[Usaco2017 Feb]Why Did the Cow Cross the Road III (Gold)
Description
给定长度为2N的序列,1~N各处现过2次,i第一次出现位置记为ai,第二次记为bi,求满足ai < aj < bi < bj的对数
Sample Input
4
3
2
4
4
1
3
2
1
Sample Output
3
HINT
N<=100000
树状数组维护,一个数出现第一次就加入树状数组,出现第二次的时候统计有多少个数出现一次,并把当前数去掉。记得倒着加入
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define lowbit(x) ((x)&(-x))
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
int x=0,f=1;char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x>=10) print(x/10);
putchar(x%10+'0');
}
const int N=2e5;
int tree[N+10],vis[N+10];
int n,ans;
void insert(int x,int v){for (;x<=n;x+=lowbit(x)) tree[x]+=v;}
int query(int x){
int res=0;
for (;x;x-=lowbit(x)) res+=tree[x];
return res;
}
int main(){
n=read()*2;
for (int i=1,x;i<=n;i++){
x=read();
if (vis[x]){
insert(vis[x],-1);
ans+=query(vis[x]);
}
else vis[x]=n-i+1,insert(vis[x],1); //倒着加入
}
printf("%d\n",ans);
return 0;
}