HDOJ 1498 —— 50 years, 50 colors 二分图匹配（最小点覆盖 = 最大二分匹配）

50 years, 50 colors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1752    Accepted Submission(s): 962


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

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。

需要知道：最小覆盖点 == 最大二分匹配 

#include <stdio.h>
#include <string.h>

bool hs[110];//统计谁最后没办法在k步内全部踩爆
int mp[110][110];//读入矩阵
int flag[110];//匈牙利算法里面记录匹配的数组
bool vis[110];//匈牙利算法里面防止已经匹配的点再次匹配的标记数组
int step[110];//记录多少步能把这个颜色的所有气球踩爆
int n;//矩阵的阶数

int dfs(int u,int c)
{
for(int i = 1;i <= n;i++)
{
if(mp[u][i] == c && !vis[i])
{
vis[i] = true;
if(!flag[i] || dfs(flag[i],c))
{
flag[i] = u;
return 1;
}
}
}
return 0;
}

int main()
{
int k;
while(scanf("%d%d",&n,&k),n || k)
{
memset(mp,0,sizeof(mp));
memset(step,0,sizeof(step));
memset(hs,false,sizeof(hs));
for(int i = 1;i <= n;i++)
{
for(int j = 1;j <= n;j++)
scanf("%d",&mp[i][j]);
}
for(int i = 1;i <= 50;i++)
{
memset(flag,0,sizeof(flag));
for(int j = 1;j <= n;j++)
{
memset(vis,false,sizeof(vis));
step[i] += dfs(j,i);//统计全部踩爆的步数
}
}
int bj = 0;
for(int i = 1;i <= 50;i++)
{
if(step[i] > k)//如果全部踩爆的步数在给的步数之上
{
hs[i] = true;//标记下来k步内没办法踩爆的气球颜色编号
bj++;
}
}
if(!bj)
printf("-1\n");
else
{
int c = 0;
for(int i = 1;i <= 50;i++)
{
if(hs[i])
{
if(!c)
printf("%d",i);
else
printf(" %d",i);
c++;
}
}
printf("\n");
}
}
return 0;
}