2010年辽宁省赛 NBUT 1218【DFS实现树的遍历与更新】
2015-08-18 01:48
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[1218] You are my brother
时间限制: 1000 ms 内存限制: 131072 K链接:NBUT 1218
问题描述
Little
A gets to know a new friend, Little B, recently. One day, they realize that they are family 500 years ago. Now, Little A wants to know whether Little B is his elder, younger or brother.
输入
There are multiple test cases.
For each test case, the first line has a single integer, n (n<=1000). The next n lines have two integers a and b (1<=a,b<=2000) each, indicating b is the father of a. One person has exactly one father, of course. Little A is numbered 1 and Little B is numbered
2.
Proceed to the end of file.
输出
For each test case, if Little B is Little A’s younger, print “You are my younger”. Otherwise, if Little B is Little A’s elder, print “You are my elder”. Otherwise, print “You are my brother”. The output for each test case occupied exactly one line.
样例输入
5 1 3 2 4 3 5 4 6 5 6 6 1 3 2 4 3 5 4 6 5 7 6 7
样例输出
You are my elder You are my brother
题意:
给定N次输入,每次输入一行a,b,表示b是a的父亲。最后求节点1与节点2的关系【即比较深度】N<=1000,1<=a,b<=2000
分析:
题意很明确,对于每个节点我都用一个值Rank保存该节点的深度,所以每次把节点a为根节点的树合并到节点b为根节点的树上面的时候,我要把从节点a以及节点a以下的节点全部的Rank值都加上节点b的Rank值。就这样,一个结构体保存着每个节点的信息,每次合并两棵树的时候DFS更新这些节点就OK了。代码实现:
#include <cmath> #include <cstdio> #include <string> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define FIN freopen("input.txt","r",stdin) struct Node { int Rank; vector<int> Son; } Nodes[2000 + 5]; void dfs(int pos, const int& val) { Nodes[pos].Rank += val; int num = Nodes[pos].Son.size(); if(num == 0) return; for(int i = 0; i < num; i++) { int &np = Nodes[pos].Son[i]; dfs(np, val); } } void init() { for(int i = 1; i <= 2000; i++) { Nodes[i].Rank = 1; Nodes[i].Son.clear(); } } int main() { // FIN; int n, a, b; while(~scanf("%d", &n)) { init(); for(int i = 1; i <= n; i++) { scanf("%d%d", &a, &b); Nodes[b].Son.push_back(a); dfs(a, Nodes[b].Rank); } if(Nodes[1].Rank == Nodes[2].Rank) printf("You are my brother\n"); else if(Nodes[1].Rank > Nodes[2].Rank) printf("You are my elder\n"); else printf("You are my younger\n"); } return 0; }
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