POJ 3067 Japan（树状数组求逆序对）

Japan

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14795		Accepted: 3968

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

Source

Southeastern Europe 2006

 

 

 

题意：日本东部和西部沿海各有n，m座城市，想通过修路来连同。给出城市之间的联线，求共有多少的交点；

 

先排序，然后求逆序对。

可以用归并排序，也可以用树状数组。

这里用树状数组实现。

注意边的范围比较大，1000*1000.

结果超int，用long long

 

/*
POJ 3067 Japan
按照第一个数从小到大排序，第一个数相同则按照第二个数从小到大排序。
然后用树状数组求逆序对。

边比较大，结果要用long long


*/

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;
const int MAXN=1010;
int c[MAXN];
struct Node
{
    int a,b;
}node[1000*1000];
int N,M,K;
bool cmp(Node a,Node b)
{
    if(a.a==b.a)return a.b<b.b;
    return a.a<b.a;
}
int lowbit(int x)
{
    return x&(-x);
}
void add(int i,int val)
{
    while(i<=M)
    {
        c[i]+=val;
        i+=lowbit(i);
    }
}
int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=c[i];
        i-=lowbit(i);
    }
    return s;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int iCase=0;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        iCase++;
        scanf("%d%d%d",&N,&M,&K);
        for(int i=1;i<=K;i++)
        {
            scanf("%d%d",&node[i].a,&node[i].b);
        }
        sort(node+1,node+1+K,cmp);
        memset(c,0,sizeof(c));
        long long ans=0;
        for(int i=1;i<=K;i++)
        {
            ans+=sum(M)-sum(node[i].b);
            add(node[i].b,1);
        }
        printf("Test case %d: %I64d\n",iCase,ans);
    }
    return 0;
}