poj3067之树状数组
来源:互联网 发布:mac怎么删除第三方软件 编辑:程序博客网 时间:2024/05/29 16:01
Japan
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 19073 Accepted: 5170
Description
Japan plans to welcome the ACM ICPC World Finals and a lot of roads must be built for the venue. Japan is tall island with N cities on the East coast and M cities on the West coast (M <= 1000, N <= 1000). K superhighways will be build. Cities on each coast are numbered 1, 2, ... from North to South. Each superhighway is straight line and connects city on the East coast with city of the West coast. The funding for the construction is guaranteed by ACM. A major portion of the sum is determined by the number of crossings between superhighways. At most two superhighways cross at one location. Write a program that calculates the number of the crossings between superhighways.
Input
The input file starts with T - the number of test cases. Each test case starts with three numbers – N, M, K. Each of the next K lines contains two numbers – the numbers of cities connected by the superhighway. The first one is the number of the city on the East coast and second one is the number of the city of the West coast.
Output
For each test case write one line on the standard output:
Test case (case number): (number of crossings)
Test case (case number): (number of crossings)
Sample Input
13 4 41 42 33 23 1
Sample Output
Test case 1: 5
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <queue>#include <algorithm>#include <map>#include <cmath>#include <iomanip>#define INF 99999999typedef long long LL;using namespace std;const int MAX=1000+10;int n,m,k;int c[MAX];struct Node{int x,y;bool operator<(const Node &a)const{if(x == a.x)return y<a.y;return x<a.x;}}s[MAX*MAX];int lowbit(int x){return x&(-x);}void Update(int x,int y){while(x<=m){c[x]+=y;x+=lowbit(x);}}int Query(int x){int sum=0;while(x>0){sum+=c[x];x-=lowbit(x);}return sum;}int main(){int t,num=0;scanf("%d",&t);while(t--){memset(c,0,sizeof c);scanf("%d%d%d",&n,&m,&k);for(int i=0;i<k;++i){scanf("%d%d",&s[i].x,&s[i].y);}sort(s,s+k);LL sum=0;for(int i=0;i<k;++i){sum+=i-Query(s[i].y);Update(s[i].y,1);}printf("Test case %d: %lld\n",++num,sum);}return 0;}
0 0
- poj3067之树状数组
- Japan poj3067--树状数组
- 树状数组POJ3067
- POJ3067树状数组
- poj3067---树状数组
- poj3067 Japan(树状数组)
- POJ3067 Japan 树状数组
- poj3067 Japan (树状数组)
- POJ3067 Japan(树状数组)
- poj3067 树状数组
- POJ3067--Japan(树状数组)
- POJ3067 Japan(树状数组)
- POJ3067-Japan(树状数组)
- poj3067 树状数组(Binary Indexed Tree)
- 树状数组专题(四)POJ3067
- poj3067 树状数组求逆序数
- poj3067 Japan 树状数组&逆序对
- POJ3067 Japan 线段树 || 树状数组
- 入门视频采集与处理(学会分析YUV数据)
- C++中public,protected,private访问
- hdu 1102
- 怪异模式(Quirks Mode)对 HTML 页面的影响w
- V4L2视频采集与H.264编码源码大放送:Linux视频采集与编码
- poj3067之树状数组
- 嵌入式LINUX环境下视频采集知识(V4L2)
- Android 之 语言 国际化
- MD5加密算法(1)
- 第二周----玩日期(2)
- Android实战技巧之七:按钮控制ViewPager的左右翻页
- 组合,Java,算法
- 【动态规划】zb的生日
- 交通灯调度系统