pat 1079. Total Sales of Supply Chain(树模拟)

偏简单的题目

注意考虑N=1的特殊情况

用BFS求根叶距离（如果暴力从每个叶结点回溯的话会超时）

#include<stdio.h>#include<math.h>#include<vector>#include<queue>using namespace std;int number[100005];int lev[100005];vector<int> vs[100005];queue<int> q;int main(){int n,num,j,i,x,y,l,sz;double p,r;scanf("%d%lf%lf",&n,&p,&r);for(i=0;i<n;i++){scanf("%d",&num);if(num){for(j=0;j<num;j++){scanf("%d",&x);vs[i].push_back(x);}}else{scanf("%d",&x);number[i]=x;}}q.push(0);while(!q.empty()){y=q.front();sz=vs[y].size();for(l=0;l<sz;l++){lev[vs[y][l]]=lev[y]+1;q.push(vs[y][l]);}    q.pop();}double sum=0,per;if(n==1){sum=p*number[0];printf("%.1lf\n",sum);return 0;}for(i=0;i<n;i++){if(number[i]){per=1+r/100;sum+=pow(per,lev[i])*number[i];}}sum*=p;printf("%.1lf\n",sum);return 0;}