PAT (Advanced Level) 1004. Counting Leaves (30)

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PAT (Advanced Level) 1004. Counting Leaves (30)

简单DFS. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<vector> using namespace std; +; vector<int>g[maxn]; int n,m; int ans[maxn]; int root; int Deep; void dfs(in…

PAT 解题报告 1004. Counting Leaves (30)

1004. Counting Leaves (30) A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100, th…

PTA (Advanced Level) 1004 Counting Leaves

Counting Leaves A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Specification: Each input file contains one test case. Each case starts with a line containing 0 < N < 100…

PAT (Advanced Level) 1049. Counting Ones (30)

数位DP.dp[i][j]表示i位,最高位为j的情况下总共有多少1. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<vector> using namesp…

【PAT甲级】1004 Counting Leaves (30 分)（BFS）

题意:给出一棵树的点数N,输入M行,每行输入父亲节点An,儿子个数n,和a1,a2,...,an(儿子结点编号),从根节点层级向下依次输出当前层级叶子结点个数,用空格隔开.(0<N<100) AAAAAccepted code: #include<bits/stdc++.h> using namespace std; vector<]; queue<int>q; ; ; ; ; ; void BFS(){ int x=q.front();//队头元素调用 if(v…

PAT 1004 Counting Leaves (30分)

1004 Counting Leaves (30分) A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Specification: Each input file contains one test case. Each case starts with a line containing 0…