poj 2391 Ombrophobic Bovines 二分+最大流

标签：poj   算法   

同poj 2112.

代码：

//poj 2391
//sep9
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long ll;

const int maxN=1024;
const int maxM=100002;
const ll MAX=(1ULL<<63)-1; 

struct Edge
{
	int v,f,nxt;
}e[maxM*2+10];

queue<int> que;
int src,sink;
int g[maxN+10];
int nume;
bool vis[maxN+10];
int dist[maxN+10];

ll d[maxN][maxN];
int in[maxN],out[maxN];
int f,p,sum;
 
void addedge(int u,int v,int c)
{
	e[++nume].v=v;e[nume].f=c;e[nume].nxt=g[u];g[u]=nume;
	e[++nume].v=u;e[nume].f=0;e[nume].nxt=g[v];g[v]=nume;
}

void init()
{
	memset(g,0,sizeof(g));	
	nume=1;
}

int bfs()
{
	while(!que.empty()) que.pop();
	memset(dist,0,sizeof(dist));
	memset(vis,0,sizeof(vis));
	vis[src]=true;
	que.push(src);	
	while(!que.empty()){
		int u=que.front();que.pop();
		for(int i=g[u];i;i=e[i].nxt)
			if(e[i].f&&!vis[e[i].v]){
				que.push(e[i].v);
				dist[e[i].v]=dist[u]+1;
				vis[e[i].v]=true; 
				if(e[i].v==sink)
					return 1;
			}
	}
	return 0;
}

int dfs(int u,int delta)
{
	if(u==sink)
		return delta;
	int ret=0;
	for(int i=g[u];ret<delta&&i;i=e[i].nxt)
		if(e[i].f&&dist[e[i].v]==dist[u]+1){
			int dd=dfs(e[i].v,min(e[i].f,delta-ret));
			if(dd>0){
				e[i].f-=dd;
				e[i^1].f+=dd;
				ret+=dd;
			}
			else
				dist[e[i].v]=-1;
		}	
	return ret;
}

int dinic()
{
	int ret=0;
	while(bfs()==1)
		ret+=dfs(src,INT_MAX);
	return ret;	
}
bool work(ll max_len)
{
	init();
	src=0,sink=2*f+1;		
	for(int i=1;i<=f;++i)
		addedge(src,i,in[i]);
	for(int i=1;i<=f;++i)
		for(int j=1;j<=f;++j)
			if(d[i][j]<=max_len)
				addedge(i,j+f,INT_MAX);
	for(int j=f+1;j<=2*f;++j)
		addedge(j,sink,out[j-f]);
	if(dinic()==sum)
		return true;
	return false;
}

int main()
{
	scanf("%d%d",&f,&p);
	for(int i=1;i<=f;++i){
		for(int j=1;j<=f;++j)
			d[i][j]=d[j][i]=MAX;
		d[i][i]=0;
	}
	sum=0;
	for(int i=1;i<=f;++i){
		scanf("%d%d",&in[i],&out[i]);
		sum+=in[i];
	}
	while(p--){
		int a,b;
		ll c;
		scanf("%d%d%I64d",&a,&b,&c);
		c=min(d[a][b],c);
		d[a][b]=d[b][a]=c;
	}
	for(int k=1;k<=f;++k)
		for(int i=1;i<=f;++i)
			for(int j=1;j<=f;++j)
				if(d[i][k]!=MAX&&d[k][j]!=MAX)
					d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
	ll l=0,r=MAX/2,mid,ans=-1;
	while(l<r){
		mid=(l+r)/2;
		if(work(mid)){
			ans=mid;
			r=mid;
		}else
			l=mid+1;
	}
	printf("%I64d",ans);
	return 0;	
}
 

poj 2391 Ombrophobic Bovines 二分+最大流

标签：poj   算法   

原文地址：http://blog.csdn.net/sepnine/article/details/45148261