ZROI #88

传送门

分析

我们考虑把每个A[i]考虑为山峰的高度，每次的B考虑为海平面

于是我们知道对于A[i]和A[i-1]，如果A[i-1]<A[i]则在A[i-1]<B<=A[i]时会使陆地总数加一

于是树状数组维护即可

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define N cnt
int D[],a[],opt[],b[],c[],d[];
int id[],cnt;
int lb(int x){return x&(-x);}
void add(int x,int k){while(x<=N)D[x]+=k,x+=lb(x);}
void update(int le,int ri,int k){add(le,k),add(ri+,-k);}
int q(int x){int res=;while(x){res+=D[x];x-=lb(x);}return res;}
int wh(int x){return lower_bound(id+,id+cnt+,x)-id;}
int main(){
    int n,m,i,j,k;
    scanf("%d%d",&n,&m);
    for(i=;i<=n;i++)scanf("%d",&a[i]),id[++cnt]=a[i];
    for(i=;i<=m;i++){
      scanf("%d",&opt[i]);
      if(opt[i]==)scanf("%d",&b[i]),id[++cnt]=b[i];
        else scanf("%d%d",&c[i],&d[i]),id[++cnt]=d[i];
    }
    sort(id+,id+cnt+);
    cnt=unique(id+,id+cnt+)-id-;
    for(i=;i<=n;i++)
      if(a[i]>a[i-])
        update(wh(a[i-])+,wh(a[i]),);
    for(i=;i<=m;i++){
      if(opt[i]==)printf("%d\n",q(wh(b[i])));
        else {
          if(a[c[i]]<a[c[i]+])update(wh(a[c[i]])+,wh(a[c[i]+]),-);
          if(a[c[i]-]<a[c[i]])update(wh(a[c[i]-])+,wh(a[c[i]]),-);
          a[c[i]]=d[i];
          if(a[c[i]]<a[c[i]+])update(wh(a[c[i]])+,wh(a[c[i]+]),);
          if(a[c[i]-]<a[c[i]])update(wh(a[c[i]-])+,wh(a[c[i]]),);
        }
    }
    return ;
}