PAT甲级 1106 Lowest Price in Supply Chain (25分) DFS/C++
2020-06-24 04:26
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1106 Lowest Price in Supply Chain (25分)
DFS深度优先遍历
题目大意:一个供应商给多个经销商或零售商供应商品,每经过一个经销商,价格将上涨 r% 。要求找到消费者能够拿到最便宜的价格,以及可以从多少家零售商拿到最便宜价格。
也就是说:给定一棵树,要求找到最短路径,并求出最短路径数量。
#include<iostream> #include<vector> using namespace std; vector<vector<int>> v;//存放每一个结点的孩子 int n,num,minDepth=100001;//结点数,最短路径条数,最短路径长度 void dfs(int root,int depth); int main(){ #ifdef ONLINE_JUDGE //如果有oj系统(在线判定),则忽略文件读入,否则使用文件作为标准输入 #else freopen("1.txt", "r", stdin); //从1.txt输入数据 #endif double p,r; cin>>n>>p>>r; v.resize(n); int *flag=new int[n]; for(int i=0;i<n;i++){ int k; cin>>k; for(int j=0;j<k;j++){ int x;cin>>x; v[i].push_back(x); flag[x]=1; } } int i=0; for(;i<n;i++){ if(flag[i]!=1) break;//找到根结点 } dfs(i,0); double minp=p; for(i=0;i<minDepth;i++){ minp+=minp*r*0.01; } printf("%.4f %d\n",minp,num); return 0; } void dfs(int root,int depth){ if(v[root].size()){ for(int i=0;i<v[root].size();i++){ dfs(v[root][i],depth+1); } } else { if(depth<minDepth) { minDepth=depth; num=1; } else if(depth==minDepth) { num++; } return ; } }
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