HDU 2066 SPFA实现0ms

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define Maxn 1010
#define Maxm 1010*1010
#define INT_MAX 1<<30
int used[Maxn],head[Maxn],outqueue[Maxn],queue[Maxn],low[Maxn],temp=-999999999,len = 0;
struct EdgeNode
{
    int to,w,next;
}edge[Maxm];

bool SPFA(int start)
{
    int i,iq;
    for(i = 0;i <= temp;i++)
    {
        low[i] = INT_MAX;
    }
    memset(used,0,sizeof(used));
    memset(outqueue,0,sizeof(outqueue));
    iq = 0;
    queue[iq++] = start;
    used[start] = 1;
    low[start] = 0;
    i = 0;
    while(i!=iq)
    {
        int top = queue[i];
        used[top] = 0;
        outqueue[top]++;
        if(outqueue[top] > temp) return false;
        for(int k = head[top];k!=-1;k = edge[k].next)
        {
            if(low[edge[k].to] > low[top] + edge[k].w)
            {
                low[edge[k].to] = low[top] + edge[k].w;
            }
            if(!used[edge[k].to])
            {
                used[edge[k].to] = 1;
                queue[iq++] = edge[k].to;
                len++;
            }
        }
        i++;
    }
    return true;
}
int main()
{
    int T,S,D;
    while(~scanf("%d%d%d",&T,&S,&D))
    {
        memset(head,-1,sizeof(head));
        int i,j,p=0;
        int a,b,w;
        int x,y;
        for(i = 0;i < T;i++)
        {
            scanf("%d%d%d",&a,&b,&w);
            edge[p].to = b;
            edge[p].w = w;
            edge[p].next = head[a];
            head[a] = p++;
            edge[p].to = a;
            edge[p].w = w;
            edge[p].next = head[b];
            head[b] = p++;
            if(a > temp) temp = a;
            if(b > temp) temp = b;
        }
        temp++;
        for(i = 0;i < S;i++)
        {
            scanf("%d",&x);
            edge[p].to = x;
            edge[p].w = 0;
            edge[p].next = head[0];
            head[0] = p++;
            edge[p].to = 0;
            edge[p].w = 0;
            edge[p].next = head[x];
            head[x] = p++;
        }
        for(i = 0;i < D;i++)
        {
            scanf("%d",&y);
            edge[p].to = y;
            edge[p].w = 0;
            edge[p].next = head[temp];
            head[temp] = p++;
            edge[p].to = temp;
            edge[p].w = 0;
            edge[p].next = head[y];
            head[y] = p++;
        }
//        for(i = 0;i <= temp;i++)
//        {
//            for(j = head[i];j != -1;j = edge[j].next)
//            {
//                cout<<i<<" "<<edge[j].to<<" "<<edge[j].w<<endl;
//            }
//        }
        SPFA(0);
        printf("%d\n",low[temp]);
    }
    return 0;
}

建图的时候一定要建双向的因为有used这个数组进行已遍历的判断所以不用担心