hdu 1498(最小点覆盖集)

50 years, 50 colors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2256    Accepted Submission(s): 1266

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

 

题意:给出一个矩阵,里面有一些颜色不同的气球,一次能够消灭一行或者一列的气球,问某一种颜色的气球能否在 k 次内消除完毕,如果不能则输出气球颜色编号,如果全部可以则输出-1.

题解:最小点覆盖问题,枚举每一种颜色的气球,然后对每一个做一次最大匹配,如果最大匹配>k，则不能够消灭.时间复杂度O（n^3） ,这都能在62MS跑过去,数据太水。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include <math.h>
using namespace std;
const int N = 105;
int n,k;
int graph[N][N],graph1[N][N];
int linker[N],result[N];
bool vis[N];
int Hash[55];
bool dfs(int u){
    for(int v=0;v<n;v++){
        if(graph1[u][v]&&!vis[v]){
            vis[v] = true;
            if(linker[v]==-1||dfs(linker[v])){
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF){
        if(n==0&&k==0) break;
        memset(Hash,0,sizeof(Hash));
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                scanf("%d",&graph[i][j]);
                Hash[graph[i][j]] = 1;
            }
        }
        int m = 0;
        for(int i=1;i<=50;i++){
            if(!Hash[i]) continue;
            memset(graph1,0,sizeof(graph1));
            memset(linker,-1,sizeof(linker));
            for(int j=0;j<n;j++){
                for(int k=0;k<n;k++){
                    if(graph[j][k]==i) {
                        graph1[j][k] = 1;
                    }
                }
            }
            int res = 0;
            for(int u=0;u<n;u++){
                memset(vis,0,sizeof(vis));
                if(dfs(u)) res++;
            }
            if(res>k) result[m++] = i;
        }
        if(m==0){
            printf("-1
");
            continue;
        }
        for(int i=0;i<m-1;i++){
            printf("%d ",result[i]);
        }
        printf("%d
",result[m-1]);
    }
    return 0;
}