Codeforces Round #401(Div. 2)E. Hanoi Factory【贪心+栈】

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. Thei-th ring has inner radius a_i, outer radius b_i and height h_i. 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 ifb_j ≤ b_i.
• Rings should not fall one into the the other. That means one can place ring j on the ring i only if b_j > a_i.
• 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 a_i,b_i andh_i (1 ≤ a_i, b_i, h_i ≤ 10⁹,b_i > a_i) — inner radius, outer radius and the height of thei-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

Note

In the first sample, the optimal solution is to take all the rings and put them on each other in order3, 2, 1.

In the second sample, one can put the ring 3 on the ring4 and get the tower of height 3, or put the ring 1 on the ring 2 and get the tower of height 4.

题目大意：

现在一共有N个零件，如果存在：bi>=bj&&bj>ai的两个零件i，j，那么此时我们就可以将零件j放在零件i上。我们现在要组成一个大零件，使得高度最高，问这个最高高度。

思路：

1、首先按照b从大到小排序，如果b相等，那么按照a从大到小排序。将数组排序完成之后，我们考虑贪心寻找这个最大高度。

2、接下来我们直接模拟即可。模拟过程的关键思路就是寻找一条最长的链，所以我们可以加入一个栈，当遇到零件i无法和i-1相连接的时候，我们将第i-1个零件扔掉，判断第i个零件能否再和之前的零件连接，如果还不能，继续扔掉零件，直到能够相连接或者零件扔空了为止（让当前零件作为基出现），让当前零件组装起来。

按照此时排序方案排序结束之后，我们只要能够保证了更多的零件相组装起来就能保证结果尽可能的大了。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<stack>using namespace std;#define ll __int64struct node{    ll a,b,h;}a[100500],now,pre;int cmp(node a,node b){    if(a.b==b.b)    {        return a.a>b.a;    }    else return a.b>b.b;}int main(){    int n;    while(~scanf("%d",&n))    {        stack<node>s;        for(int i=1;i<=n;i++)        {            scanf("%I64d%I64d%I64d",&a[i].a,&a[i].b,&a[i].h);        }        ll sum=0;        ll output=0;        sort(a+1,a+1+n,cmp);        for(int i=1;i<=n;i++)        {            if(s.empty())            {                sum=a[i].h;                s.push(a[i]);            }            else            {                now=s.top();                if(now.a<a[i].b)                {                    sum+=a[i].h;                    s.push(a[i]);                }                else                {                    output=max(output,sum);                    while(s.size()>0&&s.top().a>=a[i].b)                    {                        pre=s.top();                        s.pop();                        sum-=pre.h;                    }                    sum+=a[i].h;                    s.push(a[i]);                }            }        }        output=max(output,sum);        printf("%I64d\n",output);    }}