POJ 3067—— Japan(树状数组)

本文介绍了一种计算k条连接东西海岸城市的直线(超级高速公路)间交叉点数量的方法。通过将城市按序排列并利用逆序数的概念,文章提供了一个有效的解决方案来计算这些交叉点。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Japan
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 20991 Accepted: 5685

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

1
3 4 4
1 4
2 3
3 2
3 1

Sample Output

Test case 1: 5



————————————————分割线————————————————

题目大意:

求k条线段的交点个数


思路:

将西边的城镇从小到大排序,如果编号一样,则对东边的城镇按从小到大排序,求逆序数   (或者按从大到小排序,求比它小的数)

画画图就非常清楚了

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
#define M 1000000+10
using namespace std;
int n,m,k;
struct node
{
    int x,y;
}a[M];
int c[1010];
bool cmp(node x,node y)
{
    return (x.x<y.x||(x.x==y.x&&x.y<y.y));
}

inline void update(int x,int v)
{
    while(x<=m){
        c[x]+=v;
        x+=x&-x;
    }
}
inline ll getsum(int x)
{
    ll sum=0;
    while(x>0){
        sum+=c[x];
        x-=x&-x;
    }
    return sum;
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--){
        ll cnt=0;
        memset(c,0,sizeof(c));
        scanf("%d %d %d",&n,&m,&k);
        for(int i=0;i<k;++i){
            scanf("%d %d",&a[i].x,&a[i].y);
        }
        sort(a,a+k,cmp);
        for(int i=0;i<k;++i){
            update(a[i].y,1);
            cnt+=i-getsum(a[i].y)+1;
        }
        printf("Test case %d: %lld\n",cas++,cnt);
    }
    return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值