POJ2352 Stars 【树状数组】or【线段树】

标签：poj2352

Description

Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars. 

For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it‘s formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3. 

You are to write a program that will count the amounts of the stars of each level on a given map.

Input

Output

Sample Input

5
1 1
5 1
7 1
3 3
5 5

Sample Output

1
2
1
1
0

Hint

树状数组节点保存区间和，总区间为[0, maxn)。

树状数组：

//#define DEBUG
#include <stdio.h>
#define maxn 32002

int tree[maxn], level[maxn], n;

int lowBit(int x)
{ 
	return x & (-x); 
}

void update(int pos)
{
	while(pos < maxn){
		++tree[pos];
		pos += lowBit(pos);
	}
}

int getSum(int pos)
{
	int sum = 0;
	while(pos > 0){
		sum += tree[pos];
		pos -= lowBit(pos);
	}
	return sum;
}

int main()
{
	#ifdef DEBUG
		freopen("stdin.txt", "r", stdin);
		freopen("stdout.txt", "w", stdout);
	#endif

	int x, y, i;
	scanf("%d", &n);
	
	for(i = 0; i < n; ++i){
		scanf("%d%d", &x, &y);
		++level[getSum(++x)];
		update(x);
	}
	
	for(i = 0; i < n; ++i)
		printf("%d\n", level[i]);
	
	return 0;
}

线段树：

#include <stdio.h>
#define maxn 32002
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1

int tree[maxn << 2], level[maxn];

void pushUp(int rt)
{
	tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}

void update(int pos, int l, int r, int rt)
{
	if(l == r){
		++tree[rt]; return;
	}
	
	int mid = (l + r) >> 1;
	if(pos <= mid) update(pos, lson);
	else update(pos, rson);
	
	pushUp(rt);
}

int query(int left, int right, int l, int r, int rt)
{
	if(l == left && r == right) return tree[rt];
	
	int mid = (l + r) >> 1;
	if(right <= mid) return query(left, right, lson);
	else if(left > mid) return query(left, right, rson);
	else return query(left, mid, lson) + query(mid + 1, right, rson);
}

int main()
{
	int n, x, y, i;
	scanf("%d", &n);
	
	for(i = 0; i < n; ++i){
		scanf("%d%d", &x, &y);
		++level[query(0, x, 0, maxn - 1, 1)];
		update(x, 0, maxn - 1, 1);
	}
	
	for(i = 0; i < n; ++i)
		printf("%d\n", level[i]);
	
	return 0;
}

POJ2352 Stars 【树状数组】or【线段树】,布布扣,bubuko.com

POJ2352 Stars 【树状数组】or【线段树】

标签：poj2352

原文地址：http://blog.csdn.net/chang_mu/article/details/37737211