Codeforces 1156E Special Segments of Permutation(启发式合并)

题意：

给一个n的排列，求满足a[l]+a[r]=max(l,r)的(l,r)对数，max(l,r)指的是l到r之间的最大a[p]

n<=2e5

思路：

先用单调栈处理出每个点能扩展的l[i],r[i]

搜索以每个点为最大值时的贡献，对每个点只搜索它左边部分或右边部分最小的那个

可以证明，每个点最多被搜到logn次，类似于启发式合并的思想，

复杂度为nlogn

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
    
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 
#define LLONG_MAX 9223372036854775807

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e6+11;
const int maxm = 1e5+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int a[maxn];
int l[maxn],r[maxn];
int idx[maxn];
ll ans = 0;
void gao(int x){
    int L,R;
    int pL,pR;
    if(x-l[x]<r[x]-x){
        L=l[x];
        R=x-1;
        pL=x+1;
        pR=r[x];
    }
    else{
        pL=l[x];
        pR=x-1;
        L=x+1;
        R=r[x];
    }
    for(int i = L; i <= R; i++){
        int y = a[x]-a[i];
        if(idx[y]>=pL&&idx[y]<=pR)ans++;
    }
}
int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
        idx[a[i]]=i;
        l[i] = r[i] = i;
    }
    a[0]=a[n+1]=n+1;
    for(int i = 1; i <= n; i++){
        while(a[l[i]-1] <= a[i]){
            l[i] = l[l[i]-1];
        }
    }
    for(int i = n; i >= 1; i--){
        while(a[r[i]+1] <= a[i]){
            r[i] = r[r[i]+1];
        }
    }
    ans = 0;
    for(int i = 1; i <= n; i++){
        gao(i);
    }
    printf("%lld",ans);
    return 0;
}