NBUT 1218 You are my brother

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
  • 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
  • 提示
  • 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……
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