Round #189 (Div.2) B. Ping-Pong (Easy Version)

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.

Sample test(s)

Input

51 1 51 5 112 1 21 2 92 1 2

Output

NOYES

————————————————————Dividing Line————————————————————

思路：一开始我们以为这是区间之间有交集，就属于同一个集合，否则就不是。然后他们用并查集做了一遍，错了。后来发现不是简单的集合关系，因为有可能从区间A可以到区间B但是从区间B去无法到达区间A。比如（5，6）和（3，8）。所以要用dfs，在查询的时候，从起点dfs到终点，判断是否有路。

代码如下：

#include <cstdio>#include <cstdlib>#include <cstring>const int N = 105;struct Node{int l, r;}inter[N];bool vis[N];int cnt = 0;bool judge(int id_a, int id_b){if(inter[id_a].l > inter[id_b].l && inter[id_a].l < inter[id_b].r)return true;if(inter[id_a].r > inter[id_b].l && inter[id_a].r < inter[id_b].r)return true;return false;}bool dfs(int id_s, int id_e){if(id_s == id_e)return true;vis[id_s] = true;for(int i = 1; i <= cnt; i++) if(!vis[i]) {if(judge(id_s, i) && dfs(i, id_e))return true;}return false;}int main(){int n;int op, a, b;scanf("%d", &n);while(n--) {scanf("%d%d%d", &op, &a, &b);if(op == 1) {cnt++;inter[cnt].l = a;inter[cnt].r = b;}else if(op == 2) {memset(vis, false, sizeof(vis));if(dfs(a, b))puts("YES");elseputs("NO");}}return 0;}