洛谷P1629

题目链接：邮递员

思路：先正向存边，求出最短路后再反向建边，再跑一遍最短路即可！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<bits/stdc++.h>
using namespace std;
const int N = 520000;
const int maxn = 1200;
struct Node{
	int ne;
	int to;
	int w;
}e[N<<1];
int head[maxn],dis[maxn];
int n,m,x[N],y[N],val[N],cnt;
bool vis[maxn];
void init()
{
	memset(head,-1,sizeof(head));
	memset(dis,0x7f,sizeof(dis));
	memset(vis,0,sizeof(vis));
	cnt = 0;
}
void add(int u,int v,int val)
{
	e[cnt].to = v;
	e[cnt].ne = head[u];
	e[cnt].w = val;
	head[u] = cnt ++;
}
void SPFA(int k)
{
	queue<int>q;
	q.push(k);
	dis[k] = 0;
	vis[k] = 1;
	while(!q.empty())
	{
		int now = q.front();
		q.pop();
		for(int i= head[now];~i;i=e[i].ne)
		{
			int to = e[i].to;
			if(dis[to] > dis[now] + e[i].w)
			{
				dis[to] = dis[now] + e[i].w;
				if(!vis[to])
				{
					q.push(to);
					vis[to] = 1;
				}
			}
		}
		vis[now] = 0;
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	init();
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&x[i],&y[i],&val[i]);
		add(x[i],y[i],val[i]);
	} 
	SPFA(1);
	int sum = 0;
	for(int i=1;i<=n;i++)
	{
		sum += dis[i];
	}
	memset(e,0,sizeof(e));
	init();
	for(int i=1;i<=m;i++)
	{
		add(y[i],x[i],val[i]);
	}
	for(int i=2;i<=n;i++)
	{
		SPFA(1);
		sum += dis[i];
	}
	printf("%d\n",sum);
	return 0;
}