Codeforces 652D Nested Segments 树状数组离线处理

最新推荐文章于 2019-05-25 19:37:45 发布

原创最新推荐文章于 2019-05-25 19:37:45 发布 · 441 阅读

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该博客介绍了一种解决线性段上nested segments问题的方法，利用树状数组进行离线处理。通过输入的n个段，计算每个段内包含的完整子段数量。首先，根据左坐标对段进行降序排序，然后依次将右坐标加入数组，并使用树状数组查询当前段包含的右坐标个数，以此确定子段数量。提供了示例输入和输出，帮助理解解题思路。

You are given n segments on a line. There are no ends of some segments that coincide. For each segment find the number of segments it contains.

Input
The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of segments on a line.

Each of the next n lines contains two integers li and ri ( - 109 ≤ li < ri ≤ 109) — the coordinates of the left and the right ends of the i-th segment. It is guaranteed that there are no ends of some segments that coincide.

Output
Print n lines. The j-th of them should contain the only integer aj — the number of segments contained in the j-th segment.

Example
Input
4
1 8
2 3
4 7
5 6
Output
3
0
1
0
Input
3
3 4
1 5
2 6
Output
0
1
1

求出每个区间内包含了几个完整的子区间。按照左坐标从大到小排序，再按顺序将区间的右坐标加入数组，由于未加入的区间左坐标在已加入的左边，肯定不被包含，而已经加入数组的都是左坐标符合条件的，之后只要区间查询包含了几个右坐标，就能知道包含了几个子区间而且没有遗漏

#include <iostream>
#include <stdio.h>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <math.h>
#include <iterator>
#include <string.h>
using namespace std;
typedef long long ll;
int mo[4][2]={0,1,1,0,0,-1,-1,0};
const int MAXN=0x3f3f3f3f;
const int sz=200005;
int n;
struct node{
    int l,r,num;
}a[sz];
struct number{
    int num,v;
}hs[sz*2];
bool cmp(node x,node y){
    return x.l>y.l;
}
bool cmpnum(number x,number y){
    return x.v<y.v;
}
int tr[sz*2];
int ans[sz];
void add(int x){
    for(;x<=2*n;x+=x&(-x)){
        tr[x]++;
    }
}
int sum(int x){
    int s=0;
    for(;x>0;x-=x&(-x)){
        s+=tr[x];
    }
    return s;
}
int query(int x,int y){
    return sum(y)-sum(x-1);
}
int main()
{
    //freopen("C:\\Users\\Administrator\\Desktop\\r.txt","r",stdin);
    while(scanf("%d",&n)!=EOF){
        memset(tr,0,sizeof(tr));
        for(int i=1;i<=n;i++){
            scanf("%d%d",&a[i].l,&a[i].r);
            a[i].num=i;
            hs[i].v=a[i].l;
            hs[i].num=i;
            hs[i+n].v=a[i].r;
            hs[i+n].num=i+n;
        }
        sort(hs+1,hs+1+2*n,cmpnum);
        for(int i=1;i<=2*n;i++){
            if(hs[i].num<=n)
                a[hs[i].num].l=i;
            else
                a[hs[i].num-n].r=i;
        }
        sort(a+1,a+1+n,cmp);
        for(int i=1;i<=n;i++){
            ans[a[i].num]=query(a[i].l,a[i].r);
            add(a[i].r);
        }
        for(int i=1;i<=n;i++){
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}