1090 Highest Price in Supply Chain (25)

A supply chain is a network of retailers（零售商）, distributors（经销商）, and suppliers（供应商）-- everyone involved in moving a product from supplier to customer.

Starting from one root supplier, everyone on the chain buys products from one's supplier in a price P and sell or distribute them in a price that is r% higher than P. It is assumed that each member in the supply chain has exactly one supplier except the root supplier, and there is no supply cycle.

Now given a supply chain, you are supposed to tell the highest price we can expect from some retailers.

Input Specification:

Each input file contains one test case. For each case, The first line contains three positive numbers: N (≤10^5), the total number of the members in the supply chain (and hence they are numbered from 0 to N−1); P, the price given by the root supplier; and r, the percentage rate of price increment for each distributor or retailer. Then the next line contains N numbers, each number Si​ is the index of the supplier for the i-th member. Sroot​ for the root supplier is defined to be −1. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the highest price we can expect from some retailers, accurate up to 2 decimal places, and the number of retailers that sell at the highest price. There must be one space between the two numbers. It is guaranteed that the price will not exceed 10^10.

Sample Input:

9 1.80 1.00
1 5 4 4 -1 4 5 3 6

Sample Output:

1.85 2

题目大意：给一棵树，在树根处货物的价格为p，然后每往下走一层，价格增加r%，求这棵树的最高价格，以及这个价格的叶子结点个数。
分析：读入时，标记每个节点的父亲节点，并将这个节点的坐标存入它父亲节点的孩子数组中。读入完成后，从根节点开始BFS，每次遍历时，标记当前节点的深度。BFS完后，输出最深的节点的价值和个数即可。

这道题有点类似1079。注意开数组的时候，太大了需要用指针去分配，直接声明开不下。

#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
#include      <map>
#include      <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;

typedef struct node
{
    double val;
    int fat,depth,id;
    vector<int>child;
}node;

int main(void)
{
    #ifdef test
    freopen("in.txt","r",stdin);
    //freopen("in.txt","w",stdout);
    clock_t start=clock();
    #endif //test

    int n,root;
    double p,r;
    scanf("%d%lf%lf",&n,&p,&r);
    node *num=(node*)malloc(sizeof(node)*(n+5));
    for(int i=0;i<n;++i)
        num[i].val=0,num[i].fat=num[i].depth=-2,num[i].id=i;

    for(int i=0;i<n;++i)
    {
        int a;scanf("%d",&a);
        num[i].fat=a;
        if(a==-1)root=i,num[i].depth=0,num[i].val=p;
        else num[a].child.push_back(i);
    }
    node *sta=(node*)malloc(sizeof(node)*(n+5));sta[0]=num[root];

    int cnt=1,left=0,right=1,d=0;
    while(left<right)
    {
        d++;
        int ll=left,rr=right;
        for(int i=ll;i<rr;++i)
        {
            int len=sta[i].child.size();
            if(len>0)
            {
                for(int j=0;j<len;++j)
                {
                    int index=sta[i].child[j];
                    num[index].depth=d;num[index].val=sta[i].val*(1.0+r/100.0);
                    sta[cnt++]=num[index];
                }
            }
        }
        left=right,right=cnt;
    }
    int sum=0;
    double ans=0;
    for(int i=0;i<n;++i)
    {
        if(num[i].depth+1==d)
        {
            sum++;ans=num[i].val;
        }
    }
    printf("%.2f %d\n",ans,sum);
    free(num);
    free(sta);

    #ifdef test
    clockid_t end=clock();
    double endtime=(double)(end-start)/CLOCKS_PER_SEC;
    printf("\n\n\n\n\n");
    cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
    cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
    #endif //test
    return 0;
}