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。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 const int N=600; 6 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 7 int n,m,start,dis,path=0,shortest=1e8,allpeople=0;//path为最短路径数，allpeople为最大救援人数 8 void read() 9 { 10 memset(vis,0,sizeof(vis)); 11 fill(tu[0],tu[0]+N*N,0); 12 cin>>n>>m>>start>>dis; 13 for (int i=0;i<n;i++) 14 cin>>city[i]; 15 int a,b,c; 16 for (int i=0;i<m;i++) 17 { 18 cin>>a>>b>>c; 19 tu[a][b]+=c;//注意题目，是无向图！！ 20 tu[b][a]+=c; 21 } 22 } 23 void dfs(int step,int now,int people) 24 { 25 people+=city[now];//要先加上该城市救援人数！ 26 if (now==dis) 27 { 28 if (step<shortest) 29 { 30 shortest=step; 31 allpeople=people; 32 path=1; 33 } 34 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！！ 35 { 36 path++; 37 allpeople=max(allpeople,people); 38 } 39 return; 40 } 41 vis[now]=1; 42 for (int i=0;i<n;i++) 43 { 44 if (tu[now][i]>0&&vis[i]==0) 45 { 46 step+=tu[now][i]; 47 dfs(step,i,people); 48 step-=tu[now][i]; 49 } 50 } 51 vis[now]=0; 52 } 53 int main() 54 { 55 // freopen("in.txt","r",stdin); 56 read();//输入函数 57 dfs(0,start,0);//深搜遍历所有路径 58 cout<<path<<" "<<allpeople<<endl; 59 60 return 0; 61 }
