题意：给你一个邻接矩阵信息，某点发生火灾，告诉你一些位置有消防队，问各个消防队到火灾地点的最短时间，并输出最短路的路径，输出按最短时间由小到大排序。

就是一个最短路问题，输出路径，直接dijkstra了，1A还是挺爽的

#include<cstring>
#include<string>
#include<fstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
#include<functional>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
#define MAXN 100010
#define eps 1e-7
#define INF 0x7FFFFFFF
#define seed 131
#define long long ll;
#define unsigned long long ull;
#define lson l,m,rt<<1
#define rson r,m+1,rt<<1|1

struct node{
    int id, s, time, pathl;
    int path[25];
}ans[25];
int edge[25][25];
int vis[25];
int dist[25];
int path[25];
int n;
bool cmp(node x,node y){
    return x.time<y.time;
}
void dijkstra(int s,int e){
    int i, j;
    for(i=1;i<=n;i++){
        dist[i] = edge[s][i];
    }
    memset(vis,0,sizeof(vis));
    memset(path,0,sizeof(path));
    vis[s] = 1;
    for(i=0;i<n-1;i++){
        int temp = INF, k = -1;
        for(j=1;j<=n;j++){
            if(!vis[j]&&dist[j]<temp){
                k = j;
                temp = dist[j];
            }
        }
        if(k==-1)   break;
        vis[k] = 1;
        for(j=1;j<=n;j++){
            if(!vis[j]&&edge[k][j]!=INF&&dist[j]>dist[k]+edge[k][j]){
                dist[j] = dist[k] + edge[k][j];
                path[j] = k;
            }
        }
    }
}
int main(){
    int i, j, s, e, x;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        for(j=1;j<=n;j++){
            scanf("%d",&x);
            if(x==-1)   edge[i][j] = INF;
            else    edge[i][j] = x;
        }
    }
    scanf("%d",&e);
    int k = 0;
    printf("Org\tDest\tTime\tPath\n");
    while(scanf("%d",&s)!=EOF){
        dijkstra(s,e);
        ans[k].id = k + 1;
        ans[k].s = s;
        ans[k].time = dist[e];
        ans[k].pathl = 1;
        int ll = 1;
        ans[k].path[0] = e;
        int ee = e;
        while(path[ee]!=0){
            ans[k].path[ll++] = path[ee];
            ee = path[ee];
            ans[k].pathl++;
        }
        ans[k].path[ll++] = s;
        ans[k].pathl++;
        if(ans[k].pathl==2&&ans[k].path[0]==ans[k].path[1]) ans[k].pathl = 1;
        k++;
    }
    sort(ans,ans+k,cmp);
    for(i=0;i<k;i++){
        printf("%d\t%d\t%d\t",ans[i].s,e,ans[i].time);
        for(j=ans[i].pathl-1;j>0;j--){
            printf("%d\t",ans[i].path[j]);
        }
        printf("%d\n",ans[i].path[j]);
    }
    printf("\n");
    return 0;
}

/*
6
0  3  4 -1 -1 -1
-1 0  4  5 -1 -1
2  3  0 -1 -1  2
8  9  5  0  1 -1
7  2  1 -1  0 -1
5 -1  4  5  4  0
5 5
*/