Optimal Milking (二分图多重匹配+最短路)_optimal milking 二分图多重匹配-优快云博客

本文探讨了如何分配奶牛到挤奶机的问题，旨在最小化最远行走距离的同时确保挤奶机不过载。通过使用Floyd算法进行路径优化，并采用深度优先搜索策略分配奶牛，实现了有效的资源分配。

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FJ has moved his K (1 <= K <= 30) milking machines out into the cow pastures among the C (1 <= C <= 200) cows. A set of paths of various lengths runs among the cows and the milking machines. The milking machine locations are named by ID numbers 1..K; the cow locations are named by ID numbers K+1..K+C.

Each milking point can "process" at most M (1 <= M <= 15) cows each day.

Write a program to find an assignment for each cow to some milking machine so that the distance the furthest-walking cow travels is minimized (and, of course, the milking machines are not overutilized). At least one legal assignment is possible for all input data sets. Cows can traverse severa

l paths on the way to their milking machine.

Input

* Line 1: A single line with three space-separated integers: K, C, and M.

* Lines 2.. ...: Each of these K+C lines of K+C space-separated integers describes the distances between pairs of various entities. The input forms a symmetric matrix. Line 2 tells the distances from milking machine 1 to each of the other entities; line 3 tells the distances from machine 2 to each of the other entities, and so on. Distances of entities directly connected by a path are positive integers no larger than 200. Entities not directly connected by a path have a distance of 0. The distance from an entity to itself (i.e., all numbers on the diagonal) is also given as 0. To keep the input lines of reasonable length, when K+C > 15, a row is broken into successive lines of 15 numbers and a potentially shorter line to finish up a row. Each new row begins on its own line.

Output

A single line with a single integer that is the minimum possible total distance for the furthest walking cow.

Sample Input

Sample Output

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<vector>
const int INF=100000;
using namespace std;
int kk,c,m,n,mid;
int dis[1300][1300];
int cnt[1300];
int a[1300][1300];
int v[1300];
void floyd()
{
    int i,j,k;
    for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
            for(k=1;k<=n;k++)
                dis[j][k]=min(dis[j][i]+dis[i][k],dis[j][k]);
}
int dfs(int u)
{
    for(int i=1;i<=kk;i++)
    {
        if(v[i]) continue;
        if(dis[u][i]>mid) continue;
        v[i]=1;
        if(cnt[i]<m)
        {
            a[i][cnt[i]++]=u;
            return 1;
        }
       else
       {
           for(int j=0;j<cnt[i];j++)
           {
               if(dfs(a[i][j]))
               {
                   a[i][j]=u;
                   return 1;
               }
           }
       }

    }
    return 0;
}
int hh()
{
    memset(cnt,0,sizeof(cnt));
   for(int i=kk+1;i<=n;i++)
   {
       memset(v,0,sizeof(v));
       if(!dfs(i)) return 0;
    }
    return 1;
}
int main()
{
  while(~scanf("%d%d%d",&kk,&c,&m))
  {
        n=kk+c;
      for(int i=1;i<=n;i++)
          for(int j=1;j<=n;j++)
          {
            scanf("%d",&dis[i][j]);
              if(dis[i][j]==0)
                dis[i][j]=INF;
          }
          floyd();
        int l=0,r=100000;
        while(l<r)
        {
            mid=(l+r)/2;
            if(hh())
             r=mid;
            else
              l=mid+1;
        }
        printf("%d\n",r);
  }
}