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：

View Code

#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;

}