POJ 3067 Japan 【 树状数组 】

题意：左边有n个城市，右边有m个城市，现在修k条路，问会形成多少个交点

先按照x从小到大排，x相同的话，则按照y从小到大排，然后对于每一个y统计前面有多少个y比它大，它们就一定会相交

另外要用long long

#include<iostream>  
#include<cstdio>  
#include<cstring> 
#include <cmath> 
#include<stack>
#include<vector>
#include<map> 
#include<set>
#include<queue> 
#include<algorithm>  
using namespace std;

typedef long long LL;
const int INF = (1<<30)-1;
const int mod=1000000007;
const int maxn=1000005;

int a[maxn];
int c[maxn];//树状数组
int n,m,k;

struct node{
    int x,y;
}p[maxn];

int cmp(node n1,node n2){
    if(n1.x != n2.x) return n1.x < n2.x;
    return n1.y < n2.y;
}

int lowbit(int x){ return x & (-x);}

int sum(int x){
    int ret =0;
    while(x > 0){
        ret+=c[x];x-=lowbit(x);
    }
    return ret;
}

void add(int x,int d){
    while(x < maxn){
        c[x]+=d;x+=lowbit(x);
    } 
}

int main(){
    int T;
    scanf("%d",&T);
    int kase=0;
    while(T--){
        scanf("%d %d %d",&n,&m,&k);
        for(int i=1;i<=k;i++) scanf("%d %d",&p[i].x,&p[i].y);
        sort(p+1,p+k+1,cmp);
        
        memset(c,0,sizeof(c));
        LL ans=0;
        for(int i=1;i<=k;i++){
            add(p[i].y,1);
            ans += i- sum(p[i].y);
        }
        printf("Test case %d: %I64d
",++kase,ans);
    }
    return 0;
}