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来源：互联网发布：数据挖掘工程师就业编辑：程序博客网时间：2024/06/17 19:27

可持久化线段树+二分

简要题意：给出l,r,w,在[l,r]中找连续的长度为w的一段，最大化这一段中的最小值。

最大化最小值，想到二分，那么check就应当是判断[l,r]内是否存在一段由大于等于二分答案的数所组成的长度≥w的连续区间。如果已经明确了这些数在哪里，这种查询是可以用线段树维护的。再考虑每一次实际上都是应用大于等于二分答案的所有数，删去小于二分答案的所有数。因此只需按数值从大到小维护可持久化线段树即可。O(nlog2n)

//21:22 ~ 21:48#include<cstdio>#include<cstdlib>#include<algorithm>#define N 100005#define cmin(u,v) (u)>(v)?(u)=(v):0#define cmax(u,v) (u)<(v)?(u)=(v):0using namespace std;namespace runzhe2000{    int in()    {        int r = 0; char c = getchar();        for(; c < '0' || c > '9'; c = getchar());        for(; c >='0' && c <='9'; c = getchar())r=r*10+c-'0';        return r;    }    int n, h[N], pos[N], mx, rm;    bool cmp(int a, int b){return h[a] > h[b];}    struct node    {        node *ch[2];        int lm, rm, mx;        void pushup(int l, int r, int mid)        {            lm = (ch[0]->mx == mid-l+1 ? ch[0]->mx + ch[1]->lm : ch[0]->lm);            rm = (ch[1]->mx == r-mid ? ch[1]->mx + ch[0]->rm : ch[1]->rm);            mx = max(max(ch[0]->mx, ch[1]->mx), ch[0]->rm + ch[1]->lm);        }    }mem[N*20], *tot, *null, *root[N];    node *newnode()    {        node *p = ++tot;        return p;    }    void init()    {        null = tot = mem;        *null = (node){{null,null},0,0,0};        for(int i = 0; i <= n; i++) root[i] = newnode();        *root[0] = *null;    }    void ins(node *x, node *y, int l, int r, int p)    {        if(l == r) {y->lm = y->rm = y->mx = 1; return;}        int mid = (l+r)>>1;        if(p<=mid)         {            y->ch[1] = x->ch[1];            y->ch[0] = newnode();            ins(x->ch[0], y->ch[0], l, mid, p);        }        else        {            y->ch[0] = x->ch[0];            y->ch[1] = newnode();            ins(x->ch[1], y->ch[1], mid+1,r,p);        }        y->pushup(l,r,mid);    }    void query(node *x, int l, int r, int ql, int qr)    {        if(ql <= l && r <= qr)        {            cmax(mx, x->mx);            cmax(mx, rm + x->lm);            rm = x->mx == (r-l+1) ? rm + x->mx : x->rm;            return;        }        int mid = (l+r)>>1;        if(ql <= mid) query(x->ch[0], l, mid, ql, qr);        if(mid < qr) query(x->ch[1], mid+1, r, ql, qr);    }    void main()    {        n = in(); init();        for(int i = 1; i <= n; i++) h[i] = in(), pos[i] = i;        sort(pos+1,pos+1+n,cmp);        for(int i = 1; i <= n; i++) ins(root[i-1], root[i], 1, n, pos[i]);        for(int m = in(), i = 1, lef, rig, w; i <= m; i++)        {            lef = in(), rig = in(), w = in();            int l, r; for(l = 1, r = n; l < r; )            {                int mid = (l+r)>>1;                mx = rm = 0; query(root[mid], 1, n, lef, rig);                if(mx >= w) r = mid;                else l = mid + 1;            }            printf("%d\n",h[pos[l]]);        }    }}int main(){    runzhe2000::main();}

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