NBUT1218-You are my brother

最新推荐文章于 2019-04-03 16:09:36 发布

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本文介绍了一个基于深度优先搜索的家庭成员关系查询算法，该算法能够确定两个人之间的长幼关系。通过递归的方式，算法能有效判断出给定编号的两人是否为兄弟、兄妹或有明确的长幼之分。

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You are my brother
时间限制: 1000 ms　内存限制: 131072 K
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

样例输出
```
You are my elder
You are my brother
```
提示
```
无
```
来源
```
辽宁省赛2010
```

题意：给出一张家谱，问1号和2号谁的年龄大

解题思路：dfs

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <functional>
#include <climits>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int visit[20005];
int f[20005];
int n,a,b,k,kk;

void dfs(int x,int y)
{
    if(f[x]==x) {k=y;return ;}
    dfs(f[x],y+1);
}

void dfs1(int x,int y)
{
    if(f[x]==x) {kk=y;return ;}
    dfs1(f[x],y+1);
}

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=0;i<20000;i++) f[i]=i;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&a,&b);
            f[a]=b;
        }
        dfs(1,1);
        dfs1(2,1);
        if(kk==k) printf("You are my brother\n");
        else if(kk>k) printf("You are my younger\n");
        else printf("You are my elder\n");
    }
    return 0;
}