fafu 1568 Matrix（二分匹配+二分）

Description:

 

      

You are given a matrix which is n rows m columns (1 <= n <= m <= 100). You are supposed to choose n elements, there is no more than 1 element in the same row and no more than 1 element in the same column. What is the minimum value of the K_th largest in the n elements you have chosen.

 

Input:

 

First line is an integer T (T ≤ 15), the number of test cases. At the beginning of each case is three integers n, m, K, 1 <= K <= n <= m <= 100. 
      Then n lines are given. Each line contains m integers, indicating the matrix. 
      All the elements of the matrix is in the range [1, 10^9]. 

 

Output:

 

For each case, output Case #k: one line first, where k means the case number count from 1. Then output the answer.

 

Sample Input:

 

2
2 3 1
1 2 4
2 4 1

3 4 2
1 5 6 6
8 3 4 3
6 8 6 3

 

Sample Output:

 

Case #1: 1 
Case #2: 3

二分枚举答案，如果a[i][j]<=mid则建一条xi->yj的边，然后对二分图跑最大匹配，如果最大匹配数>=n-k+1,则ans<=mid，否则ans>mid

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 106
#define inf 1e10
int n,m,k;
int mp[N][N];
int cnt[N][N];
int vis[N];
int link[N];
vector<int> v;

void build_map(int mid){
   memset(cnt,0,sizeof(cnt));
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            if(mp[i][j]<=mid){
               cnt[i][j]=1;
            }
        }
    }
}

bool dfs(int u){
   for(int i=0;i<m;i++){
       if(!vis[i] && cnt[u][i]){
           vis[i]=1;
           if(link[i]==-1 || dfs(link[i])){
              link[i]=u;
              return true;
           }
       }
   }
   return false;
}
bool solve(int mid){
   build_map(mid);
      memset(link,-1,sizeof(link));
      int ans=0;
      for(int i=0;i<n;i++){
          memset(vis,0,sizeof(vis));
          if(dfs(i)){
             ans++;
          }
      }

      if(ans>=(n-k+1)) return true;
      return false;



}
int main()
{
   int ac=0;
   int t;
   scanf("%d",&t);
   while(t--){
       memset(mp,0,sizeof(mp));
       scanf("%d%d%d",&n,&m,&k);
       for(int i=0;i<n;i++){
           for(int j=0;j<m;j++){
               scanf("%d",&mp[i][j]);
           }
       }
       int low=0;
       int high=inf;
       while(low<high){
           int mid=(low+high)>>1;
           if(solve(mid)){
               high=mid;
               //ans=mid;
           }else{
               low=mid+1;
           }
       }
       printf("Case #%d: ",++ac);
       printf("%d
",low);



   }
    return 0;
}