50 years, 50 colors

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.

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

-1121 2 3 4 5-1

题意，给出n*n的矩阵，每个点上一个气球，颜色有的不同，对每一种颜色可以打k次，每次可以打掉一行或者一列的相同颜色的色球。如果可以消灭完，就不输出，否则输出气球的颜色，如果全部消灭完 输出-1

就是对每一种颜色进行一次二分匹配，判断可不可以在k次内消灭完。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int mm[120][120] , c[100] ;struct node{    int v ;    node *next ;} *head[120];int link[120] , temp[120] , top , p[120];int f(int i){    for( node *q = head[i] ; q != NULL ; q = q->next )    {        int v = q->v ;        if( !temp[v] )        {            temp[v] = 1 ;            if( link[v] == -1 || f( link[v] ) )            {                link[v] = i ;                return 1;            }        }    }    return 0;}int main(){    int i , j , k , n , m , ans  ;    while(scanf("%d %d", &n, &m) && n+m != 0)    {        memset(c,0,sizeof(c));        for(i = 1 ; i <= n ; i++)            for(j = 1 ; j <= n ; j++)            {                scanf("%d", &mm[i][j]);                c[ mm[i][j] ]++ ;            }        top = 0 ;        for(i = 1 ; i <= 50 ; i++)        {            ans = 0 ;            memset(head,NULL,sizeof(head));            memset(link,-1,sizeof(link));            if( !c[i] )                continue ;            for(j = 1 ; j <= n ; j++)                for(k = 1 ; k <= n ; k++)                {                    if( mm[j][k] == i )                    {                        node *q = new node ;                        q->v = k ;                        q->next = head[j] ;                        head[j] = q ;                    }                }            for(j = 1 ; j <= n ; j++)            {                memset(temp,0,sizeof(temp));                if(  f(j) )                    ans++ ;            }            if( ans > m )                p[top++] = i ;        }        if(top == 0)            printf("-1\n");        else        {            for(i = 0 ; i < top ; i++)                if(i == top-1)                    printf("%d\n", p[i]);                else                    printf("%d ", p[i]);        }    }    return 0;}