CodeForces D. Nested Segments【逆序对类型】

最新推荐文章于 2020-08-01 15:55:33 发布

DTL66

最新推荐文章于 2020-08-01 15:55:33 发布

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分类专栏：逆序对 22.树状数组

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逆序对

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本文介绍了一种解决区间嵌套问题的有效算法，通过离散化处理和利用树状数组来统计每个区间被包含的次数。输入为一系列不重合区间的端点坐标，输出每个区间内所含有的其他区间数量。

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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.

Examples

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

AC代码：

//离散化之后 将结束时间看成id 按开始时间降序插入统计插入元素之前已插入的元素个数即为该元素的贡献 

#include<cstdio>
#include<algorithm>

using namespace std;

const int MAXN=2e5+11;

struct Node{
	int l,r,id;
}b[MAXN<<1];
int C[MAXN<<1],a[MAXN<<1],ans[MAXN];

int Lowbit(int x) {
	return x&(-x);
} 
void Add(int x,int d,int N) {
	while(x<=N) {
		C[x]+=d; x+=Lowbit(x);
	}
}
int Sum(int x) {
	int ret=0;
	while(x>0) {
		ret+=C[x]; x-=Lowbit(x); 
	}
	return ret;
}

bool cmp(Node a,Node b) {
	if(a.l!=b.l) return a.l>b.l;
	return a.r<b.r;
}

int main() {
	int N;
	while(~scanf("%d",&N)) {
		int cnt=0;
		for(int i=0;i<N;++i) {
			scanf("%d%d",&b[i].l,&b[i].r);
			a[cnt++]=b[i].l; a[cnt++]=b[i].r; 
			b[i].id=i;
		}
		sort(a,a+cnt);
		int top=unique(a,a+cnt)-a;
		for(int i=0;i<N;++i) {
			int p=lower_bound(a,a+top,b[i].l)-a;
			b[i].l=p+1;
			p=lower_bound(a,a+top,b[i].r)-a;
			b[i].r=p+1;
		}
		top++;
		for(int i=0;i<=top;++i) C[i]=0;
		sort(b,b+N,cmp);
		for(int i=0;i<N;++i) {
			Add(b[i].r,1,top); 
			ans[b[i].id]=Sum(b[i].r-1); 
		}
		for(int i=0;i<N;++i) printf("%d\n",ans[i]);
	} 
	return 0;
}