codeforces 777E. Hanoi Factory （线段树）

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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

Note

In the first sample, the optimal solution is to take all the rings and put them on each other in order 3, 2, 1.

In the second sample, one can put the ring 3 on the ring 4 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：给出n个盘子，求最多能叠多高。

第i个盘子可以叠在第j个盘子上，当且仅当第i个盘子的外径<=第j个盘子的外径，第i个盘子的外径>第j个盘子的内径。

题解：线段树

对于盘子的内外径进行离散化，以线段数的下标，表示当且内径为i的高度。

将盘子以外径为第一关键字（降序），内径为第二关键字（升序），每次不断将盘子加入，选取可以放置的最大高度。

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#define N 300003#define LL long long using namespace std;struct data{int a,b,h;}t[N];LL tr[N*4],delta[N*4];int a[N],b[N],n;int cmp(int x,int y){return a[x]<a[y];}int cmp1(data a,data b){return a.b>b.b||a.b==b.b&&a.a>b.a;}LL qjmax(int now,int l,int r,int ll,int rr){if (ll<=l&&r<=rr) return tr[now];int mid=(l+r)/2;LL ans=0;if (ll<=mid) ans=max(ans,qjmax(now<<1,l,mid,ll,rr));if (rr>mid) ans=max(ans,qjmax(now<<1|1,mid+1,r,ll,rr));return ans;}int update(int now){tr[now]=max(tr[now<<1],tr[now<<1|1]);}void pointchange(int now,int l,int r,int x,LL v){if (l==r) {tr[now]=max(tr[now],v);return;}int mid=(l+r)/2;if (x<=mid) pointchange(now<<1,l,mid,x,v);else pointchange(now<<1|1,mid+1,r,x,v);    update(now);}int main(){freopen("a.in","r",stdin);scanf("%d",&n);for (int i=1;i<=n;i++) scanf("%d%d%d",&t[i].a,&t[i].b,&t[i].h),a[i]=t[i].a,b[i]=i,a[i+n]=t[i].b,b[i+n]=i+n;sort(b+1,b+n*2+1,cmp);    int cnt=0;    for (int i=1;i<=n*2;i++)     if (a[b[i]]!=a[b[i-1]]) {    if (b[i]<=n) t[b[i]].a=++cnt;    else t[b[i]-n].b=++cnt;   }      else{     if (b[i]<=n) t[b[i]].a=cnt;    else t[b[i]-n].b=cnt;  }    sort(t+1,t+n+1,cmp1);    for (int i=1;i<=n;i++) {    //cout<<t[i].a<<" "<<t[i].b<<" "<<t[i].h<<endl;    LL high=qjmax(1,1,cnt,1,t[i].b-1);    pointchange(1,1,cnt,t[i].a,(LL)high+t[i].h);}printf("%I64d\n",tr[1]);}

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