POJ 3067—— Japan（树状数组）

最新推荐文章于 2022-12-07 11:35:26 发布

fa茶

最新推荐文章于 2022-12-07 11:35:26 发布

阅读量536

点赞数

CC 4.0 BY-SA版权

分类专栏： # 数据结构

本文链接：https://blog.youkuaiyun.com/u014141559/article/details/38824781

数据结构专栏收录该内容

76 篇文章

订阅专栏

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

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;
}