ZOJ1298(单源最短路径)

这道题因为循环判断语句WA了三十多发,生活完全无法自理.
卒

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF 100000100         //无穷大
#define MAXN 550             //顶点个数的最大值
int n,m;                      //顶点个数
int Edge[MAXN][MAXN];       //邻接矩阵
int S[MAXN];                //Dijkstra算法用到的三个数组
int dist[MAXN];
int caseno = 1;
void Dijkstra(int v0)
{
    int i,j;
    for(i = 1;i <= n;i++){
        dist[i] = Edge[v0][i];
        S[i] = 0;
    }
    S[v0] = 1; //顶点v0加入到顶点集合S
    for(i = 1;i < n;i++){
        int min = INF,u;
                //选择当前集合T中具有最短路径的顶点u
        for(j = 1;j <= n;j++){
            if(!S[j]&&dist[j] < min){
                u = j; min = dist[j];
            }
        }
        S[u] = 1;
        for(j = 1;j <= n;j++){
            if(!S[j] && Edge[u][j] != INF&&dist[u]+Edge[u][j] < dist[j]){
                dist[j] = dist[u] + Edge[u][j];
            }
        }
    }
    double maxtime1 = -INF,maxtime2 = -INF; 
    int pos,pos1,pos2; //最后倒下的关键时间及位置
    for(i = 1;i <= n;i++){
        if(dist[i] > maxtime1){
            maxtime1 = dist[i]; pos = i;
        }
    }
       //每一行中间普通牌倒下的时间最大值及位置
    for(i = 1;i <= n;i++){
        for(j = i+1;j <= n;j++){
            double t = (dist[i]+dist[j]+Edge[i][j])/2.0;
            if(Edge[i][j] < INF&&t > maxtime2){
                maxtime2 = t;
                pos1 = i;
                pos2 = j;
            }
        }
    }
    if(maxtime2 > maxtime1)
        printf("The last domino falls after %.1f seconds, between key dominoes %d and %d.\n\n",maxtime2,pos1,pos2);
    else printf("The last domino falls after %.1f seconds, at key domino %d.\n\n",maxtime1,pos);
}

int main()
{
    while (~scanf("%d%d", &n, &m) && (n||m)) {
    memset(Edge,0,sizeof(Edge));

    int i,j;//读入顶点个数n,边数
    int u,v,w;              //边的起点和终点及权值

    for(i = 1;i <= m ;i++){
        scanf("%d%d%d",&u,&v,&w);       //读入边的起点和终点,权值
        Edge[u][v] = Edge[v][u] = w;                 //构造邻接矩阵
    }
    for(i = 1;i <= n;i++){
        for(j = i;j <= n;j++){
            if(j == i) Edge[i][j] = 0;
            else if(Edge[i][j] == 0)Edge[i][j] = Edge[j][i] = INF;

        }
    }
    printf("System #%d\n",caseno++);
    Dijkstra(1);            //求顶点0到其他顶点的最短路径
    }
    return 0;
}