poj2112 二分+floyd+dinic

Optimal Milking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 12936		Accepted: 4679
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

题意：K个挤奶器，C头牛，每个挤奶器最多能挤M头牛的奶，如何分配使得所有牛都被挤一次奶，并且使得这些牛到挤奶器所走路程中最长的最小化。

此题和poj2391类似，都是二分距离，然后最大流。

首先预处理各点之间的最短路，然后源点连向所有牛一条边，边权为1，所有挤奶器连向汇点，边权为M，对于二分的距离，若牛i与挤奶器j的最短路<=该距离，则加一条边(i,j),边权为1.

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#define Maxn 250
using namespace std;

int K,C,M;
int dist[Maxn][Maxn];
void floyd(){
    for(int k=1;k<=K+C;k++)
        for(int i=1;i<=K+C;i++)
            for(int j=1;j<=K+C;j++)
                dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
}
const int inf=0x3f3f3f3f;
struct line{
    int to,next,cap;
}p[Maxn*Maxn*2];
int head[Maxn];
int q[Maxn];
int d[Maxn];
int tot;
int src,t;
int n,m;
void addedge(int a,int b,int c){
    p[tot].to=b;
    p[tot].next=head[a];
    p[tot].cap=c;
    head[a]=tot++;
}
void insert(int a,int b,int c){
    addedge(a,b,c);
    addedge(b,a,0);
}
bool bfs(){
    memset(d,-1,sizeof d);
    int s=0,e=-1;
    q[++e]=src;
    d[src]=0;
    while(s<=e){
        int u=q[s++];
        for(int i=head[u];i!=-1;i=p[i].next){
            int v=p[i].to;
            if(d[v]==-1&&p[i].cap){
                d[v]=d[u]+1;
                q[++e]=v;
            }
        }
    }
    return d[t]!=-1;
}
int dfs(int u,int alpha){
    if(u==t) return alpha;
    int w,used=0;
    for(int i=head[u];i!=-1&&used<alpha;i=p[i].next){
        int v=p[i].to;
        if(p[i].cap&&d[v]==d[u]+1){
            w=dfs(v,min(alpha-used,p[i].cap));
            used+=w;
            p[i].cap-=w;
            p[i^1].cap+=w;
        }
    }
    if(!used) d[u]=-1;
    return used;
}
int dinic(){
    int ans=0;
    src=0,t=K+C+1;
    while(bfs())
        ans+=dfs(src,inf);
    return ans;
}
void add(int mid){
    memset(head,-1,sizeof head);
    tot=0;
    for(int i=1;i<=C;i++) insert(0,K+i,1);
    for(int i=K+1;i<=K+C;i++)
        for(int j=1;j<=K;j++)
            if(dist[i][j]<=mid) insert(i,j,1);
    for(int i=1;i<=K;i++) insert(i,K+C+1,M);
}
int main()
{
    while(cin>>K>>C>>M){
        for(int i=1;i<=K+C;i++)
            for(int j=1;j<=K+C;j++){
                scanf("%d",&dist[i][j]);
                if(i!=j&&!dist[i][j]) dist[i][j]=inf;
            }
        floyd();
        int l=0,r=inf;
        while(l<r){
            int mid=l+r>>1;
            add(mid);
            if(dinic()<C) l=mid+1;
            else r=mid;
        }
        printf("%d\n",l);
    }
	return 0;
}