Codeforces Round #401 (Div. 2)：E. Hanoi Factory

E. Hanoi Factory

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Of course you have heard the famous task about Hanoi Towers, but did you know that there is a special factory producing the rings for this wonderful game? Once upon a time, the ruler of the ancient Egypt ordered the workers of Hanoi Factory to create as high tower as possible. They were not ready to serve such a strange order so they had to create this new tower using already produced rings.

There are n rings in factory's stock. The i-th ring has inner radius ai, outer radius bi and height hi. The goal is to select some subset of rings and arrange them such that the following conditions are satisfied:

• Outer radiuses form a non-increasing sequence, i.e. one can put the j-th ring on the i-th ring only if bj ≤ bi.
• Rings should not fall one into the the other. That means one can place ring j on the ring i only if bj > ai.
• The total height of all rings used should be maximum possible.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of rings in factory's stock.

The i-th of the next n lines contains three integers ai, bi and hi (1 ≤ ai, bi, hi ≤ 109, bi > ai) — inner radius, outer radius and the height of the i-th ring respectively.

Output

Print one integer — the maximum height of the tower that can be obtained.

Examples

input

31 5 12 6 23 7 3

output

6

input

41 2 11 3 34 6 25 7 1

output

4

http://codeforces.com/contest/777/problem/E

比赛的时候码到一半时间到了，但思路是没错的，用栈模拟一下就好了，所有圆环按大R排序，相同则按小r排序

（都是从大到小），如果一个圆环放不上去，那么后面的圆环会全部都放不上去，这个时候要将汉诺塔顶端的圆环

一个一个撤下，直到能放为止继续放，中间最大的高度即是答案

#include<stdio.h>#include<stack>#include<algorithm>#define LL long longusing namespace std;typedef struct Res{int h;int R, r;bool operator < (const Res &b) const{if(R>b.R || R==b.R && r>b.r)return 1;return 0;}}Res;Res s[100005];stack<Res> st;int main(void){int i, n;LL sum, ans;scanf("%d", &n);for(i=1;i<=n;i++)scanf("%d%d%d", &s[i].r, &s[i].R, &s[i].h);sort(s+1, s+n+1);sum = ans = s[1].h;st.push(s[1]);for(i=2;i<=n;i++){while(st.empty()==0 && s[i].R<=st.top().r){sum -= st.top().h;st.pop();}st.push(s[i]);sum += s[i].h;ans = max(ans, sum);}printf("%I64d\n", ans);return 0;}