POJ 题目2481 Cows(树状数组)
来源:互联网 发布:永无止境 电影 知乎 编辑:程序博客网 时间:2024/05/16 04:42
Cows
Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 13920 Accepted: 4607
Description
Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good.
Farmer John has N cows (we number the cows from 1 to N). Each of Farmer John's N cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval [S,E].
But some cows are strong and some are weak. Given two cows: cowi and cowj, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cowi is stronger than cowj.
For each cow, how many cows are stronger than her? Farmer John needs your help!
Farmer John has N cows (we number the cows from 1 to N). Each of Farmer John's N cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval [S,E].
But some cows are strong and some are weak. Given two cows: cowi and cowj, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cowi is stronger than cowj.
For each cow, how many cows are stronger than her? Farmer John needs your help!
Input
The input contains multiple test cases.
For each test case, the first line is an integer N (1 <= N <= 105), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 105) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge.
The end of the input contains a single 0.
For each test case, the first line is an integer N (1 <= N <= 105), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 105) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge.
The end of the input contains a single 0.
Output
For each test case, output one line containing n space-separated integers, the i-th of which specifying the number of cows that are stronger than cowi.
Sample Input
31 20 33 40
Sample Output
1 0 0
Hint
Huge input and output,scanf and printf is recommended.
Source
POJ Contest,Author:Mathematica@ZSU
题目大意:
给出n头牛,他们有两个属性x和y,当A牛的x比B牛的x小于或等于,A牛的y比B牛的y大于或等于,但x和y不能同时等于,则称为A牛比B牛更强大
问对每头牛,有多少头牛比他强大
ac代码
#include<stdio.h>#include<string.h>#include<stdlib.h>struct ss{int e,s,id;}b[100100];int val[100100],a[100100];int cmp(const void *a,const void *b){if((*(struct ss *)a).s==(*(struct ss *)b).s)return (*(struct ss *)a).e-(*(struct ss *)b).e;return (*(struct ss *)b).s-(*(struct ss *)a).s;}int n;int lowbit(int x){return x&(-x);}void update(int p,int q){while(p<=n){a[p]+=q;p+=lowbit(p);}}int sum(int p){int ans=0;while(p>0){ans+=a[p];p-=lowbit(p);}return ans;}int main(){//int n;while(scanf("%d",&n)!=EOF,n){int i;memset(val,0,sizeof(val));memset(a,0,sizeof(a));for(i=1;i<=n;i++){scanf("%d%d",&b[i].e,&b[i].s);b[i].id=i;b[i].e++;b[i].s++;}qsort(b+1,n,sizeof(b[0]),cmp);val[b[1].id]=sum(b[1].e);update(b[1].e,1);for(i=2;i<=n;i++){if(b[i].e==b[i-1].e&&b[i].s==b[i-1].s){val[b[i].id]=val[b[i-1].id];}else{val[b[i].id]=sum(b[i].e);}update(b[i].e,1);}printf("%d",val[1]);for(i=2;i<=n;i++){printf(" %d",val[i]);}printf("\n");}}
0 0
- POJ 题目2481 Cows(树状数组)
- POJ 题目2182 Lost Cows(树状数组+二分)
- POJ 2481 Cows(树状数组)
- poj 2481 Cows (树状数组)
- POJ 2481 Cows(树状数组)
- poj 2481 Cows(树状数组)
- POJ - 2481 - Cows (树状数组+排序!!)
- POJ 2481 Cows(树状数组)
- POJ 2481 Cows(树状数组)
- POJ 2481Cows(树状数组)
- POJ 2481 Cows (树状数组)
- (POJ 2481)Cows 树状数组
- poj 2481 Cows(树状数组)
- POJ 2481 Cows (树状数组)
- POJ 2481 Cows(树状数组)
- POJ 2481 Cows 树状数组
- POJ 2481 Cows【树状数组】
- POJ 2481Cows 树状数组
- Spring的@Autowired需要注意的地方
- 用js判断浏览器类型以及版本
- GitHub简单入门教程
- 生物芯片 完全平方数
- 百度电话面试PHP职位
- POJ 题目2481 Cows(树状数组)
- 指针(一)
- JAVA多线程实现的三种方式
- [cacti]nginx+php+cacti+mysql+php-fpm 安装小记
- 第十二周阅读程序一
- Reservoir Sampling - 蓄水池抽样
- 用Python计算北京地铁的两站间最短换乘路线
- 你应当知道的7个Java工具
- nodejs多线程,真正的非阻塞