POJ2135: Farm Tour 题解

这题可以十分巧妙的转化

一来一回可以转化成寻找两条没用重边的去的路

把每条边的流量都设为1，费用是时间

然后就可以跑一个最大流量为2的费用流

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <utility>
#include <map>
#include <stack>
#include <set>
#include <vector>
#include <queue>
#include <deque>
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define LL long long
#define Pair pair<int,int>
#define LOWBIT(x) x & (-x)
using namespace std;

const int MOD=1e9+7;
const int INF=0x7ffffff;
const int magic=348;

int n,e;
priority_queue<Pair> q;
int dist[100048],h[100048];

int preve[100048],prevv[100048];
int head[100048],to[200048],nxt[200048],f[200048],w[200048],tot=1;
inline void addedge(int s,int t,int cap,int cost)
{
	to[++tot]=t;nxt[tot]=head[s];head[s]=tot;f[tot]=cap;w[tot]=cost;
	to[++tot]=s;nxt[tot]=head[t];head[t]=tot;f[tot]=0;w[tot]=-cost;
}

void dijkstra()
{
	int i,x,y,dd;
	for (i=1;i<=n;i++) dist[i]=INF;
	dist[1]=0;
	q.push(mp(0,1));
	while (!q.empty())
	{
		x=q.top().y;dd=q.top().x;
		dd=-dd;q.pop();
		//cout<<x<<endl;
		if (dd>dist[x]) continue;
		for (i=head[x];i;i=nxt[i])
		{
			y=to[i];
			if (f[i] && dist[y]>dist[x]+w[i]+h[x]-h[y])
			{
				dist[y]=dist[x]+w[i]+h[x]-h[y];
				prevv[y]=x;preve[y]=i;
				q.push(mp(-dist[y],y));
			}
		}
	}
}

int min_cost_flow()
{
	int i,x,y,u,minf,res=0;
	for (i=1;i<=n;i++) h[i]+=dist[i];
	//for (i=1;i<=n;i++) cout<<h[i]<<' ';
	//cout<<endl;
	minf=INF;
	for (u=n;u!=1;u=prevv[u])
		minf=min(minf,f[preve[u]]);
	res=minf*h[n];
	for (u=n;u!=1;u=prevv[u])
	{
		f[preve[u]]-=minf;
		f[preve[u]^1]+=minf;
	}
	return res;
}

int main ()
{
	//freopen ("shit.out","w",stdout);
	int i,x,y,c;
	scanf("%d%d",&n,&e);
	for (i=1;i<=e;i++)
	{
		scanf("%d%d%d",&x,&y,&c);
		addedge(x,y,1,c);
		addedge(y,x,1,c);
	}
	int ans=0;
	//for (i=head[1];i;i=nxt[i]) cout<<to[i]<<' ';
	//cout<<endl;
	for (i=1;i<=2;i++)
	{
		dijkstra();//cout<<i<<endl;
		ans+=min_cost_flow();
		//cout<<"_______________________"<<endl;
	}
	printf("%d\n",ans);
	return 0;
}