POJ 2481 Cows (树状数组)

题意：给定n个区间，问每个区间所覆盖的区间的个数。

思路：第三道树状数组，终于有感觉了，要不就傻X了，对区间排序，然后树状数组查询，思路和Stars那道差不多，不过还要处理重合区间的情况，只需排序后O(N)扫描一遍就好了。

/* Cows 2481
 * 树状数组第三题，对牛排序即可 ，注意要
 * 排除体型相等的牛.
 */
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
using namespace std;

const int BORDER = (1<<20)-1;
const int MAXSIZE = 37;
const int MAXN = 100200;
const int INF = 1000000000;
#define CLR(x,y) memset(x,y,sizeof(x))
#define ADD(x) x=((x+1)&BORDER)
#define IN(x) scanf("%d",&x)
#define OUT(x) printf("%d\n",x)
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))

typedef struct{
    int s,e;
    int id;
}Node;

Node node[MAXN];
int tre[MAXN],n_tre;
int n;
int ans[MAXN],same[MAXN];

int lowbit(int x)
{
    return x&(-x);
}
void modify(int ind,int delta)
{
    while( ind <= n_tre)
    {
        tre[ind] += delta;
        ind += lowbit(ind);
    }
}
int get_sum(int ind)
{
    int sum = 0;
    while(ind > 0)
    {
        sum += tre[ind];
        ind -= lowbit(ind);
    }
    return sum;
}
bool cmp(const Node& a,const Node& b)
{
    if(a.e == b.e)
        return a.s < b.s;
    return a.e > b.e;
}
int init()
{
    n_tre = 100400;
    CLR(same,0);
    CLR(tre,0);
    CLR(ans,0);
    return 0;
}
int input()
{
    for(int i = 0; i < n; ++i)
    {
        scanf("%d%d",&node[i].s,&node[i].e);
        ++node[i].s;
        ++node[i].e;
        node[i].id = i;
    }
    return 0;
}
int _find_same()
{
    int i,j,tmp;
    for(i = 1; i < n; ++i)
    {
        tmp = 0;
        if(node[i-1].s == node[i].s && node[i-1].e == node[i].e)
            same[i] = same[i-1] + 1;
    }
    return 0;
}
int work()
{
    int i,j,s,e,tmp;
    sort(node,node+n,cmp);
    _find_same();
    for(i = 0; i < n; ++i)
    {
        ans[node[i].id] = get_sum(node[i].s)-same[i];
        modify(node[i].s,1);
    }
    printf("%d",ans[0]);
    for(i = 1; i < n; ++i)
        printf(" %d",ans[i]);
    printf("\n");
    return 0;
}
int main()
{
    while(IN(n))
    {
        if(!n)
            break;
        init();
        input();
        work();
    }
    return 0;
}