One day Polycarp decided to rewatch his absolute favourite episode of well-known TV series "Tufurama". He was pretty surprised when he got results only for season 7 episode 3 with his search query of "Watch Tufurama season 3 episode 7 online full hd free". This got Polycarp confused — what if he decides to rewatch the entire series someday and won't be able to find the right episodes to watch? Polycarp now wants to count the number of times he will be forced to search for an episode using some different method.
TV series have n seasons (numbered 1 through n), the i-th season has ai episodes (numbered 1 through ai). Polycarp thinks that if for some pair of integers x and y (x < y) exist both season x episode y and season y episode x then one of these search queries will include the wrong results. Help Polycarp to calculate the number of such pairs!
The first line contains one integer n (1 ≤ n ≤ 2·105) — the number of seasons.
The second line contains n integers separated by space a1, a2, ..., an (1 ≤ ai ≤ 109) — number of episodes in each season.
Print one integer — the number of pairs x and y (x < y) such that there exist both season x episode y and season y episode x.
5 1 2 3 4 5
0
3 8 12 7
3
3 3 2 1
2
第一反应是主席树:
注意unique的使用
/**求出满足a[i]>=j&&a[j]>=i的有序对< i , j >的对数(i<j)。**/
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn = 200000;
typedef long long int ll;
int n,a[maxn+5],b[maxn+5],th[maxn+5];
int tot,sz,root[maxn+5],ls[maxn*50+5],rs[maxn*50+5],sum[maxn*50+5];
void build(int &v,int l,int r)
{
v=++tot;
sum[v] = 0;
if(l==r) return;
int mid = (l+r)>>1;
build(ls[v],l,mid);
build(rs[v],mid+1,r);
}
void updata(int &v,int last,int l,int r,int q)
{
v = ++tot;
ls[v] = ls[last];
rs[v] = rs[last];
sum[v] = sum[last]+1;
if(l==r) return;
int mid = (l+r)>>1;
if(q<=mid) updata(ls[v],ls[last],l,mid,q);
if(q>mid) updata(rs[v],rs[last],mid+1,r,q);
}
int query(int v,int last,int l,int r,int L,int R)
{
if(r<L||l>R) return 0;
int mid = (l+r)>>1;
if(L<=l&&r<=R) return sum[v] - sum[last];
int cost = 0;
if(mid>=L) cost+=query(ls[v],ls[last],l,mid,L,R);
if(mid<R) cost+=query(rs[v],rs[last],mid+1,r,L,R);
return cost;
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]), b[i] = a[i];
sort(b+1,b+n+1);//*********
sz = unique(b + 1, b + n + 1) - (b + 1);// unique返回一个个重复的位置
for(int i=1;i<=n;i++)
{
th[i] = lower_bound(b+1,b+sz+1,a[i]) - b;
}
build(root[0],1,sz);
for(int i=1;i<=n;i++) updata(root[i],root[i-1],1,sz,th[i]);
ll ans = 0;
for(int i=1;i<=n;i++)
{
int l = i, r = min(a[i],n), q;
if(l >= r) continue;
q = lower_bound(b+1,b+sz+1,i) - b;
ans +=query(root[r],root[l],1,sz,q,sz);
}
cout<<ans<<endl;
return 0;
}
树状数组：
/**
求出满足a[i]>=j&&a[j]>=i的有序对< i , j >的对数(i<j)。
换一个突破口改变就这么大：
对j,满足a[j]>=i，找到最大的i，此时只要在 1~i 中找到a[i] > j的数有多少就行了，
**/
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<vector>
using namespace std;
const int maxn = 200000;
typedef long long int ll;
int n,a[maxn+5];
int C[maxn+5];
vector<int>g[maxn+5];
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 updata(int x,int d)
{
while(x<=n){
C[x]+=d;x+=lowbit(x);
}
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
a[i] = min(n,a[i]);
g[min(i-1,a[i])].push_back(i);
}
ll ans = 0;
for(int i=1;i<=n;i++)
{
updata(a[i],1);
for(int j=0;j<g[i].size();j++)
{
ans+=Sum(n) - Sum(g[i][j]-1);
}
}
cout<<ans<<endl;
return 0;
}