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 = (<<)-;
 const int mod=;
 const int maxn=;

 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 =;
     while(x > ){
         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=;
     while(T--){
         scanf("%d %d %d",&n,&m,&k);
         for(int i=;i<=k;i++) scanf("%d %d",&p[i].x,&p[i].y);
         sort(p+,p+k+,cmp);

         memset(c,,sizeof(c));
         LL ans=;
         for(int i=;i<=k;i++){
             add(p[i].y,);
             ans += i- sum(p[i].y);
         }
         printf("Test case %d: %I64d\n",++kase,ans);
     }
     return ;
 }