poj1511

题目大意：

起点为1号点，求1到所有点最短路径及所有点到1的最短路径来回总和

思路：

因为题目规模太大，不能用二维数组来存储点与点之间的权值，而且考虑到也会超时，所以选用spfa加邻接表的方式来做这道题

1-所有其他点的最短路径很好求，而其他点到1的的最短距离可以通过把起点与终点反向从而求出的便是其他点到1的距离了

代码如下：

#include <iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxs = 1000000+5;
const __int64 INF = 0xFFFFFFFF;
__int64 dist1[maxs],dist2[maxs],ans=0;
struct Edge
{
    int e,next,weight;
}edge1[maxs],edge2[maxs];
int head1[maxs],head2[maxs];
int n,m;
void spfa(Edge edge[],__int64 dist[],int head[maxs])
{
    bool vis[maxs];
    memset(vis,false,sizeof(vis));
    for(int i=1;i<=n;i++)
        dist[i]=INF;
    dist[1]=0;
    queue<int> q;
    q.push(1);
    vis[1]=true;
    while(!q.empty())
    {
        int cur = q.front();
        q.pop();
        vis[cur]=false;
        for(int i=head[cur];i!=-1;i=edge[i].next)
        {
            if(dist[edge[i].e]>dist[cur]+edge[i].weight)
            {
                dist[edge[i].e]=dist[cur]+edge[i].weight;
                if(!vis[edge[i].e])
                {
                    q.push(edge[i].e);
                    vis[edge[i].e]=true;
                }
            }

        }
    }
}
int main()
{
    freopen("in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        memset(head1,-1,sizeof(head1));
        memset(head2,-1,sizeof(head2));
        memset(edge1,0,sizeof(edge1));
        memset(edge2,0,sizeof(edge2));
        for(int i=1;i<=m;i++)
        {
            int a,b,w;
            scanf("%d%d%d",&a,&b,&w);
            edge1[i].e=b;
            edge1[i].weight=w;
            edge1[i].next=head1[a];
            head1[a]=i;
            edge2[i].e=a;
            edge2[i].weight=w;
            edge2[i].next=head2[b];
            head2[b]=i;
        }
        ans=0;
        spfa(edge1,dist1,head1);
        spfa(edge2,dist2,head2);
        for(int i=1;i<=n;i++)
            ans+=(dist1[i]+dist2[i]);
        printf("%I64d\n",ans);
    }
    return 0;
}