NBUT 1218 You are my brother_you are my brothers csdn-优快云博客

本文详细阐述了如何通过家族关系图判断两个个体之间的年龄关系，包括输入格式、算法流程及输出结果，适用于解决家庭历史记录分析问题。

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[1218] 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

题目意思很简单，输入N对数a，b，表示b是a的父亲，最后求的是1和2的关系，也就是树的深度。（注意，若a和b不在同一棵树上，则他们的关系为brother）

代码一：

#include <cstring>
#include <cstdio>
int main() {
    int n,x,y;
    int ans[2005];
    while(~scanf("%d",&n)) {
        memset(ans,0,sizeof(ans));
        for(int i=0; i<n; i++) {
            scanf("%d%d",&x,&y);
            ans[x]=y;
        }
        int temp=1,a=0,b=0;
        while(ans[temp]){
            a++;
            temp=ans[temp];
        }
        temp=2;
        while(ans[temp]!=0) {
            b++;
            temp=ans[temp];
        }
        if(a>b)
            printf("You are my elder\n");
        else if(a<b)
            printf("You are my younger\n");
        else
            printf("You are my brother\n");
    }
    return 0;
}

代码二：

#include <cstdio>
using namespace std;
int main() {
    int n,x,y;
    int la,lb,a,b;
    while(~scanf("%d",&n)) {
        la=1;   lb=2;    a=0;    b=0;
        while(n--){
            scanf("%d%d",&x,&y);
            if(la!=lb) {
                if(x==la) {
                    la=y;   a++;
                }
                if(x==lb) {
                    lb=y;   b++;
                }
            }
        }
        if(a>b)
            printf("You are my elder\n");
        else if(a<b)
            printf("You are my younger\n");
        else
            printf("You are my brother\n");
    }
    return 0;
}

代码三：

#include <cstdio>
int fa[2010];
int n,x,y;
int Find(int x,int &H) {
    while(x != fa[x]) {
        x = fa[x];
        H++;
    }
    return x;
}
int main() {
    while(~scanf("%d",&n)) {
        for(int i = 1; i <= 2000; i++)      //用并查集时，初始化要注意上限，如果用n会TLE
            fa[i] = i;
        while(n--) {
            scanf("%d %d",&x,&y);
            fa[x] = y;
        }
        int a = 0 , b = 0 ;
        int f1 = Find( 1 , a );
        int f2 = Find( 2 , b );
        if(f1 != f2)                  //此时 1 和 2 并不在同一棵树上
            printf("You are my brother\n");
        else {
            if(a > b)
                printf("You are my elder\n");
            else if(a < b)
                printf("You are my younger\n");
            else
                printf("You are my brother\n");
        }
    }
    return 0;
}

以上代码有一部分是根据网上题解思路所写的，据说这题还可以用 DFS+线段树来做，只是目前还没搞懂，等待更新……