HDU——1498 50 years, 50 colors

Posted 2020-10-04

　　　　　　　　　　50 years, 50 colors

　　　　　　　　　　Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
　　　　　　　　　　　　　　　　Total Submission(s): 2683    Accepted Submission(s): 1538

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

 

 

题目大意：

给你一个 n*n 的矩阵，每个格子上对应着相应颜色的气球，每次你可以选择一行或一列的同种颜色的气球进行踩破，问你在K次这样的操作后，哪些颜色的气球是不可能被踩破完的。

思路：

本题要求的是求出什么样的气球是根本就不可能被踩破的，而不是求的我们经过k次操作后剩下的气球的情况。

这样的话我们可以这样来考虑：现在我们一共有k次操作，也就是说对于每一种颜色的球，我们有k次机会都可以选择这一种颜色来把它全部清除。这样的话我们只需要判断我们在进行k次操作以后，我们当前选的颜色是否可以被完全清除就好了！

怎么判断？！我们仍然可以采用二分图来做，我们对于每一种颜色的球单独处理，将其横纵坐标看成两类，每一次只能选择一行或者是一列，这就跟上一个题是一样的了，然后进行二分图匹配，这样我们匹配出来的是我们可以用几次来将这种颜色的求全部消灭。

我们判断我们构建出来的图的最大匹配数是否大于k，若不是，那么说明我们可以将这种球消灭。反之，则不可以。记录。

代码：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 110
using namespace std;
bool vis[N],vist[N];
int n,k,sum,tot,ans[N],girl[N],a[N][N],map[N][N];
int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1; ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();}
    return x*f;
}
int find(int x)
{
    for(int i=1;i<=n;i++)
    {
        if(!vis[i]&&map[x][i])
        {
            vis[i]=true;
            if(girl[i]==-1||find(girl[i])) {girl[i]=x; return 1;}
        }
    }
    return 0;
}
int col()
{
    int s=0;
    memset(girl,-1,sizeof(girl));
    for(int i=1;i<=n;i++)
    {
        memset(vis,0,sizeof(vis));
        if(find(i)) s++;
    }
    return s;
}
void begin()
{
    sum=0,tot=0;
    memset(a,0,sizeof(a));
    memset(ans,0,sizeof(ans));
    memset(vist,0,sizeof(vist));
}
int main()
{
    while(1)
    {
        n=read(),k=read();
        if(n==0&&k==0) break;
        begin();
        for(int i=1;i<=n;i++)
         for(int j=1;j<=n;j++)
          a[i][j]=read();
        for(int i=1;i<=n;i++)
         for(int j=1;j<=n;j++)
          if(!vist[a[i][j]])
          {
               vist[a[i][j]]=true;
               memset(map,0,sizeof(map));
               for(int u=1;u<=n;u++)
                for(int v=1;v<=n;v++) 
                 if(a[u][v]==a[i][j])
                  map[u][v]=1;
               if(col()>k) ans[++sum]=a[i][j];
          }
       if(sum==0) {printf("-1\n"); continue;}
       sort(ans+1,ans+1+sum);
       for(int i=1;i<sum;i++) printf("%d ",ans[i]);
       printf("%d\n",ans[sum]);
    }
    return 0;
}