zoj3129----------------Japan 逆序对~~

最新推荐文章于 2021-05-14 12:42:34 发布

原创最新推荐文章于 2021-05-14 12:42:34 发布 · 515 阅读

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#merge #numbers #construction #each #n2 #input

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本文介绍了一种算法，用于解决 ACM ICPC 世界总决赛中超级高速公路建设的问题，具体为计算不同高速公路之间的交叉点数量。通过排序和归并排序的方法，有效地解决了该问题，并提供了完整的 C++ 代码实现。

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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 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

先对左元素按升序排序，相等的再按有元素升序排

最后求右边元素序列的逆序对即可，

注意，结果用long long存，否则WA

代码：

#include <iostream>
#include <climits>
#include <algorithm>
using namespace std;


typedef struct Road 
{
	int n;
	int m;	
}NM;

int cmp(Road a,Road b)
{
	if(a.n!=b.n)
		return a.n<b.n;
	else
		return a.m<b.m;		
}

long long merge(long long A[],int p,int q,int r)
{
	int n1=q-p+1;
	int n2=r-q;
	long long count=0;
	long long *L=new long long[n1+2];
	long long *R=new long long[n2+2];
	for(int i=1;i<=n1;i++)
		L[i]=A[p+i-1];	
	for(int i=1;i<=n2;i++)
		R[i]=A[q+i];
	L[n1+1]=INT_MAX;
	R[n2+1]=INT_MAX;
	int i,j; 
	i=1;
	j=1;
	for(int k=p;k<=r;k++)
	{
		if(L[i]<=R[j])
		{
			count=count+j-1;
			A[k]=L[i];
			i++;	
		}
		else
		{
			A[k]=R[j];
			j++;	
		}	
	}
	delete[]L;
	delete[]R;
	return count;
}


long long merge_sort(long long A[],int p,int r)
{
	long long count=0;
	if(p<r)
	{
		int q=(p+r)/2;
		count=count+merge_sort(A,p,q);	
		count=count+merge_sort(A,q+1,r);
		count=count+merge(A,p,q,r);
	}
	return count;
}

int main()
{
	int T;
	cin>>T;
	for(int j=1;j<=T;j++)
	{
		int N,M,K;
		cin>>N>>M>>K;
		Road NM[K];
		for(int i=0;i<K;i++)
			cin>>NM[i].n>>NM[i].m;
		sort(NM,NM+K,cmp);
		long long A[K+1];
		for(int i=1;i<=K;i++)
			A[i]=NM[i-1].m;
		long long count;
		count=merge_sort(A,1,K);
		cout<<"Test case "<<j<<": "<<count<<endl;	
	}	
	return 0;
}