1106 Lowest Price in Supply Chain

1. 本题和1090 Highest Price in Supply Chain适成对比，都是先构建一棵树，但本题是求最小层数和个数，链接题是求最大层数和个数。在极值更换和个数更新方面，两道题是一样的，但要注意，如果是求最小，一开始要初始化成最大，求最大，一开始要初始化成最小。

2. 我一开始尝试用链接题的DFS修改来通过这道，发现行不通，原因是DFS只会让层数增大，不可能越增大反而更容易满足，于是改成层次遍历，结点也加上层的属性，就做出来了。

AC代码

#include<cstdio>
#include<map>
#include<set>
#include<string>
#include<cstring>
#include<iostream>
#include<vector>
#include<stack>
#include<queue>
#include<algorithm>
#include<cmath>
typedef long long LL;using namespace std;const int maxn = 100010;int n;//结点总数
double initPrice,rate;//出厂价和利率
int endDepth = maxn,deepNum= 0;struct Node{vector<int> children;int layer;
}node[maxn]; bool appear[maxn] = {false};int findRoot(){for(int i=0;i<n;i++){if(!appear[i])return i;}
} void layerOrder(int root){queue<int> que;node[root].layer = 0;que.push(root);while(!que.empty()){int top = que.front();que.pop();int size = node[top].children.size();if(!size){//到了叶子结点了，看能不能替换最小层数 if(node[top].layer<endDepth){endDepth = node[top].layer;deepNum = 1;}else if(node[top].layer==endDepth)deepNum++;}else{for(int i=0;i<size;i++){int childI = node[top].children[i];node[childI].layer = node[top].layer+1;que.push(childI);}}}
}int main(){scanf("%d %lf %lf",&n,&initPrice,&rate);  int root,childNum,nodeIdx;for(int i=0;i<n;i++){scanf("%d",&childNum);while(childNum--){scanf("%d",&nodeIdx);appear[nodeIdx] = true;node[i].children.push_back(nodeIdx);}}root = findRoot();layerOrder(root);double hPrice = initPrice*pow(1+rate/100,endDepth);printf("%.4f %d",hPrice,deepNum); return 0;
}

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