POJ3067

解析复杂道路网络：超级高速公路交叉问题的高效解决策略

最新推荐文章于 2022-12-07 11:35:26 发布

原创最新推荐文章于 2022-12-07 11:35:26 发布 · 413 阅读

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#numbers #tree #construction #each #input #output

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Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15350		Accepted: 4129

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

Source

Southeastern Europe 2006

现在纸上建图，可以发现，对于点（x1,y1）和（x2,y2）,当x2>x1&&y2<y1时，会有crossing，所以将输入的数据先按前坐标从大到小，再按后坐标从大到小排序，用树状数组求解即可。注意用long long型

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=1005;
long long tree[maxn];
int N,M,K;
struct node
{
    int x,y;
} highway[maxn*maxn];
bool cmp(node a,node b)
{
    if(a.x==b.x) return a.y>b.y;
    else return a.x>b.x;
}
inline int Lowbit(int x)
{
    return x&(-x);
}
void Update(int x)
{
    for(int i=x; i<maxn; i+=Lowbit(i))
        tree[i]++;
}
long long Getsum(int x)
{
    long long s=0;
    for(int i=x; i>0; i-=Lowbit(i))
        s+=tree[i];
    return s;
}
int main()
{
    int T,ca=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&N,&M,&K);
        memset(tree,0,sizeof(tree));
        for(int i=0; i<K; i++)
            scanf("%d%d",&highway[i].x,&highway[i].y);
        sort(highway,highway+K,cmp);
        long long ans=0;
        for(int i=0; i<K; i++)
        {
            ans+=Getsum(highway[i].y-1);
            Update(highway[i].y);
        }
        printf("Test case %d: %lld\n",ca++,ans);
    }
    return 0;
}