poj_2481Cows

Cows

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 9871		Accepted: 3233

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 <= 10⁵), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10⁵) 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

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
#pragma warning(disable : 4996)
const int MAXN = 100005;
typedef struct Node
{
	int s;
	int e;
	int pos;
}Node;
Node node[MAXN];
int tree[MAXN];
int level[MAXN];

bool cmp(Node x,Node y)
{
	if(x.e == y.e) return x.s < y.s;
	else return x.e > y.e;
}

int LowBit(int t)
{
	return t&(-t);  
}

void Update(int pos, int num)
{
	while(pos <= MAXN)
	{
		tree[pos] += num;
		pos += LowBit(pos);
	}
}

int GetSum(int end)
{
	int sum = 0;
	while(end > 0)
	{
		sum += tree[end];
		end -= LowBit(end);
	}
	return sum;
}

int main()
{
	freopen("in.txt", "r", stdin);
	int n, x, y;
	while(scanf("%d", &n) != EOF)
	{
		memset(tree, 0, sizeof(tree));
		if(n==0)
		{
			break;
		}
		for(int i = 1; i <= n; i++)
		{
			scanf("%d %d", &x, &y);
			node[i].pos = i;
			node[i].s = x + 1;
			node[i].e = y + 1;
		}
		sort(node + 1, node + n + 1, cmp);
		for (int i = 1; i <= n; i++)
		{
			cout << node[i].pos << " " << node[i].s << " " << node[i].e << endl;
		}
		for(int i = 1; i <= n; i++)
		{
			Update(node[i].s, 1);
			level[node[i].pos] = GetSum(node[i].s) - 1;
		}
		for (int i = 1; i < n; i++)
		{
			printf("%d ", level[i]);
		}
		printf("%d\n", level[n]);
	}
}

9
5 5
5 6
1 4
4 5
1 4
4 6
2 3
3 5
1 4

4 1 0 2 0 0 3 0 0

没有考虑相同的点，导致WA，后来想一想，这样也不能满足Ei - Si > Ej - Sj。也就是说排除两组数据相同的情况，任何满足 Si <= Sj and Ej <= Ei 的必然会满足Ei - Si > Ej - Sj。

AC代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
#pragma warning(disable : 4996)
const int MAXN = 100005;
typedef struct Node
{
	int s;
	int e;
	int pos;
}Node;
Node node[MAXN];
int tree[MAXN];
int level[MAXN];

bool cmp(Node x,Node y)
{
	if(x.e == y.e) return x.s < y.s;
	else return x.e > y.e;
}

int LowBit(int t)
{
	return t&(-t);  
}

void Update(int pos, int num)
{
	while(pos <= MAXN)
	{
		tree[pos] += num;
		pos += LowBit(pos);
	}
}

int GetSum(int end)
{
	int sum = 0;
	while(end > 0)
	{
		sum += tree[end];
		end -= LowBit(end);
	}
	return sum;
}

int main()
{
	freopen("in.txt", "r", stdin);
	int n, x, y;
	while(scanf("%d", &n) != EOF)
	{
		memset(tree, 0, sizeof(tree));
		if(n == 0)
		{
			break;
		}
		for(int i = 1; i <= n; i++)
		{
			scanf("%d %d", &x, &y);
			node[i].pos = i;
			node[i].s = x + 1;
			node[i].e = y + 1;
		}
		sort(node + 1, node + n + 1, cmp);
		/*for (int i = 1; i <= n; i++)
		{
			cout << node[i].pos << " " << node[i].s << " " << node[i].e << endl;
		}*/
		Update(node[1].s, 1);
		level[node[1].pos] = GetSum(node[1].s) - 1;
		x = node[1].s;
		y = node[1].e;
		for(int i = 2; i <= n; i++)
		{
			if(node[i].s == x && node[i].e == y)
			{
				level[node[i].pos] = level[node[i-1].pos];
				Update(node[i].s, 1);
				continue;
			}
			x = node[i].s;
			y = node[i].e;
			Update(node[i].s, 1);
			level[node[i].pos] = GetSum(node[i].s) - 1;
		}
		for (int i = 1; i < n; i++)
		{
			printf("%d ", level[i]);
		}
		printf("%d\n", level[n]);
	}
}