japan（树状数组求逆序数）

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本文介绍了一个算法问题，涉及计算日本岛屿上建设的高速公路之间的交叉点数量。通过将城市配对视为坐标点，并利用树状数组求解逆序数的方法，实现了高效计算。

http://poj.org/problem?id=3067

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

题意：日本岛东海岸与西海岸分别有N和M个城市，现在修高速公路连接东西海岸的城市，求交点个数。记每条告诉公路为(x,y), 即东岸的第x个城市与西岸的第y个城市修一条路。当两条路有交点时，满足（x1-x2）*（y1-y2） < 0。所以，将每条路按x从小到达排序，若x相同，按y从小到大排序。然后按排序后的公路用树状数组在线更新，求y的逆序数之和即为交点个数。

上面说的可能有点难理解，详细说明如下。

记第i条边的端点分别为xi,yi。

方法：

1.就是将x坐标，由小到大排序后，求y坐标的逆序数。但是要注意，排序方法是在当x相同的情况下，y要按照有小到大排序，这样才可以。（可以画一下）

2.maxn之前设成了1000，但是要知道，最多有1000*1000个连线。否则RE

3.逆序数这样想。比如2 2 3 4 1.想到，当我们从头开始，遇到第一个2，此时要看后边比2小的有几个。但是树状数组我们没法知道未来的，只能算已经扫进去的，所以倒过来，就看自己能否跟前边的来一个贡献。就看到了当前，比他大的数有几个。

代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<memory.h>
using namespace std;
typedef long long LL;
const int maxn=1023004;
LL c[maxn];
struct node{
    int x,y;
}a[maxn];
int cmp(node a,node b){
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;
}
int lowbit(int x){
    return x&(-x);
}
void add(int x){
    int i=x;
    for(;i<=maxn;i+=lowbit(i)){
        c[i]+=1;
    }
}
long long sum(int x){
    int i=x;long long sum=0;
    for(;i>0;i-=lowbit(i)){
        sum+=c[i];
    }
    return sum;
}
int main()
{
    int t,n,m,k;
    int cas=0;
    scanf("%d",&t);
    while(t--){
        memset(c,0,sizeof(c));
        memset(a,0,sizeof(a));
        cas++;
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=k;i++){
            scanf("%d%d",&a[i].x,&a[i].y);
        }
        sort(a+1,a+1+k,cmp);
        /*for(int i=1;i<=k;i++){
            cout<<a[i].x<<" "<<a[i].y<<endl;
        }
        */
        long long ans=0;
        for(int i=1;i<=k;i++){
            ans+=( sum(maxn) - sum(a[i].y) );
            add(a[i].y);
        }
        printf("Test case %d: %lld\n",cas,ans);
    }
}
/*
30
3 3 3
1 3
2 2
3 1

3 4 4
1 4
2 3
3 2
3 1

2 3 2
1 2
2 1

4 4 5
1 2
3 2
3 3
3 4
4 1


*/