首页 > 代码库 > POJ2112_Optimal Milking(网洛流最大流Dinic+最短路Flody+二分)

POJ2112_Optimal Milking(网洛流最大流Dinic+最短路Flody+二分)

2024-07-19 00:36:13   225人阅读

解题报告

农场有k个挤奶机和c头牛，每头牛到每一台挤奶机距离不一样，每台挤奶机每天最多挤m头牛的奶。

寻找一个方案，安排每头牛到某一挤奶机挤奶，使得c头牛须要走的全部路程中的最大路程的最小值。

要使每一头牛都去挤奶，那么建完模型就要推断是否满流。

因为是多源多点的网络，如果源点0,汇点n+1（n=k+c）

源点到每一头牛的容量为1,每一台机器到汇点的容量为m;用flody求出随意一头牛到随意一台机器的最短路;

对于取最大距离的最小值能够用二分来找。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define inf 99999999
#define K 33
#define C 210
using namespace std;
int k,c,m,mmap[K+C][K+C],l[K+C],edge[K+C][K+C],n;
void flody()
{
    for(int l=1; l<=n; l++)
    {
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(mmap[i][j]>mmap[i][l]+mmap[l][j])
                    mmap[i][j]=mmap[i][l]+mmap[l][j];
            }
        }
    }
}
int bfs()
{
    memset(l,-1,sizeof(l));
    queue<int>Q;
    Q.push(0);
    l[0]=0;
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        for(int i=0; i<=n+1; i++)
        {
            if(edge[u][i]&&l[i]==-1)
            {
                l[i]=l[u]+1;
                Q.push(i);
            }
        }
    }
    if(l[n+1]>0)return 1;
    else return 0;
}
void G(int mid)
{
    int i,j;
    memset(edge,0,sizeof(edge));
    for(i=1; i<=k; i++)
        edge[i][n+1]=m;
    for(i=k+1; i<=n; i++)
        edge[0][i]=1;
    for(i=k+1; i<=n; i++)
    {
        for(j=1; j<=k; j++)
        {
            if(mmap[i][j]<=mid)
                edge[i][j]=1;
            else edge[i][j]=0;
        }
    }
}
int dfs(int x,int f)
{
    int a;
    if(x==n+1)return f;
    for(int i=0; i<=n+1; i++)
    {
        if(edge[x][i]&&l[i]==l[x]+1&&(a=dfs(i,min(f,edge[x][i]))))
        {
            edge[x][i]-=a;
            edge[i][x]+=a;
            return a;
        }
    }
    return 0;
}
int dinic(int mid)
{
    int ans=0,a;
    G(mid);
    while(bfs())
        while(a=dfs(0,inf))
            ans+=a;
    return ans;
}
int main()
{
    int i,j;
    while(~scanf("%d%d%d",&k,&c,&m))
    {
        n=k+c;
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                scanf("%d",&mmap[i][j]);
                if(!mmap[i][j])
                    mmap[i][j]=inf;
            }
        }
        flody();
        int L=0,R=20000;
        while(L<R)
        {
            int mid=(L+R)/2;
            int ans=dinic(mid);
            if(ans>=c)R=mid;
            else L=mid+1;
        }
        printf("%d\n",R);
    }
    return 0;
}




Optimal Milking
Time Limit: 2000MS Memory Limit: 30000K
Total Submissions: 11664 Accepted: 4238
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


POJ2112_Optimal Milking(网洛流最大流Dinic+最短路Flody+二分)


联系
我们