codeforces 320B bfs

题目：

B. Ping-Pong (Easy Version)

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In this problem at each moment you have a set of intervals. You can move from interval(a, b) from our set to interval (c, d) from our set if and only if c < a < d orc < b < d. Also there is a path from intervalI₁ from our set to intervalI₂ from our set if there is a sequence of successive moves starting fromI₁ so that we can reachI₂.

Your program should handle the queries of the following two types:

1. "1 x y" (x < y) — add the new interval(x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals.
2. "2 a b" (a ≠ b) — answer the question: is there a path froma-th (one-based) added interval to b-th (one-based) added interval?

Answer all the queries. Note, that initially you have an empty set of intervals.

Input

The first line of the input contains integer n denoting the number of queries,(1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed10⁹ by their absolute value.

It's guaranteed that all queries are correct.

Output

For each query of the second type print "YES" or "NO" on a separate line depending on the answer.

Examples

Input

51 1 51 5 112 1 21 2 92 1 2

Output

NOYES

题意：区间(a,b)和(c,d)若a<c<b或c<b<d则两个区间相通。有两种操作：1、增加一个区间，2、检查两个点是否相通。

方法：广度优先搜索。

代码：

//By Sean Chen#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;struct road{    int u,v;};road Map[110];int cnt,visit[110];int check(int x,int y){    int a=Map[x].u,b=Map[x].v;    int c=Map[y].u,d=Map[y].v;    if((c<a && a<d) || (c<b && b<d))        return 1;    return 0;}void bfs(int x,int y){    queue <int> Q;    Q.push(x);    while(Q.size())    {        int a=Q.front();        Q.pop();        for(int i=0;i<cnt;i++)            if(!visit[i] && check(a,i))            {                Q.push(i);                visit[i]=1;            }    }    return;}int main(){    int T,a,x,y;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&a,&x,&y);        if(a==1)        {            Map[cnt].u=x;            Map[cnt].v=y;            cnt++;        }        else        {            memset(visit,0,sizeof(visit));            bfs(x,y);            if(visit[y])                printf("YES\n");            else                printf("NO\n");        }    }    return 0;}