POJ 2352 Stars

树状数组入门题，题设比较简单

#include<stdio.h>
#define MAXC 32010
int C[MAXC], R[MAXC];
//////////////////树状数组模板
int lowbit(int t)
{
    return t & (t ^ (t - 1));      //计算最小幂
}
int add(int i,int v)
{
	while(i<MAXC)      //取与C[i]有关的值添上增加的值
	{
	     C[i]+=v;
		 i+=lowbit(i);
	}
	return 0;
}
int sum(int i)
{
	int s=0;
	while(i>0)                //取与C[i]有关的值求和
	{
	     s+=C[i];
	     i-=lowbit(i);
	}
	return s;
}
//////////////////树状数组模板
int main()
{
    int n,i,x,y;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
	   scanf("%d %d",&x,&y); 
           ++R[sum(++x)];       
//因为输入是有序的，所以可以立刻确定结果。又因为线段数组是从1开始的，所以x要总体右移一个单位长度          
           add(x,1);  
 //add和sum这两个操作先后没关系,因为sum中肯定不包括本身
}

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

再附上线段树解法，但是速度要慢好多

#include <stdio.h>
#define L(t) ((t) << 1)
#define R(t) ((t) << 1 | 1)
#define MAXC 32010  

struct SegTree
{
    int l,r;
    long long add, sum;    
    int getMid(){
        return ( l + r) >>1;
    }
    int getDis(){
        return r - l + 1;
    }
}       tree[MAXC << 2];

int arr[MAXC];      //存放X坐标
int R[MAXC];        //存放每个等级对应的结果数

void build(int left, int right, int t){           //递归构造
    tree[t].l = left;
    tree[t].r = right;
    tree[t].add = 0;
	tree[t].sum=0;
    if(left == right){
        return;
    }
    int mid = tree[t].getMid();
    build(left, mid, L(t));
    build(mid + 1, right, R(t));
}

void update(int left, int right, int a, int t){         
	
    if( left <= tree[t].l && right >= tree[t].r ){           
		
        tree[t].add += a;            //若当前子树被目标区间覆盖 更新子树的sum和增量（在本次查询中，增量下行到此为之）
        tree[t].sum += a * tree[t].getDis();    
		
        return;
    }
    if( tree[t].add ){                //向子结点传递增量，并更新其sum,最后清空自己的增量
        tree[L(t)].sum += tree[L(t)].getDis() * tree[t].add;
        tree[R(t)].sum += tree[R(t)].getDis() * tree[t].add;
        tree[L(t)].add += tree[t].add;
        tree[R(t)].add += tree[t].add;
        tree[t].add = 0;        
    }
    int mid = tree[t].getMid();
	
    if(right <= mid ){
		
        update(left, right, a, L(t));       //目标区间仅在左子树上
    } else if (left > mid ){
		
        update(left, right, a, R(t));           //目标区间仅在右子树上
    } else {
		
        update(left, right, a, L(t));                          //目标区间同时在左右子树上
        update(left, right, a, R(t));
    }    
    tree[t].sum = tree[L(t)].sum + tree[R(t)].sum ;    //更新父结点的sum
	
}

long long query(int left, int right, int t){
	
    if(left <= tree[t].l && right >= tree[t].r ){    
		
        return tree[t].sum;
    }
    if( tree[t].add ){               //这一段和update函数一样，是一个pushDown
        tree[L(t)].sum += tree[L(t)].getDis() * tree[t].add;
        tree[R(t)].sum += tree[R(t)].getDis() * tree[t].add;
        tree[L(t)].add += tree[t].add;
        tree[R(t)].add += tree[t].add;
        tree[t].add = 0;
    }
    int mid = tree[t].getMid();
	
    if(right <= mid ){
	
        return query(left, right, L(t));    
    } else if( left > mid ){
		
        return query(left, right, R(t));
    }  else {
		
        return query(left, mid ,L(t)) + 
        query(mid + 1, right, R(t));
    }
}

int main()  
{  
    int n,i,x,y;  
    scanf("%d",&n);  

     build(1,MAXC,1);
	  
	 int k;
	 for(i=1;i<=n;i++)
    {
        scanf("%d%d",&x,&y);
        x++;
		k=query(1,x,1);
		
        R[k]++;
        update(x,x,1,1);
    }
  

   for (i=0;i<n;i++)
      printf("%d\n",R[i]);

    return 0;  
}