POJ-2352：Stars

來源：POJ

標籤：樹狀陣列

參考資料：

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題目

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.

輸入

The first line of the input file contains a number of stars N (1<=N<=15000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0<=X,Y<=32000). There can be only one star at one point of the plane. Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate.

輸出

The output should contain N lines, one number per line. The first line contains amount of stars of the level 0, the second does amount of stars of the level 1 and so on, the last line contains amount of stars of the level N-1.

輸入樣例

5
1 1
5 1
7 1
3 3
5 5

輸出樣例

1
2
1
1
0

提示

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

解題思路

注意題目按序輸入（y升序，若y相等再x升序）。假設當前點是(x1,y1)（那麼其實已經確定它的等級了，原因還是按序輸入），我們只需要求當前x小於等於x1的點的個數（不包括點(x1,y1)）。採用樹狀陣列（注意x>=0，而樹狀陣列索引從1開始）或線段樹都可以解決此題。

參考程式碼

#include<stdio.h>
#include<string.h>
#define MAXN 40000
int c[MAXN+10];//樹狀陣列
int level[MAXN/2];//記錄每個等級的個數

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

void update(int x){
    while(x<=MAXN){
        c[x]++;
        x+=lowbit(x);
    }
}

int sum(int x){
    int ret=0;
    while(x>0){
        ret+=c[x];
        x-=lowbit(x);
    }
    return ret;
}

int main(){
    int n;
    while(~scanf("%d",&n)){
        memset(c,0,sizeof(c));
        memset(level,0,sizeof(level));
        for(int i=0;i<n;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            update(x+1);//將x座標加1，滿足樹狀陣列索引要求，下同。
            level[sum(x+1)-1]++;//減1是因為不包括當前的點。
        }
        for(int i=0;i<n;i++)
            printf("%d\n",level[i]);
    }
    return 0;
}