【分析】1021 Deepest Root (25 分)【DFS解法】

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.

Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤10^4) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N−1 lines follow, each describes an edge by given the two adjacent nodes’ numbers.

Output Specification:
For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print Error: K components where K is the number of connected components in the graph.

Sample Input 1:
5
1 2
1 3
1 4
2 5
Sample Output 1:
3
4
5

Sample Input 2:
5
1 3
1 4
2 5
3 4
Sample Output 2:
Error: 2 components

题意：给出N，表示N个点，接下来有N-1对数，表示某两个点间有连接。判断是否为无环图，如果是无环图，则将它看做一棵树，求在哪个节点为根节点的情况下，树最深。如有并列，则升序输出根节点。

本题是一道细节题。

如果一幅图是一个无环连通图，则该图能构成一棵树。所以要判断给出的图是否是无环连通图，当然这道题降低了难度，因为题目指定了图中有N个节点，且只有N-1条边，那么如果这幅图是连通的，则必然无环；如果这幅图不连通，则必然有环。

因此，只需判断连通分量(连通块)是否大于1，就可知是否Error。下面贴出代码，请读者自行体会。

#include<bits/stdc++.h>
using namespace std;const int INF = 1e4+5;    //结点的最大数量
vector<vector<int>>graph(INF);
bool visit[INF];
int maxlevel[INF], level[INF];  //最大深度、当前深度void DFS(int v, int start) { //深度优先遍历 visit[v] = true;for(int i : graph[v])if(!visit[i]) {level[i] = level[v] + 1;    //当前节点深度为父节点深度+1maxlevel[start] = max(maxlevel[start], level[i]);DFS(i, start); }
} int main() {int n; cin >> n;for(int i = 1; i < n; i++) {int a, b; cin >> a >> b;graph[a].push_back(b);graph[b].push_back(a);}int num = 0;        //记录连通分量数量for(int i = 1; i <= n; i++) { //深搜求连通分量数量 if(!visit[i]) {++num;      //每进行一次遍历连通分量数+1DFS(i, i); } } if(num > 1) printf("Error: %d components\n", num);else {for(int i = 1; i <= n; i++) {memset(level, 0, sizeof(level));   //将level数组元素置为0memset(visit, false, sizeof(visit)); //将visit数组元素置为falseDFS(i, i); } vector<int>v; //存储结果int deepLevel = 0;for(int i = 1; i <= n; i++) if(maxlevel[i] > deepLevel) {v.clear();v.push_back(i);deepLevel = maxlevel[i];} else if(maxlevel[i] == deepLevel)v.push_back(i);for(int i = 0; i < v.size(); i++) printf("%d\n", v[i]);} return 0;
}

耗时：

【分析】1021 Deepest Root (25 分)【DFS解法】相关推荐

【PAT甲级】1021 Deepest Root (25 分)（暴力，DFS）
题意: 输入一个正整数N(N<=10000),然后输入N-1条边,求使得这棵树深度最大的根节点,递增序输出.如果不是一棵树,输出这张图有几个部分. trick: 时间比较充裕数据可能也不是很极限 ...
1021 Deepest Root (25 分) 【难度: 中 / 知识点: 树的直径连通块】
https://pintia.cn/problem-sets/994805342720868352/problems/994805482919673856 方法一: 数组模拟邻接表第一步: 爆搜df ...
1021. Deepest Root (25)
题目链接:http://www.patest.cn/contests/pat-a-practise/1021 题目: 1021. Deepest Root (25) 时间限制 1500 ms 内存限制 ...
浙大PAT 1021. Deepest Root (25)
1021. Deepest Root (25) 时间限制 1500 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A graph ...
【PAT - 甲级1021】Deepest Root (25分)（并查集，暴力枚举）
题干: A graph which is connected and acyclic can be considered a tree. The height of the tree depends ...
PTA Deepest Root (25分)
释放无限光明的是人心,制造无边黑暗的也是人心,光明和黑暗交织着,厮杀着,这就是我们为之眷恋又万般无奈的人世间. A graph which is connected and acyclic can b ...
PAT甲级-1021 Deepest Root（25分）
题目:1021 Deepest Root(25分) 分析:找出以某个节点为根时,最深的根,这题可能会超时要用vector来表示二维数组疑问:代码一是第二次写的超时了,代码二是第一次写的AC了,找不出 ...
PAT甲级1021 Deepest Root ：[C++题解]树的最大深度、并查集、dfs求树的深度
文章目录题目分析题目链接题目分析分析: 考察知识点:并查集.dfs.树的深度给定n个结点,n-1条边,只要能保证只有1个连通分量,就是一棵树.否则的话就不是树,它是不连通的. 用并查集来看是 ...
7-121 深入虎穴 (25 分)(dfs,bfs)
7-121 深入虎穴 (25 分) 著名的王牌间谍 007 需要执行一次任务,获取敌方的机密情报.已知情报藏在一个地下迷宫里,迷宫只有一个入口,里面有很多条通路,每条路通向一扇门.每一扇门背后或者是一 ...

【分析】1021 Deepest Root (25 分)【DFS解法】

立志用最少的代码做最高效的表达

耗时：

【分析】1021 Deepest Root (25 分)【DFS解法】相关推荐

最新文章

热门文章