PAT1079 Total Sales of Supply Chain

传送门

给一颗销售供应的树。在树根初货物价格为P，从树根每往子节点走一层，价格会上涨R%。给出叶节点的货物量，求出价格之和。

思路：dfs树遍历。

#include <iostream>
#include<bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define scl(x) scanf("%lld",&x)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define prs(x) printf("%s\n",(x))
#define prl(x) printf("%lld\n",x)
#define ll long long
#define PII pair<int,int>
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=1e6+5;
struct node{
    vector<int>child;
    int level;
    int cnt;
}tree[maxn];
int n;
double p,r;
int m,cnt,son;
int flag[maxn];
double ans;
void dfs(int root){
    if(tree[root].child.size()==0){
        double price=p*pow((1+r*0.01),tree[root].level);
        ans+=tree[root].cnt*price;
        return;
    }
    rep(i,0,tree[root].child.size()){
        tree[tree[root].child[i]].level=tree[root].level+1;
        dfs(tree[root].child[i]);
    }

}
int main(){
    cin>>n>>p>>r;
    rep(t,0,n){
        cin>>m;
        if(m==0){
            cin>>cnt;
            tree[t].cnt=cnt;
        }
        else{
            rep(i,0,m){
                cin>>son;
                tree[t].child.push_back(son);
                flag[son]++;
            }
        }
    }
    int root=0;
    while(flag[root]) root++;
    dfs(root);
    printf("%.1lf\n",ans);
}