CodeForces 56E-Find the Path（技巧）

Description

Vasya is interested in arranging dominoes. He is fed up with common dominoes and he uses the dominoes of different heights. He putn dominoes on the table along one axis, going from left to right. Every domino stands perpendicular to that axis so that the axis passes through the center of its base. Thei-th domino has the coordinatex_i and the heighth_i. Now Vasya wants to learn for every domino, how many dominoes will fall if he pushes it to the right. Help him do that.

Consider that a domino falls if it is touched strictly above the base. In other words, the fall of the domino with the initial coordinatex and heighth leads to the fall of all dominoes on the segment[x + 1, x + h - 1].

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) which is the number of dominoes. Then follown lines containing two integers x_i andh_i ( - 10⁸ ≤ x_i ≤ 10⁸, 2 ≤ h_i ≤ 10⁸) each, which are the coordinate and height of every domino. No two dominoes stand on one point.

Output

Print n space-separated numbers z_i — the number of dominoes that will fall if Vasya pushes thei-th domino to the right (including the domino itself).

Sample Input

Input

416 520 510 1018 2

Output

3 1 4 1 

Input

40 101 59 1015 10

Output

4 1 2 1 

                                                                                                    

题意：给出多米诺骨牌所在的x 轴和高h，求出每一个多米诺骨牌向右倒下时能使多少的牌倒下。

思路：使用数组将每一个牌能够使右边的牌倒下的的数目记录下来，在操作的时候要使用一点小技巧才能不T，具体看代码

CODE：

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <string>#include <cstring>#include <queue>#include <stack>#include <vector>#include <set>#include <map>const int inf=0xfffffff;typedef long long ll;using namespace std;const int Max=100005;struct node{    int x,h,d;    int sum;    int id;}dom[Max];bool cmp(node a,node b){    return a.x<b.x;}bool idd(node a,node b){    return a.id<b.id;}int main(){    //freopen("in.in","r",stdin);    int n;    while(~scanf("%d",&n)){        for(int i=0;i<n;i++){            scanf("%d%d",&dom[i].x,&dom[i].h);            dom[i].id=i;            dom[i].d=dom[i].x+dom[i].h;        }        sort(dom,dom+n,cmp);        dom[n-1].sum=1;        for(int i=n-2;i>=0;i--){            int t=i+1;            dom[i].sum=1;            while(dom[i].d>dom[t].x && t<n){                dom[i].sum+=dom[t].sum;                t+=dom[t].sum; //不T 的关键~            }        }        sort(dom,dom+n,idd);        printf("%d",dom[0].sum);        for(int i=1;i<n;i++)            printf(" %d",dom[i].sum);        printf("\n");    }    return 0;}