【HDU 1498 50 years, 50 colors】+ 二分图

50 years, 50 colors
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2419 Accepted Submission(s): 1358

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

~选择一种颜色~可任选一行或者一列踩爆当前行或者列的气球~如果不能在K次以内所有气球全部踩爆~输出该颜色~~如果能踩爆所有的气球输出 -1~

二分图匹配~~

AC代码：

#include<cstdio>#include<cstring>using namespace std;int mp[110][110],color[52],vis[110],pa[110],ans[110],N,K;int DFS(int x,int y){    for(int i = 1 ; i <= N; i++)        if(mp[y][i] == x && pa[i] == -1){            pa[i] = 1;        if(vis[i] == -1 || DFS(x,vis[i])){            vis[i] = y;            return 1;        }    }    return 0;}int solve(int x){    int cut = 0;    memset(vis,-1,sizeof(vis));    for(int i = 1 ; i <= N; i++){        memset(pa,-1,sizeof(pa));        cut += DFS(x,i);    }    return cut;}int main(){    while(scanf("%d %d",&N,&K) != EOF && (N || K)){        memset(color,-1,sizeof(color));        for(int i = 1 ; i <= N; i++)            for(int j = 1 ; j <= N ; j++){                scanf("%d",&mp[i][j]);                if(color[mp[i][j]] == -1)                    color[mp[i][j]] = 1;            }    int nl = 0;    for(int i = 1 ; i <= 50 ; i++){        int sum = 0;        if(color[i] == 1){            sum += solve(i);        if(sum > K)            ans[++nl] = i;        }    }    if(nl == 0)        printf("-1\n");    else{        for(int i = 1 ; i < nl ;i++)            printf("%d ",ans[i]);        printf("%d\n",ans[nl]);    }    }    return 0;}