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poj2352(线段树)
/1.题目（theme）
Stars
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 32652 | Accepted: 14263 |
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.
You are to write a program that will count the amounts of the stars of each level on a given map.
You are to write a program that will count the amounts of the stars of each level on a given map.
Input
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.
Output
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.
Sample Input
51 15 17 13 35 5
Sample Output
12110
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
2.题目大意（The meaning of the questions）：
求星星的等级，然后输出个等级的数目从0-（n-1）；
星星等级的计算如下：先画个坐标轴xy，在坐标轴上画出每个星星（根据它的坐标），在它的左下方有多少个星星（包括正下方），那它就有多少级。
3.解决方式（solution）：
1.线段树：把x轴作为处理对象，因为输入的坐标是随着y轴递增，这就为题目用线段树查询提供方便，这道题用线段树就不用管y，只需要处理x就行，就化简为区间线段树，不用担心因为y的问题出错，这就是这道题的有趣之处。
举个例子好了，6
1 1
2 1
3 1
2 2
3 5
1 7
化简为（去y留x）：1 2 3 2 3 1（这些就是建好树之后，要加进去的点，加完一个查询一个就行）
4.代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <algorithm> 5 #include <string.h> 6 using namespace std; 7 struct st{ 8 int value; 9 }root[32005*4];10 int level[32005*2];11 void build(int rt,int l,int r)12 {13 int mid;14 if (l==r) {root[rt].value=http://www.mamicode.com/0;return ;}15 mid=(l+r)/2;16 build(2*rt,l,mid);17 build(2*rt+1,mid+1,r);18 root[rt].value=http://www.mamicode.com/root[rt*2].value+root[rt*2+1].value;19 }20 int query(int rt,int l,int r,int left,int x)21 {22 int mid;23 if (left==l&&r==x)return root[rt].value;24 mid=(l+r)/2;25 if (x<=mid){26 return query(rt*2,l,mid,left,x);27 }28 if (left>mid){29 return query(rt*2+1,mid+1,r,left,x);30 }31 return (query(rt*2,l,mid,left,mid)+query(rt*2+1,mid+1,r,mid+1,x));32 }33 void updata(int rt,int l,int r,int x)34 {35 int mid;36 if (l==r){root[rt].value++;return;}37 mid=(l+r)/2;38 if (x>mid){39 updata(rt*2+1,mid+1,r,x);40 }else {41 updata(rt*2,l,mid,x);42 }43 root[rt].value=http://www.mamicode.com/root[rt*2].value+root[rt*2+1].value;44 }45 int main()46 {47 int n,x,y,sum,maxx=32005,i;48 scanf("%d",&n);49 for (i=0;i<n;i++){50 level[i]=0;51 }52 build(1,0,maxx);53 for (i=0;i<n;i++){54 scanf("%d%d",&x,&y);55 sum=query(1,0,maxx,0,x);56 level[sum]++;57 updata(1,0,maxx,x);58 }59 for (i=0;i<n;i++){60 printf("%d\n",level[i]);61 }62 return 0;63 }
poj2352(线段树)
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