HDU——T 1498 50 years, 50 colors

http://acm.hdu.edu.cn/showproblem.php?pid=1498

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2687    Accepted Submission(s): 1540

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

 

 

Source

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

 

 

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题意：给你一个n*n的矩阵，k次可以将每行或每列同颜色消去的操作，求一定不能消去的颜色，若没有输出-1；

n<=100；；；枚举每次想要消去的颜色，以同种颜色的格子的坐标构图，得出最大匹配。

#include <algorithm>
#include <cstring>
#include <cstdio>

const int N(110);
bool map[N][N],vis[N],linked[N];
int n,k,col[N][N],ans[N],match[N];

bool find(int u)
{
    for(int v=1;v<=n;v++)
        if(map[u][v]&&!vis[v])
        {
            vis[v]=1;
            if(!match[v]||find(match[v]))
            {
                match[v]=u;
                return true;
            }
        }
    return false;
}
bool link()
{
    int ret=0;
    for(int i=1;i<=n;i++)
    {
        if(find(i)) ret++;
        memset(vis,0,sizeof(vis));
    }
    memset(match,0,sizeof(match));
    return ret>k;
}

int main()
{
    for(;scanf("%d%d",&n,&k)&&n&&k;)
    {
        int cnt=0;
        for(int i=1;i<=n;i++)
          for(int j=1;j<=n;j++)
            scanf("%d",&col[i][j]);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
              {
                  if(linked[col[i][j]]) continue;
                  linked[col[i][j]]=true;
                  for(int u=1;u<=n;u++)
                      for(int v=1;v<=n;v++)
                          if(col[u][v]==col[i][j])
                              map[u][v]=1;
                  if(link()) ans[++cnt]=col[i][j];
                  memset(map,0,sizeof(map));
            }
        memset(linked,0,sizeof(linked));
        if(!cnt) puts("-1");
        else
        {
            std:: sort(ans+1,ans+cnt+1);
            for(int i=1;i<cnt;i++)
                printf("%d ",ans[i]);
            printf("%d\n",ans[cnt]);
        }
        memset(ans,0,sizeof(ans));
    }
    return 0;
}