poj2352 树状数组
2015-03-12 14:56
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如题:http://poj.org/problem?id=2352
Stars
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
Sample Output
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
Source
Ural Collegiate Programming Contest 1999
题目大意:有很多星星,星星的等级是星星左下所有星星的和,求各个等级星星有多少个。
思路:C【】是个树状数组,记录的是x坐标前包含x的星星的个数。每一个x,y更新树状数组,level[sum(x)-1]++.x从0开始,所有x处理时+1.
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define n 32005
int N;
int c
;
int level15005];
int lowbit(int x)
{
return x&(x^(x-1));
}
void add(int x,int num)
{
while(x<=n)
{
c[x]+=num;
x+=lowbit(x);
}
}
int sum(int x)
{
int s=0;
while(x>0)
{
s+=c[x];
x-=lowbit(x);
}
return s;
}
int main()
{
freopen("C:\\1.txt","r",stdin);
cin>>N;
int i;
for(i=0;i<N;i++)
{
int x,y;
scanf("%d%d",&x,&y);
x+=1;
add(x,1);
level[sum(x)-1]++;
}
for(i=0;i<N;i++)
printf("%d\n",level[i]);
}
Stars
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 34943 | Accepted: 15175 |
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
5 1 1 5 1 7 1 3 3 5 5
Sample Output
1 2 1 1 0
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
Source
Ural Collegiate Programming Contest 1999
题目大意:有很多星星,星星的等级是星星左下所有星星的和,求各个等级星星有多少个。
思路:C【】是个树状数组,记录的是x坐标前包含x的星星的个数。每一个x,y更新树状数组,level[sum(x)-1]++.x从0开始,所有x处理时+1.
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define n 32005
int N;
int c
;
int level15005];
int lowbit(int x)
{
return x&(x^(x-1));
}
void add(int x,int num)
{
while(x<=n)
{
c[x]+=num;
x+=lowbit(x);
}
}
int sum(int x)
{
int s=0;
while(x>0)
{
s+=c[x];
x-=lowbit(x);
}
return s;
}
int main()
{
freopen("C:\\1.txt","r",stdin);
cin>>N;
int i;
for(i=0;i<N;i++)
{
int x,y;
scanf("%d%d",&x,&y);
x+=1;
add(x,1);
level[sum(x)-1]++;
}
for(i=0;i<N;i++)
printf("%d\n",level[i]);
}
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