求左下角星星之和 树状数组或线段树 poj 2352 Stars

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http://poj.org/problem?id=2352

Description

Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars. 

For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it‘s formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3. 

You are to write a program that will count the amounts of the stars of each level on a given map.

Input

Output

Sample Input

5
1 1
5 1
7 1
3 3
5 5

Sample Output

1
2
1
1
0

Hint

Source

思路：树状数组
分析：
1 题目是要求出每一个点的左下(正左+正下)有几个星星，那个这个点就是第几层，最后输出0～n-1层的点的个数。比如样列编号为5的星星，左下有3个星星那么5就处于第三层
2 利用树状数组，我们知道树状数组C中，C[i]表示的是原先数组A中的某一段和。题目明确指出输入的时候是按照y值增大的顺序（y相同是x增大），那么我们应该要用什么做为原先的数组A呢，很显然就是X轴，就是A[2]表示x = 2的点的个数。那么当新加入一个点的x值为2的时候，A[2]++，这个时候就是要更新C[2]，C[2+lowbit(2)]...

3 那怎么求某个点的左下有几个星星呢，很显然处在左下的点的x值肯定小于等于当前点的，那么就是相当与树状数组求和，那么整道题的思路就完成

4 注意更新树状数组的时候，注意x<=MAXN，不是输入的n。因为更改了A[x]，就要更新C[x] , C[x+lowbit(x)]...

5 题目输入的x的坐标可能为0，所以这边我们把所有的x+1，这就避免了TLE

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int MAXN = 32010;

int n , ans[MAXN];
int treeNum[MAXN];

int lowbit(int x){
    return x&(-x);
}

int getSum(int x){
    int sum = 0;
    while(x){
        sum += treeNum[x]; 
        x -= lowbit(x);
    }
    return sum;
}

void add(int x , int val){
    while(x < MAXN){
        treeNum[x] += val; 
        x += lowbit(x);
    }
}

int main(){
    int x , y;
    while(scanf("%d" , &n) != EOF){ 
        memset(treeNum , 0 , sizeof(treeNum));
        memset(ans , 0 , sizeof(ans));
        for(int i = 0 ; i < n ; i++){
            scanf("%d%d" , &x , &y); 
            x++;
            ans[getSum(x)]++; 
            add(x , 1);
        }
        for(int i = 0 ; i < n ; i++)
            printf("%d\n" , ans[i]);
    }
    return 0;
}

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

#define MAXN 32010

struct Node{
  int left;
  int right;
  int sum;
};
Node node[4*MAXN];
int vis[MAXN];

/*建立一颗线段树*/
void buildTree(int left , int right , int pos){
   node[pos].left = left;
   node[pos].right = right;
   node[pos].sum = 0;
   if(left == right)
     return;
   int mid = (left+right)>>1;
   buildTree(left , mid , pos<<1);
   buildTree(mid+1 , right , (pos<<1)+1);
}

/*单点更新*/
void update(int index , int pos){
   if(node[pos].left == node[pos].right){
      node[pos].sum++;
      return;
   }
   int mid = (node[pos].left+node[pos].right)>>1;
   if(index <= mid)
     update(index , pos<<1);
   else
     update(index , (pos<<1)+1);
   node[pos].sum = node[pos<<1].sum + node[(pos<<1)+1].sum;
}

/*区间查询*/
int query(int left , int right , int pos){
   if(node[pos].left == left && node[pos].right == right)
     return node[pos].sum;
   int mid = (node[pos].left+node[pos].right)>>1;
   if(right <= mid)
     return query(left , right , pos<<1);
   else if(left > mid)
     return query(left , right , (pos<<1)+1);
   else
     return query(left , mid , pos<<1)+query(mid+1 , right , (pos<<1)+1);
}

int main(){
   int Case , n;
   int x , y;
   scanf("%d" , &Case);
   n = Case;
   buildTree(0 , MAXN , 1);
   memset(vis , 0 , sizeof(vis));
   while(Case--){
      scanf("%d%d" , &x , &y);
      vis[query(0 , x , 1)]++;
      update(x , 1);
   }
   for(int i = 0 ; i < n ; i++)
      printf("%d\n" , vis[i]);
   return 0;
}

求左下角星星之和 树状数组或线段树 poj 2352 Stars

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原文地址：http://blog.csdn.net/acm_10000h/article/details/48030033