POJ 2352 Stars
2015-08-10 19:48
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B - Stars
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
SubmitStatusPracticePOJ
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
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.
树状数组。交了一次RE,原因是因为树状数组的大小开小了。题目的N<=15000.把树状数组的大小开到32005后过了。
这题的意思是求 每个点左方下方有多少个比他小的点(包括本行本列)定义为level,求level为0,1,2.。。N-1的对应的点的个数。
开一个树状数组。 因为题目说Y已经是递增的了,所以每次输一个X,求一次1到X的和。插入更新树状数组,把sum放在ans[]里记录。相同的话就ans[sum]++;最后从0到n-1输出ans数组就行了。
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
SubmitStatusPracticePOJ
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
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.
树状数组。交了一次RE,原因是因为树状数组的大小开小了。题目的N<=15000.把树状数组的大小开到32005后过了。
这题的意思是求 每个点左方下方有多少个比他小的点(包括本行本列)定义为level,求level为0,1,2.。。N-1的对应的点的个数。
开一个树状数组。 因为题目说Y已经是递增的了,所以每次输一个X,求一次1到X的和。插入更新树状数组,把sum放在ans[]里记录。相同的话就ans[sum]++;最后从0到n-1输出ans数组就行了。
#include <stdio.h> #include <string.h> #define N 32005 int n; int c ; int ans ; int lowbit(int x) { return x&-x; } int sum(int x) { int ret=0; while(x>0) ret+=c[x],x-=lowbit(x); return ret; } void add(int x,int d) { while(x<=N) { c[x]+=d;x+=lowbit(x); } } int main() { while(scanf("%d",&n)>0) { int x,y; memset(ans,0,sizeof(ans)); memset(c,0,sizeof(c)); for(int i=0;i<n;i++) { scanf("%d%d",&x,&y); int num=sum(x+1); ans[num]++; add(x+1,1); } for(int i=0;i<n;i++) printf("%d\n",ans[i]); } return 0; }
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