Codeforces gym 100971 D. Laying Cables 单调栈

D. Laying Cables

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

One-dimensional country has n cities, the i-th of which is located at the point xi and has population pi, and all xi, as well as all pi, 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 xi and pi (1 ≤ xi,  pi ≤ 109) — the coordinate and the population of thei-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.

Examples

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

思路：找到左边和右边比他大的值，最后比较左右；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 100000007
#define esp 0.00000000001
const int N=2e5+10,M=1e6+10,inf=1e9+10;
struct is
{
int x,val;
int pos;
}a
;
int cmp(is x,is y)
{
return x.x<y.x;
}
int d
,k;
int l
,r
;
int ans
;
int check(int x,int len)
{
if(x<1||x>len)
return 0;
return 1;
}
int main()
{
int x,y,z,i,t;
scanf("%d",&x);
for(i=1;i<=x;i++)
scanf("%d%d",&a[i].x,&a[i].val),a[i].pos=i;
sort(a+1,a+1+x,cmp);
a[0].val=a[x+1].val=inf;
k=0;
d[++k]=0;
for(i=1;i<=x;i++)
{
while(a[d[k]].val<a[i].val)k--;
l[i]=d[k];
d[++k]=i;
}
k=0;
d[++k]=x+1;
for(i=x;i>=1;i--)
{
while(a[d[k]].val<a[i].val)k--;
r[i]=d[k];
d[++k]=i;
}
for(i=1;i<=x;i++)
{
int u=check(l[i],x),v=check(r[i],x);
if(u&&v)
{
if(a[i].x-a[l[i]].x>a[r[i]].x-a[i].x)
ans[a[i].pos]=a[r[i]].pos;
else if(a[i].x-a[l[i]].x<a[r[i]].x-a[i].x)
ans[a[i].pos]=a[l[i]].pos;
else
{
if(a[l[i]].val>a[r[i]].val)
ans[a[i].pos]=a[l[i]].pos;
else
ans[a[i].pos]=a[r[i]].pos;
}
}
else if(u)
ans[a[i].pos]=a[l[i]].pos;
else if(v)
ans[a[i].pos]=a[r[i]].pos;
else
ans[a[i].pos]=-1;
}
for(i=1;i<=x;i++)
printf("%d ",ans[i]);
return 0;
}