POJ 2112 Optimal Milking （二分+最短路径+网络流）

POJ  2112 Optimal Milking （二分+最短路径+网络流）

Optimal Milking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 10176		Accepted: 3698
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

题目大意：这题的主要意思就是，K台机器，C头牛，每台机器可供M个牛，然后K+C行以及列 告诉它们间的路径，问所有牛中，走得最远的那头牛，走了多远？

解题思路：二分最远距离，floyd最短路径算出牛到每台机器的距离，根据二分的距离来构造图的连通性，构造网络图后用网络流计算最大流量是否为牛的总数，一步步二分得到结果。

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstdlib>
using namespace std;

const int maxn=300;
const int inf=1<<28;
struct edge{
	int u,v,next,f;
	edge(int u0=0,int v0=0,int f0=0,int next0=0){
		u=u0,v=v0,f=f0,next=next0;
	}	
}e[maxn*maxn];
int head[maxn*2],visited[maxn*2],path[maxn*2],a[maxn][maxn];
int cnt,from,to,marked,K,C,M;

void ini(){
	from=0;to=K+C+1;
}

void initial(){
	cnt=0;marked=1;
	for(int i=0;i<=to;i++){
		head[i]=-1;
		visited[i]=0;
	}
}

void adde(int u,int v,int f){
	e[cnt].u=u,e[cnt].v=v,e[cnt].f=f,e[cnt].next=head[u],head[u]=cnt++;
	e[cnt].u=v,e[cnt].v=u,e[cnt].f=0,e[cnt].next=head[v],head[v]=cnt++;
}

void input(){
	int c0,n=K+C;
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++){
		scanf("%d",&c0);
		if(c0>0) a[i][j]=c0;
		else a[i][j]=inf;
	}
	for(int k=1;k<=n;k++)
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++){
		if(a[i][k]+a[k][j]<a[i][j]) a[i][j]=a[i][k]+a[k][j];
	}
}

void build(int x){
	initial();
	for(int i=1;i<=K;i++) adde(from,i,M); //机器 1 - K 编号 
	for(int i=K+1;i<=K+C;i++) adde(i,to,1); //牛 K+1 - K+C 编号
	for(int i=1;i<=K;i++){
		for(int j=K+1;j<=K+C;j++){
			if(a[i][j]<=x) adde(i,j,1);
		}
	}
} 

bool bfs(){
     int s=from,d;
     queue <int> q;
     q.push(s);
     marked++;
     visited[s]=marked;
     while(!q.empty()){
         s=q.front();
         q.pop();
         for(int i=head[s];i!=-1;i=e[i].next){
              d=e[i].v;
              if(visited[d]!=marked && e[i].f>0){
                   visited[d]=marked;
                   path[d]=i;
                   q.push(d);
                   if(d==to) return true;
              }
         }
     }
     return false;
}

int maxf(int x){
     build(x);	  
	 int maxflow=0; 
     while(bfs() ){
     	int offflow=inf;   	
     	for(int i=to;i!=from;i=e[path[i]].u){
     		offflow=min(e[path[i]].f,offflow);
        }
        for(int i=to;i!=from;i=e[path[i]].u){
     		e[path[i]].f-=offflow;
         	e[path[i]^1].f+=offflow;
        }
        maxflow+=offflow;
     }
     return maxflow;
}

void computing(){
	int l=1,r=200000;
	while(l<r){
		int mid=(l+r)/2;
		if(maxf(mid)>=C) r=mid;
		else l=mid+1;
	}
	cout<<r<<endl;
}

int main(){
	while(cin>>K>>C>>M){
		ini();
		input();
		computing();
	}
	return 0;
}