POJ 2426 Interesting Housing Problem (最大费用最大流)

最新推荐文章于 2019-12-18 21:58:47 发布

转载最新推荐文章于 2019-12-18 21:58:47 发布 · 84 阅读

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原文链接：http://www.cnblogs.com/toyking/p/3797397.html

本文通过最大费用最大流算法解决了一个有趣的住房分配问题，详细介绍了算法的应用过程及其实现步骤。

POJ 2426 Interesting Housing Problem (最大费用最大流)

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

const int maxn=30000;
const int inf=200000000;
struct edge{
	int u,v,next,f,c;
	edge(int u0=0,int v0=0,int f0=0,int c0=0,int next0=0){
		u=u0,v=v0,f=f0,c=c0,next=next0;
	}	
}e[maxn*10];
int head[maxn*2],visited[maxn*2],path[maxn*2],dist[maxn*2];
int cnt,from,to,marked,maxflow,offflow,n,r,m,casen;

void initial(){
	to=2*n+2*r+1;casen++;
	cnt=0;marked=1;
	maxflow=0;from=0;
	for(int i=0;i<=to;i++){
		head[i]=-1;
		visited[i]=0;
	}
}

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

void input(){
	for(int i=1;i<=n;i++){
		adde(0,i,1,0);
		adde(i,i+n,1,0);
	}
	for(int i=1;i<=r;i++){
		adde(2*n+i,2*n+r+i,1,0);
		adde(2*n+r+i,to,1,0);
	}
	while(m-- >0){
		int u0,v0,c0;
		scanf("%d%d%d",&u0,&v0,&c0);
		if(c0>=0) adde(u0+1+n,2*n+v0+1,1,c0);
	}
	
}

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

void computing(){
	printf("Case %d: ",casen);
	if(r<n){
		cout<<"-1"<<endl;
		return;
	}
	for(int k=0;k<n;k++){ 
     	bfs();
     	/*cout<<to;
     	for(int i=to;i!=from;i=e[path[i]].u){
     		cout<<"->"<<e[path[i]].u;
     	}
     	cout<<"  ans:"<<dist[to]<<endl;*/
     	if(dist[to]<=-inf){
     		cout<<"-1"<<endl;
			return;
     	}
		maxflow+=dist[to];
		for(int i=to;i!=from;i=e[path[i]].u){
     		e[path[i]].f-=1;
         	e[path[i]^1].f+=1;
		}
	}
    printf("%d\n",maxflow);
}

int main(){
	while(cin>>n>>r>>m){
		initial();
		input();
		computing();
	}
	return 0;
}

转载于:https://www.cnblogs.com/toyking/p/3797397.html