poj1135Domino Effect——最短路

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
queue<int>q;
int n,m,head[505],ct,dis[505],t;
double ans;
bool in[505];
struct N{
    int to,next,w;
    N(int t=0,int n=0,int o=0):to(t),next(n),w(o) {}
}edge[250005];
int main()
{
    while(scanf("%d%d",&n,&m)==2)
    {
        t++;
        if(!n&&!m)return 0;
        ct=0;
        memset(head,0,sizeof head);
        memset(dis,2,sizeof dis);
        memset(in,0,sizeof in);
        for(int i=1;i<=m;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            edge[++ct]=N(y,head[x],z);head[x]=ct;
            edge[++ct]=N(x,head[y],z);head[y]=ct;
        }
        while(q.size())q.pop();
        dis[1]=0;q.push(1);in[1]=1;
        while(q.size())
        {
            int x=q.front();q.pop();
            in[x]=0;
            for(int i=head[x];i;i=edge[i].next)
            {
                int u=edge[i].to;
                if(dis[x]+edge[i].w<dis[u])
                {
                    dis[u]=dis[x]+edge[i].w;
                    if(!in[u])in[u]=1,q.push(u);
                }
            }
        }
        int k=0,dk=0;ans=0;
        for(int i=1;i<=n;i++)
            if(dis[i]>=ans)//>=以处理n=1的情况 
            {
                ans=dis[i];
                k=i;
            }
        for(int i=head[k];i;i=edge[i].next)
        {
            int u=edge[i].to;
            if(dis[u]+edge[i].w>dis[k]&&ans<1.0*(edge[i].w-dis[k]+dis[u])/2+dis[k])
                ans=1.0*(edge[i].w-dis[k]+dis[u])/2+dis[k],dk=u;
        }
        printf("System #%d \n",t);
        if(dk>k)swap(dk,k);
        if(ans==dis[k])
            printf("The last domino falls after %.1lf seconds, at key domino %d.\n",ans,k);
        else
            printf("The last domino falls after %.1lf seconds, between key dominoes %d and %d.\n",ans,dk,k);
        printf("\n");
    }
    return 0;
}