poj 1511 Invitation Cards

http://poj.org/problem?id=1511

题意：在一个图中，求从1号节点到2-n号节点的最短距离和和从2-n节点到1节点的最短距离和；

思路：因为数据太大，用spfa,且要建立双向的；

代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;
#define max 1000010
#define inf 1000000050
struct node
{
    int v;
    int data;
    int next;
}qian[max],hou[max];
int qq1[max];
int qq2[max];
int n,m;
int a,b;
int len;
long long mm[max];
bool visit[max];
int que[max];
void spfa(node *s,int *link)
{
     for(int i = 0;i <= n;++i)
     {
         mm[i] = inf;
         visit[i] = 0;
     }
     mm[1] = 0;
     visit[1] = 1;
     que[0] = 1;
     int front = 0;
     int rear = 1;
     int temp;
     while(front != rear)
     {
         temp = que[front];
         visit[temp] = 0;
         front++;
         if(front == max)
         front = 0;
         for(int i = link[temp];i != -1; i = s[i].next)
         {
             if(mm[s[i].v] > mm[temp] + s[i].data)
             {
             mm[s[i].v] = mm[temp] + s[i].data;
             if(!visit[s[i].v])
             {
                 que[rear++] = s[i].v;
                 if(rear == max)
                 rear = 0;
                 visit[s[i].v] = 1;
             }
             }
         }
     }
}
int main()
{
    int w = 0;
    scanf("%d",&w);
    while(w--)
    {
        scanf("%d%d",&n,&m);
        fill_n(qq1,max,-1);
        fill_n(qq2,max,-1);

        for(int i = 0;i < m; ++i)
        {
            scanf("%d%d%d",&a,&b,&len);
            qian[i].v = a;
            qian[i].data = len;
            qian[i].next = qq1[b];
            qq1[b] = i;
            hou[i].v = b;
            hou[i].data = len;
            hou[i].next = qq2[a];
            qq2[a] = i;
        }
        long long ans = 0;
        spfa(qian,qq1);
        for(int i = 1;i <= n; ++i)
        ans = ans + mm[i];
        spfa(hou,qq2);
        for(int i = 1;i <= n; ++i)
        ans = ans + mm[i];
        printf("%lld\n",ans);
    }
    return 0;
}