POJ2481Cows(树状数组）

题意：就是给出N个区间，问这个区间是多少个区间的真子集。

分析：关键在于排序的策略。一边固定类型的题目。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 100005;
int C[maxn], ans[maxn];
struct Cow
{
    int low, high, index;
}cow[maxn];
bool cmp(Cow p, Cow q)
{
    if(p.high == q.high)
        return p.low < q.low;
    return p.high > q.high;
}
int lowbit(int lo)
{
    return lo & (-lo);
}
void modify(int pos, int value)
{
    while(pos < maxn)
    {
        C[pos] += value;
        pos += lowbit(pos);
    }
}
int getsum(int pos)
{
    int sum = 0;
    while(pos > 0)
    {
        sum += C[pos];
        pos -= lowbit(pos);
    }
    return sum;
}
int main()
{
    int n;
    while(cin >> n && n)
    {

        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d", &cow[i].low, &cow[i].high);
            cow[i].low++;       //注意可能为0，所以加1
            cow[i].index = i;
        }
        sort(cow + 1, cow + n + 1, cmp);
        memset(C, 0, sizeof(C));
        ans[cow[1].index] = 0;
        modify(cow[1].low, 1);
        for(int i = 2; i <= n; i++)
        {
            if(cow[i-1].low == cow[i].low && cow[i].high == cow[i-1].high)
                ans[cow[i].index] = ans[cow[i-1].index];
            else
                ans[cow[i].index] = getsum(cow[i].low);
            modify(cow[i].low, 1);
        }
        for(int i = 1; i < n; i++)
            printf("%d ", ans[i]);
        printf("%d\n", ans[n]);
    }
    return 0;
}