POJ 1511 Invitation Cards （Spfa)

题意：一群学生去城市的各个车站发邀请函，每个车站恰有一个学生。任意两个车站之间都可达，并且有一个特定的车费。假如学生们早上都从车站1出发，晚上从各个车站回到车站1，求总的最小花费。
题解：人生第一道Spfa。题意很明确，求带权最短路径。回来的时候反向建图即可。

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

#define N 1000001
#define INF 1999999999

struct Edge
{
	int v, w, next;
} edgeGo[N], edgeBack[N];

int P, Q;
int dis[N];
int headGo[N], headBack[N];
bool mark[N];
queue<int>que;

void spfa ( Edge edge[], int head[] )
{
	memset(mark,0,sizeof(mark));
	while ( ! que.empty() ) que.pop();

	int i, u, v;
	for ( i = 1; i <= P; i++ )
		dis[i] = INF;
	dis[1] = 0;
	mark[1] = 1;
	que.push(1);

	while ( !que.empty () )
	{
		u = que.front();
		que.pop();
		mark[u] = 0;
		for ( i = head[u]; i != 0; i = edge[i].next )
		{
			v = edge[i].v;
			if ( dis[v] > dis[u] + edge[i].w )
			{
				dis[v] = dis[u] + edge[i].w;
				if ( ! mark[v] )
				{
					que.push ( v );
					mark[v] = 1;
				}
			}
		}
	}
}


int main()
{
	int t, k, u, v, w;
	scanf("%d",&t);
	while ( t-- )
	{
		k = 0;
		memset(headGo,0,sizeof(headGo));
		memset(headBack,0,sizeof(headBack));

		scanf("%d%d",&P,&Q);
		while ( Q-- )
		{
			scanf("%d%d%d",&u,&v,&w);
			k++;
			edgeGo[k].v = v;
			edgeGo[k].w = w;
			edgeGo[k].next = headGo[u];
			headGo[u] = k;

			edgeBack[k].v = u;
			edgeBack[k].w = w;
			edgeBack[k].next = headBack[v];
			headBack[v] = k;
		}

		int i;
	    __int64 ans = 0;

		spfa ( edgeGo, headGo );
		for ( i = 1; i <= P; i++ )
			ans += dis[i];

		spfa ( edgeBack, headBack );
		for ( i = 1; i <= P; i++ )
			ans += dis[i];
		printf("%I64d\n",ans);
	}
	return 0;
}