目錄
1,題目描述
Sample Input:
9 1.80 1.00
1 5 4 4 -1 4 5 3 6
Sample Output:
1.85 2
題目大意
求一個供應鏈中,貨物可以賣出的最高價,以及賣出最高價的零售商數目
- each number Si is the index of the supplier for the i-th member.:編號爲0到N-1,Si即供應商,下標即商家的編號,比如3th=4,供應商4爲商家3提供貨物;
2,思路
和這一題很像@&再見螢火蟲&【PAT_甲級_1079 Total Sales of Supply Chain (25point(s)) (C++)【DFS/scanf接受double】】
簡單的DFS搜索,記錄遍歷的層數即可:
3,AC代碼
#include<bits/stdc++.h>
using namespace std;
double P, r;
int N, maxDep = 0, num = 0;
vector<int> data[100003];
void dfs(int r, int dep){
if(data[r].size() == 0){
if(dep > maxDep){
maxDep = dep;
num = 1;
}else if(dep == maxDep)
num++;
}
for(int i = 0; i < data[r].size(); i++)
dfs(data[r][i], dep + 1);
}
int main(){
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE
scanf("%d %lf %lf", &N, &P, &r);
int id, root;
for(int i = 0; i < N; i++){
scanf("%d", &id);
if(id == -1)
root = i;
else
data[id].push_back(i);
}
dfs(root, 0);
printf("%.2f %d", P * pow(1 + r / 100, maxDep), num);
return 0;
}
4,解題過程
一發入魂