2010年辽宁省赛 NBUT 1218【DFS实现树的遍历与更新】

• [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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