poj1135Domino Effect

/*poj1135Domino Effect
有n个关键的多米诺骨牌，两个多米诺骨牌中间有一排骨牌，一块骨牌倒下需要1s，从第一个关键的骨牌开始翻到，问全部骨牌翻到需要多少时间&&最后一块骨牌的位置
思路：
求每块关键骨牌翻到的时间（以为源点最小路径），每一条边全部倒下需要时间(d[u]+d[v]+val)/2;
注意：1,0->1.0s 1
*/
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 1000000000
#define MAXN 506
struct E
{
    int u,v,val;
}e[MAXN*MAXN];
int w[MAXN][MAXN],vis[MAXN],d[MAXN];
int n,m;
void dijkstra(int u)
{
    memset(vis,0,sizeof(vis));
    for(int i = 1 ; i <= n ; i++)d[i]=(i==u ? 0 : INF);
    for(int i = 1 ; i <= n ; i++)
    {
        int x,mi = INF;
        for(int y = 1 ; y <= n ; y++)if(!vis[y]&&d[y]<=mi)mi=d[x=y];
        vis[x]=1;
        for(int y = 1 ; y <= n ; y++)if(!vis[y]&&w[x][y])d[y]=min(d[y],d[x]+w[x][y]);
    }
}
double ans;
int ku,kv;
void DominoEffect()
{
    double det,tmp;
    ans=0;
    for(int i=1;i<=m;i++)
    {
        int u=e[i].u,v=e[i].v,val=e[i].val;
        det=abs(d[u]-d[v]);
        //tmp=min(d[u],d[v])+det+(val-det)*0.5;
        tmp=(d[u]+d[v]+val)*0.5;
        if(ans<tmp)
        {
            ans=tmp;
            if(det==val)
            {
                if(d[u]<d[v])kv=ku=v;
                else kv=ku=u;
            }
            else 
            {
                ku=min(u,v);
                kv=max(u,v);
            }
        }
    }
}
int main()
{
    int u,v,val,cas=0;
    while(scanf("%d%d",&n,&m)==2)
    {
        if(!n&&!m)break;
        memset(w,0,sizeof(w));
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&u,&v,&val);
            e[i].u=u,e[i].v=v,e[i].val=val;
            w[u][v]=val;
            w[v][u]=val;
        }
        dijkstra(1);
        DominoEffect();
        printf("System #%d\n",++cas);
        
        printf("The last domino falls after %.1lf seconds,",ans);
        if(n==1)printf(" at key domino 1.\n\n");
        else if(kv==ku)printf(" at key domino %d.\n\n",kv);
        else printf(" between key dominoes %d and %d.\n\n",ku,kv);
    }
}