PAT (Advanced Level) Practice 1003 Emergency

思路：用深搜遍历出所有可达路径，每找到一条新路径时，对最大救援人数和最短路径数进行更新。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 const int N=;
 int tu[N][N],city[N],vis[N];//tu存道路，city存各城市救援人数，vis为标记数组
 int n,m,start,dis,path=,shortest=1e8,allpeople=;//path为最短路径数，allpeople为最大救援人数
 void read()
 {
     memset(vis,,sizeof(vis));
     fill(tu[],tu[]+N*N,);
     cin>>n>>m>>start>>dis;
     for (int i=;i<n;i++)
         cin>>city[i];
     int a,b,c;
     for (int i=;i<m;i++)
     {
         cin>>a>>b>>c;
         tu[a][b]+=c;//注意题目，是无向图！！
         tu[b][a]+=c;
     }
 }
 void dfs(int step,int now,int people)
 {
     people+=city[now];//要先加上该城市救援人数！
     if (now==dis)
     {
         if (step<shortest)
         {
             shortest=step;
             allpeople=people;
             path=;
         }
         else if (step==shortest)//这里要用else if不能用if！！如果用if会在第一次更新时满足这两个if判断，导致path多加了一次！！
         {
             path++;
             allpeople=max(allpeople,people);
         }
         return;
     }
     vis[now]=;
     for (int i=;i<n;i++)
     {
         if (tu[now][i]>&&vis[i]==)
         {
             step+=tu[now][i];
             dfs(step,i,people);
             step-=tu[now][i];
         }
     }
     vis[now]=;
 }
 int main()
 {
 //    freopen("in.txt","r",stdin);
     read();//输入函数
     dfs(,start,);//深搜遍历所有路径
     cout<<path<<" "<<allpeople<<endl;

     return ;
 }