POJ - 2481 Cows(树状数组)

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3
1 2
0 3
3 4
0

1 0 0

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)

标签：des   style   blog   color   io   os   ar   for   strong   

原文地址：http://blog.csdn.net/u011345136/article/details/40224515