Gym 100971D 单调栈

D - Laying Cables

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice Gym 100971D

Description

standard input/output 
Announcement

 

• Statements
  

  One-dimensional country has n cities, the i-th of which is located at the point x_i and has population p_i, and all x_i, as well as all p_i, are distinct. When one-dimensional country got the Internet, it was decided to place the main server in the largest city, and to connect any other city j to the city k that has bigger population than j and is the closest to it (if there are many such cities, the largest one should be chosen). City k is called a parent of city j in this case.

  Unfortunately, the Ministry of Communications got stuck in determining from where and to where the Internet cables should be laid, and the population of the country is suffering. So you should solve the problem. For every city, find its parent city.

Input

The first line contains a single integer n(1 ≤ n ≤ 200000) — the number of cities.

Each of the next n lines contains two space-separated integers x_i and p_i(1 ≤ x_i,  p_i ≤ 10⁹) — the coordinate and the population of the i-th city.

Output

Output n space-separated integers. The i-th number should be the parent of the i-th city, or  - 1, if the i-th city doesn't have a parent. The cities are numbered from 1 by their order in the input.

Sample Input

Input

4
1 1000
7 10
9 1
12 100

Output

-1 4 2 1

Input

3
1 100
2 1
3 10

Output

-1 1 1

Input

3
1 10
3 100
2 1

Output

2 -1 2

题意： 给你n个点在x轴上的位置x和权值pos 

　　对于一个第i点 他的父亲定义为 和他最近并且 权值大于p[i]的 为点

　　输出每个点父亲，没有满足的作其父亲的点输出-1；

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f;
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
const int maxn=200000+10;
struct node{
   int x,id,p,f;
}ne[maxn];
int l[maxn],r[maxn];

bool cmpx(node a,node b)
{
   return a.x stl,str;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        FOR(i,n) scanf("%d%d",&ne[i].x,&ne[i].p),
                 ne[i].id=i;
        while(stl.size()) stl.pop();
        while(str.size()) str.pop();

        sort(ne+1,ne+n+1,cmpx);
        FOR(i,n){
           int id=ne[i].id;
           while(stl.size()&&stl.top().p<=ne[i].p) stl.pop();
           l[id]=stl.empty()?-1:stl.top().id;
           stl.push(ne[i]);
        }

        for(int i=n;i>=1;i--){
           int id=ne[i].id;
           while(str.size()&&str.top().p<=ne[i].p) str.pop();
           r[id]=str.empty()?-1:str.top().id;
           str.push(ne[i]);
        }

        sort(ne+1,ne+n+1,cmpid);
        FOR(i,n){
           if(l[i]==-1&&r[i]==-1) {PF("-1 ");CT;}
           else if(l[i]==-1) {PF("%d ",r[i]);CT;}
           else if(r[i]==-1) {PF("%d ",l[i]);CT;}

           if(ne[r[i]].x-ne[i].x>ne[i].x-ne[l[i]].x) PF("%d ",l[i]);
           else if(ne[r[i]].x-ne[i].xne[l[i]].p) PF("%d ",r[i]);
               else PF("%d ",l[i]);
           }
        }
        PF("\n");
    }
    return 0;
}

　　错因分析：刚开始想用lower_bound的，先按x从小到大排好序后，再利用lower_bound查找到枚举

的点的右侧，比当前枚举点人口数大的第一个点，后来写了后才想起lower_bound是基于二分查找的，

但是按x排好序后的人口数并不呈现有序性，当然不能用lower_bound；

解决：竟然忘了单调栈这么典型的算法，左右各扫一遍，记录大于当前点的最左和最右的点的id，

转载于:https://www.cnblogs.com/smilesundream/p/5701522.html