Language:
Stars
Time Limit:1000MSMemory Limit:65536K
Total Submissions:35169Accepted:15259
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
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.
Source
题意：有n个星星的坐标，如果某个星星坐标为(x, y), 它的左下位置为：(x0,y0)，x0<=x 且y0<=y。如果左下位置有a个星星，就表示这个星星属于level x按照y递增，如果y相同则x递增的顺序给出n个星星，求出所有level水平的数量。
思路：最典型的树状数组，，第一次做。。
代码：
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 32005#define MAXN 2005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b) for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i–)#define FRL(i,a,b) for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i–)#define mem(t, v) memset ((t) , v, sizeof(t))#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pfprintf#define DBGpf("Hi\n")typedef long long ll;using namespace std;int bit[maxn];int ans[maxn/2];int lowbit(int k){return k&(-k);}int sum(int i){int s=0;while (i>0){s+=bit[i];i-=lowbit(i);}return s;}void add(int i,int x){while (i<maxn){bit[i]+=x;i+=lowbit(i);}}int main(){int i,j,x,y,n;while (~sf(n)){mem(bit,0);mem(ans,0);for (int i=0;i<n;i++){sff(x,y);ans[sum(++x)]++;//这里要++x，不然当x=0在add的时候会死循环的。。add(x,1);}for (int i=0;i<n;i++)pf("%d\n",ans[i]);}return 0;}
今日的执着，会造成明日的后悔。
