881
技術社區[雲棲]
NYOJ282-You are my brother
You are my brother
時間限製：1000 ms | 內存限製：65535 KB
難度：3
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.
輸入
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.
輸出
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.
樣例輸入
5
1 3
2 4
3 5
4 6
5 6
6
1 3
2 4
3 5
4 6
5 7
6 7
樣例輸出
You are my elder
You are my brother
來源
遼寧省10年省賽
AC代碼:
#include<stdio.h>
#include<string.h>
struct node
{
int father;
}bin[3000];
int main()
{
int i,j,n,m,sum1,sum2,flag1,flag2,x,y;
while(scanf("%d",&n)!=EOF)
{
flag1=0;flag2=0;
sum1=0;sum2=0;
for(i=0;i<n;i++)
{
scanf("%d %d",&x,&y);
if(flag1==0&&x==1)//第一次記錄1的父輩
{
flag1=1;bin[1].father=y;
}
if(flag2==0&&x==2)//第一次記錄2的父輩
{
flag2=1;bin[2].father=y;
}
if(flag1==1&&x==bin[1].father)//如果幾錄過1的父輩,1的父輩有祖先,那麼更新1的父輩,並且輩分加1
{
sum1++;bin[1].father=y;
}
if(flag2==1&&x==bin[2].father)//如果幾錄過2的父輩,2的父輩有祖先,那麼更新2的父輩,並且輩分加1
{
sum2++;bin[2].father=y;
}
}
//根據1和2的輩分大小來判斷兩人關係
if(sum1<sum2)
printf("You are my younger\n");
else if(sum1>sum2)
printf("You are my elder\n");
else
printf("You are my brother\n");
}
return 0;
}
最後更新：2017-04-03 12:56:03