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POJ - 2481 Cows(树状数组)

2024-07-27 22:44:50   221人阅读

Description

Farmer John‘s cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good. 

Farmer John has N cows (we number the cows from 1 to N). Each of Farmer John‘s N cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval [S,E]. 

But some cows are strong and some are weak. Given two cows: cow i and cow j, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cow i is stronger than cow j. 

For each cow, how many cows are stronger than her? Farmer John needs your help!

Input

The input contains multiple test cases. 
For each test case, the first line is an integer N (1 <= N <= 10 5), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10 5) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge. 

The end of the input contains a single 0.

Output

For each test case, output one line containing n space-separated integers, the i-th of which specifying the number of cows that are stronger than cow i. 

Sample Input

3
1 2
0 3
3 4
0

Sample Output

1 0 0
题意：有n头牛，每头牛有s和e值，对于i和j两头牛来说，如果满足si<=sj，ei>=ej且ei-si>ej-sj，则说明i比j壮，求每头牛有几头比它壮。

思路：首先按照e值从大到小排序，然后为了保证ei-si>ej-sj，所以在排序函数那边注意一下，然后就是典型的树状数组的应用，每次插入的时候查找

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100005;

struct Cow {
	int s, e, num;
	bool operator <(const Cow &tmp) const {
		if (e != tmp.e)
			return e > tmp.e;
		return s < tmp.s;
	}
} cow[maxn];
int c[maxn], total[maxn], n, Max;

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

void Add(int x) {
	while (x <= Max) {
		c[x] += 1;	
		x += lowbit(x);
	}
}

int sum(int x) {
	int ans = 0;
	while (x > 0) {
		ans += c[x];
		x -= lowbit(x);
	}
	return ans;
}

int main() {
	while (scanf("%d", &n) != EOF && n) {
		memset(c, 0, sizeof(c));	
		memset(total, 0, sizeof(total));
		Max = 0;
		for (int i = 1; i <= n; i++) {
			scanf("%d%d", &cow[i].s, &cow[i].e);
			cow[i].s++, cow[i].e++;
			cow[i].num = i;
			Max = max(Max, cow[i].s);
		}

		sort(cow+1, cow+1+n);
		for (int i = 1; i <= n; i++) {
			Add(cow[i].s);
			if (i > 1 && cow[i].s == cow[i-1].s && cow[i].e == cow[i-1].e)
				total[cow[i].num] = total[cow[i-1].num];
			else total[cow[i].num] = sum(cow[i].s) - 1;
		}

		for (int i = 1; i < n; i++)
			printf("%d ", total[i]);
		printf("%d\n", total[n]);
	}	
	return 0;
}




POJ - 2481 Cows(树状数组)


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