fafu 1568 Matrix（二分匹配+二分）

Description:

 

You are given a matrix which is n rows m columns ( <= n <= m <= ). You are supposed to choose n elements, there is no more than  element in the same row and no more than  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 ≤ ), the number of test cases. At the beginning of each case is three integers n, m, K,  <= K <= n <= m <= .
      Then n lines are given. Each line contains m integers, indicating the matrix.
      All the elements of the matrix is in the range [, ^]. 

 

Output:

 

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

 

Sample Input:

 

 

Sample Output:

 

Case #:
Case #: 

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

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

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

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

    }
     return ;
 }