POJ 2122 Optimal Milking

Optimal Milking

Description

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 several 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

2 3 2
0 3 2 1 1
3 0 3 2 0
2 3 0 1 0
1 2 1 0 2
1 0 0 2 0

Sample Output

2

题意：有k个挤奶器c头奶牛，k个挤奶机位置用编号1到k表示，奶牛位置用k+1到k+c编号来表示。输入是一个矩阵的形式，对角线表示与自身的位置距离为0，其他为0的位置表示没有边相互连接。问怎么安排每个奶牛到某个挤奶器挤奶，使得c头奶牛需要走的所有路程中最大路程最小。

分析：先用floyd算法求出任意两点之间的最短路，然后在用Dinic求出最大流；在搜索最大距离确定最小值时使用二分法。

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#define INF 9999999
#define N 333
using namespace std;
int k,m,c;
int n;
int dis[N][N];//记录两点之间的最短路径
int mp[N][N];//容量网络
int qu[N*100];
int vis[N];//标记数组
int sign[N][N];//层次网络

void build_map(int x)
{
    memset(mp,0,sizeof mp);
    for(int i=k+1;i<=n;i++) mp[0][i]=1;
    for(int i=1;i<=k;i++) mp[i][n+1]=m;

    for(int i=k+1;i<=n;i++)
        for(int j=1;j<=k;j++)
        {
            if(dis[i][j]<=x) mp[i][j]=1;
        }
}

int bfs()//构建层次网络
{
    memset(vis,0,sizeof vis);
    memset(sign,0,sizeof sign);

    int qs=0,qe=1;
    vis[0]=1;
    qu[0]=0;

    while(qs<qe)
    {
        int v=qu[qs++];
        for(int i=0;i<=n+1;i++)
        {
            if(vis[i]==0 && mp[v][i])
            {
                vis[i]=1;
                qu[qe++]=i;
                sign[v][i]=1;
            }
        }
    }
    if(vis[n+1]) return 1;
    else return 0;
}

int dfs(int v,int sum)//dfs寻找增广路
{
    if(v==n+1) return sum;
    int s=sum;
    int t;

    for(int i=0;i<=n+1;i++)
    {
        if(sign[v][i])
        {
            t=dfs(i,min(sum,mp[v][i]));
            mp[v][i]-=t;
            mp[i][v]+=t;
            sum-=t;
        }
    }
    return s-sum;
}

int main()
{
    while(~scanf("%d%d%d",&k,&c,&m))
    {
        n=k+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;
            }

        for(int t=1;t<=n;t++)//floyd求解最短路
            for(int i=1;i<=n;i++)
            {
                if(dis[i][t]!=INF)
                    for(int j=1;j<=n;j++)
                    dis[i][j]=min(dis[i][j],dis[i][t]+dis[t][j]);
            }

        int le=0,ri=10009;
        int ans;
        while(le<ri)//二分找出最小的最大值
        {
            int mid=(le+ri)/2;
            ans=0;
            build_map(mid);//Dinic算法求最大流

            while(bfs()) ans+=dfs(0,INF);

            if(ans>=c) ri=mid;
            else le=mid+1;
        }

        printf("%d\n",ri);//满足条件的最小的mid用于更新ri
    }
    return 0;
}

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