HDOJ 1498-50 years, 50 colors water bipartite graph .. bipartite graph minimum vertex Overwrite

Question:

In a matrix of n * n (n <= 100) .. each lattice is filled with a color balloon (the color is measured in numbers 1 ~ 50)... each operation is to remove one row or one column of balloons with the same color. After k operations, ask which balloons of the same color still exist...

Question:

Separate each color... side (x, y ).... because for a balloon (x, y ).. as long as x is used or y is used .. then this balloon will be eliminated... how many points should I use to convert a question .. overwrite all edges .. that is, the classic bipartite graph least Point Coverage problem... it is equivalent to finding the maximum number of matching for a bipartite graph...

Program:

#include<iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<stack>#include<queue>#define ll long long#define MAXN 105using namespace std;int n,match[MAXN];bool arc[MAXN][MAXN],inmap[MAXN],used[MAXN],s[MAXN][MAXN]; bool dfs(int x){       int i;       for (i=1;i<=n;i++)          if (arc[x][i] && !used[i])          {                 used[i]=true;                 if (!match[i] || dfs(match[i]))                 {                       match[i]=x;                       return true;                 }          }              return false; }int getmax(){       int sum=0;       memset(match,0,sizeof(match));       for (int i=1;i<=n;i++)       {               memset(used,false,sizeof(used));               sum+=dfs(i);       }       return sum;}int main(){       int i,j,k,ans;       bool f;        while (~scanf("%d%d",&n,&k) && n)       {               memset(inmap,false,sizeof(inmap));               for (i=1;i<=n;i++)                  for (j=1;j<=n;j++)                     scanf("%d",&s[i][j]),inmap[s[i][j]]=true;               f=false;               for (int t=1;t<=50;t++)                  if (inmap[t])                  {                         memset(arc,false,sizeof(arc));                         for (i=1;i<=n;i++)                            for (j=1;j<=n;j++)                                if (s[i][j]==t) arc[i][j]=true;                         if (getmax()>k)                          {                                 if (!f) printf("%d",t);                                    else printf(" %d",t);                                 f=true;                         }                  }               if (!f) printf("-1");               printf("\n");        }       return 0;}