50 years, 50 colors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1638    Accepted Submission(s): 894

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

1 112 11 11 22 11 22 25 41 2 3 4 52 3 4 5 13 4 5 1 24 5 1 2 35 1 2 3 43 350 50 5050 50 5050 50 500 0

 

Sample Output

-1121 2 3 4 5-1

 

<span style="font-size:18px;color:#993399;">这道题的题意有点绕，刚开始一直没明白过来，求教的大神。题意是：给你两个数n和k,接下来输入n行n列的矩阵，</span>

<span style="font-size:18px;color:#993399;">每个数字代表一种气球，你每次可以扎一行或一列中的一种。k表示可以扎的次数，最后求的是最后剩的最少种数的</span>

<span style="font-size:18px;color:#993399;">气球代号。</span>

<span style="font-size:18px;color:#993399;">也就是看每一种颜色的最大匹配与k的关系，若是k小于最大匹配则输出。</span>

<span style="font-size:18px;"></span>

<span style="font-size:18px;">#include <iostream>#include<stdio.h>#include<math.h>#include<stdlib.h>#include<string.h>using namespace std;int map[101][101],a[101][101];int vis[10001];int pre[101];int n,k;int hungary(int x){    int i;    for(i=1; i<=n; i++)    {        if(map[x][i]==1&&vis[i]==0)        {            vis[i]=1;            if(pre[i]==-1||hungary(pre[i]))            {                pre[i]=x;                return 1;            }        }    }    return 0;}int main(){    int i,j,l,sum,t,b[101],flag;    while(scanf("%d%d",&n,&k)!=EOF)    {        t=0;flag=0;//标记变量        if(n==0&&k==0)            break;        memset(map,0,sizeof(map));        for(i=1; i<=n; i++)        {            for(j=1; j<=n; j++)            {                scanf("%d",&a[i][j]);            }        }        for(i=1; i<=50; i++)        {            sum=0;            memset(pre,-1,sizeof(pre));//因为要对每种气球求最大匹配，每次都要先初始化            for(l=1; l<=n; l++)                for(j=1; j<=n; j++)                {                    if(a[l][j]==i)                        map[l][j]=1;                    else                        map[l][j]=0;                }            for(l=1; l<=n; l++)            {                memset(vis,0,sizeof(vis));                sum+=hungary(l);            }            if(sum>k)            {                b[t++]=i;                flag=1;            }        }        if(flag==0)            printf("-1\n");        else        {            for(i=0; i<t-1; i++)                printf("%d ",b[i]);            printf("%d\n",b[t-1]);        }    }    return 0;}</span>