You are my brother

You are my brother

时间限制: 1 Sec 内存限制: 128 MB

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题目描述

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

题意概括

给出数字之间的关系，问1和2的辈分关系；

解题思路

根据所给数据，搜索计算出1和2分别到祖先的距离，根据到祖先的距离然后判断出1和2的辈分关系、

代码

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<ctype.h>
#include<algorithm>
using namespace std;

int line[2010][2010];
int dfs(int x,int num)
{
    int i,j;
    num++;
    for(i=0;i<=2000;i++){
        if(line[x][i]){
            return dfs(i,num);
        }
    }
    return num;
}
int main()
{
    int n;
    int i,j,a,b;
    while(scanf("%d",&n)!=EOF)
    {
        memset(line,0,sizeof(line));
        for(i=0;i<n;i++){
            scanf("%d %d",&a,&b);
            line[a][b]=1;
        }
        a=dfs(1,0);
        b=dfs(2,0);
        if(a<b){
            printf("You are my younger\n");
        }else if(a==b){
            printf("You are my brother\n");
        }else{
            printf("You are my elder\n");
        }
    }
    return 0;
}