Optimal Milking 分类： 图论 POJ 最短路 查找 2015-08-10 10:38 3人阅读 评论(0) 收藏

Optimal Milking

Time Limit: 2000MS Memory Limit: 30000K

Total Submissions: 13968 Accepted: 5044

Case Time Limit: 1000MS

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

Source

USACO 2003 U S Open

题意:有c头奶牛和k个挤奶器,奶牛之间,挤奶器之间,奶牛与挤奶器之间都有一定的距离,问使c头奶牛都能挤奶的最大距离的最小值;

做法:距离问题,怎样知道任意点之间的距离,而且是最短距离,最短路算法floyd.

怎样求最大距离的最小值:由于数据量比较大,不能枚举,采用二分;

要确定二分的值是不是正确,就要用Dinic算法求出最大流看看是否>=c

#include <map>
#include <list>
#include <climits>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#define WW freopen("output.txt","w",stdout)

const int Max = 300;

int Map[Max][Max];

int Dis[Max][Max];

bool vis[Max];

bool sign[Max][Max];

int K,c,m;

int n;

void BuildGraph(int MaxNum)//建立残留网络
{
    memset(Map,0,sizeof(Map));
    for(int i=K+1;i<=n;i++)
    {
        Map[0][i]=1;//将每一头奶牛与源点建立弧,对于每一头奶牛只去一个挤奶器
    }
    for(int i=1;i<=K;i++)
    {
        Map[i][n+1]=m;//将每个挤奶器与汇点建立弧,对于每个挤奶器最多接受m头奶牛
    }
    for(int i=K+1;i<=n;i++)
    {
        for(int j=1;j<=K;j++)
        {
            if(Dis[i][j]<=MaxNum)//限制条件
            {
                Map[i][j]=1;
            }
        }
    }
}
bool BFS()//Dinic算法,建立层次网络;
{
    memset(vis,false,sizeof(vis));
    memset(sign,false,sizeof(sign));
    int a;
    queue<int>Q;
    vis[0]=true;
    Q.push(0);
    while(!Q.empty())
    {
        a=Q.front();
        Q.pop();
        for(int i=0;i<=n+1;i++)
        {
            if(!vis[i]&&Map[a][i])
            {
                vis[i]=true;
                sign[a][i]=true;
                Q.push(i);
            }
        }
    }
    if(vis[n+1])//当标记不到汇点的时候说明残留网络中没有增广路,已经求得最大流,算法结束
    {
        return true;
    }
    return false;
}
int DFS(int s,int num)//
{
    if(s==n+1)
    {
        return num;
    }
    int sum = num;
    for(int i=0;i<=n+1;i++)
    {
        if(sign[s][i])
        {
            int ans=DFS(i,min(Map[s][i],num));//进行多次的增广,
            Map[s][i]-=ans;
            Map[i][s]+=ans;
            num-=ans;
            if(!num)//如果残余容量为零,则不必再增广
            {
                break;
            }
        }
    }
    return sum-num;
}
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])
                {
                    Dis[i][j]=INF;
                }
            }
        }
        for(int k=1;k<=n;k++)//通过floyd算法计算任意点之间的距离
        {
            for(int i=1;i<=n;i++)
            {
                if(Dis[i][k]!=INF)
                {
                    for(int j=1;j<=n;j++)
                    {
                        Dis[i][j]=min(Dis[i][j],Dis[i][k]+Dis[k][j]);
                    }

                }
            }
        }
        int L=0,R=10000;
        while(L<R)//通过不断的二分找到满足条件的最小值
        {
            int mid = (L+R)/2;
            int sum = 0;
            BuildGraph(mid);
            while(BFS())//Dinic算法,直到不能增广时结束
            {
                sum+=DFS(0,INF);//计算有多少的奶牛挤奶
            }
            if(sum>=c)
            {
                R=mid;
            }
            else
            {
                L=mid+1;
            }
        }
        printf("%d\n",R);//R就是所要求的最大距离的最小值
    }
    return 0;
}

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