​​http://codeforces.com/problemset/problem/574/D​​

最后被拿掉的方块肯定在最底层 对于底层的每一个方块就看它外面包着几层方块 比如1234321 最中间第四个最下面的方块外面有三层 算上自己四层 而左右两边的底层方块就只有三层

关键就是怎么求这个层数 对每个位置i的方块数减去i 看左面最小值 与当前位置作差 就是被破坏的层数 右面同理 这个方法和之前codechef上做的一个主席树很像

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
const int N=0x3f3f3f3f;

int lef[4*maxn],rgt[4*maxn];
int ary[maxn],prel[maxn],prer[maxn];
int n;

void pushup(int *minn,int cur)
{
    minn[cur]=min(minn[2*cur],minn[2*cur+1]);
}

void build(int *pre,int *minn,int l,int r,int cur)
{
    int m;
    if(l==r){
        minn[cur]=pre[l];
        return;
    }
    m=(l+r)/2;
    build(pre,minn,l,m,2*cur);
    build(pre,minn,m+1,r,2*cur+1);
    pushup(minn,cur);
}

int query(int *minn,int pl,int pr,int l,int r,int cur)
{
    int res,m;
    if(pl<=l&&r<=pr) return minn[cur];
    res=N,m=(l+r)/2;
    if(pl<=m) res=min(res,query(minn,pl,pr,l,m,2*cur));
    if(pr>m) res=min(res,query(minn,pl,pr,m+1,r,2*cur+1));
    return res;
}

int main()
{
    int i,ans,res,minnl,minnr;
    scanf("%d",&n);
    for(i=1;i<=n;i++) scanf("%d",&ary[i]);
    for(i=1;i<=n;i++) prel[i]=ary[i]-i;
    build(prel,lef,1,n,1);
    for(i=1;i<=n;i++) prer[i]=ary[i]-(n-i+1);
    build(prer,rgt,1,n,1);
    ans=0;
    for(i=1;i<=n;i++){
        if(i>1){
            res=query(lef,1,i-1,1,n,1);
            if(res<prel[i]) minnl=min(i,ary[i]-(prel[i]-res));
            else minnl=min(i,ary[i]);
        }
        else minnl=1;
        if(i<n){
            res=query(rgt,i+1,n,1,n,1);
            if(res<prer[i]) minnr=min(n-i+1,ary[i]-(prer[i]-res));
            else minnr=min(n-i+1,ary[i]);
        }
        else minnr=1;
        ans=max(ans,min(minnl,minnr));
    }
    printf("%d\n",ans);
    return 0;
}