poj3067树状数组求逆序数

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
题意：给一组成对的点求相交的线段
题解：逆序数（虽然我也不知道这是个啥），先排序按右边小的排，右边相同按左边小的排，保证不会因为右边点相同而计算重复，然后求sum(N)-sum(s[i].b)（好像是逆序数）;
一开始思路都是对的，结果写太搓了，TLE了，无奈之下看题解，居然和我写的一模一样。。。。（代码能力太差》。《）

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f3f;

struct edge{
    int a,b;
}s[N*N];
int c[N];

bool comp(const edge &x,const edge &y)
{
    if(x.a!=y.a)return x.a<y.a;
    return x.b<y.b;
}
void add(int i)
{
    while(i<N){
        c[i]++;
        i+=i&(-i);
    }
}
ll sum(int i)
{
    ll ans=;
    while(i>){
        ans+=c[i];
        i-=i&(-i);
    }
    return ans;
}
int main()
{
 /*   ios::sync_with_stdio(false);
    cin.tie(0);*/
 //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    int t,cnt=,n,m,k;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&m,&k);
        memset(c,,sizeof c);
        for(int i=;i<k;i++)scanf("%d%d",&s[i].a,&s[i].b);
        sort(s,s+k,comp);
        ll ans=;
        for(int i=;i<k;i++)
        {
            add(s[i].b);
            ans+=(sum(N)-sum(s[i].b));
        }
        printf("Test case %d: %lld\n",++cnt,ans);
    }
    return ;
}