return res;
}
int main()
{
	int k,u,v;
	bool f;
	while(scanf("%d%d",&n,&k)&&n&&k)
	{
		memset(color,false,sizeof(color));
        for(int i=1; i<=n; i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&map[i][j]),color[map[i][j]]=true;
        f=false;
        for(int t=1; t<=50; t++)
            if(color[t])
            {
                memset(g,false,sizeof(g));
                for(int i=1; i<=n; i++)
                	for(int j=1; j<=n; j++)
                        if(map[i][j]==t) g[i][j]=true;
                if(hungary()>k) 
                {
                    if(!f) printf("%d",t);
                    else printf(" %d",t);
                    f=true;
                }
            }
        if (!f) printf("-1");
        printf("\n"); 
	}
	return 0;
}