HDU1498_50 years, 50 colors(二分图/最小点覆盖=最大匹配)

解题报告

题目传送门

题意：

给你一个矩阵，矩阵里面是气球，气球有1-50种颜色，问你在k次之内能不能把哪种存在的颜色消掉（每种颜色k次机会），不能消掉的颜色按升序输出。

思路：

白想一上午了，理解错了题意，原来每种有k次可以消除的机会，还以为是总共k次机会消气球。

理解对了就很好做，类似POJ3041

求最小点覆盖。用最少的点覆盖最多的边。

每次枚举颜色看是否操作次数超过k次。

英语。。。。。。。。。。。。。。。。。。。。

。。。。。。。。。。。。。。。。。。。。。。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;int k,n,m,mmap[1100][1100],vis[1100],pre[1100],num[1100];int dfs(int x,int c) {    for(int i=1; i<=n; i++) {        if(!vis[i]&&mmap[x][i]==c) {            vis[i]=1;            if(pre[i]==-1||dfs(pre[i],c)) {                pre[i]=x;                return 1;            }        }    }    return 0;}int main() {    int i,j;    while(~scanf("%d%d",&n,&k)) {        memset(num,0,sizeof(num));        if(!n&&!k)break;        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)                scanf("%d",&mmap[i][j]);        int ans[100];        for(int c=1; c<=50; c++) {            memset(pre,-1,sizeof(pre));            for(i=1; i<=n; i++) {                memset(vis,0,sizeof(vis));                num[c]+=dfs(i,c);            }        }        int m=0,f=0;        for(i=1; i<=51; i++) {            if(num[i]&&num[i]>k) {                ans[m++]=i;                f=1;            }        }        if(!f) {            printf("-1\n");            continue;        }        sort(ans,ans+m);        for(i=0; i<m-1; i++) {            printf("%d ",ans[i]);        }        printf("%d\n",ans[m-1]);    }}

50 years, 50 colors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1607    Accepted Submission(s): 875

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