NBUT 1218 You are my brother

[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");elseprintf("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");elseprintf("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TLEfa[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");elseprintf("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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NBUT 1218 You are my brother

[1218] You are my brother

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