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原创最新推荐文章于 2026-06-18 15:24:53 发布 9.9k 阅读 AI 写同款· GEO 优化 ›

第一次打USACO和AtCoder Japan（后简称ABC）比赛总结其次，谈谈本次Cu组比赛的具体情况，一共3题，分别讲了Farmer John的3个不同的故事，然后用算法帮他解决问题，满分1000分。这两场比赛，看到了自己在算法和英语方面的不足，英语我已经给她买了一年期ReadingPlus的课程（需要的朋友请直接加复福教育蓝老师微信号fdfzjxxx07），一周内读了5000字的原版，集了好多medal，有点“成瘾”的味道，放学一到家就打开电脑开始做，希望我宝也能像其他用户一样，一年内追平到美国同年级水平，就达到了“妈妈再也不用担心我的英语学习啦”。阅读详情

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

[Submit] [Go Back] [Status]

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)

Sample Input

Sample Output

Test case 1: 5

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#define maxn 1010
using namespace std;
#define in __int64
int n,m,k;
in a[maxn];

struct node 
{
	in e,w;
}ww[maxn*maxn];

in cmp(node a,node b)
{
	if(a.e==b.e)
		return a.w > b.w;
	return 
		a.e>b.e;
}

int lowbit(int x)  //求出二进制中末尾第一个不为0的数的位置k,x+=2^k. 
{
        return  x&(-x);  //return x&(x^(x-1));
}

void insert(int p)   //构建树状数组 
{
        while(p<=n)
        {
                a[p]+=1;
                p+=lowbit(p);
        }
}

in sum(int end)  //求1到end区间的和 
{
        in s=0;
        while(end>0)
        {
                s+=a[end];
                end-=lowbit(end);
        }
        return s;
}

int main()
{
	int t;
	int V=0;
	int i;
	scanf("%d",&t);
	while(t--)
	{
		V++;
		in ans=0;
		memset(a,0,sizeof(a));
		scanf("%d%d%d",&n,&m,&k);		
		for(i=1;i<=k;i++)  
			scanf("%I64d %I64d",&ww[i].e,&ww[i].w);
		sort(ww+1,ww+k+1,cmp);
		
		for(i=1;i<=k;i++)
		{
			insert(ww[i].w);
			ans+=sum(ww[i].w-1);
		}
		printf("Test case %d: %I64d\n",V,ans);
	}
	return 0;
}