POJ 2559 Largest Rectangle in a Histogram RMQ || 单调栈

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题目链接：点击打开链接

题意就是求最大面积

枚举每个柱子作为起点

然后二分两边长度。求个区间最值。

#include<stdio.h>#include<iostream>#include<math.h>using namespace std;#define ll long long#define N 100100inline bool rd(int &n){        int x = 0, tmp = 1;        char c = getchar();        while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();        if(c == EOF) return false;        if(c == '-') c = getchar(), tmp = -1;        while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();        n = x*tmp;        return true;  }    inline void pt(ll n){        if(n < 0)          putchar('-'), n = -n;        int len = 0,data[20];        while(n)            data[len++] = n%10, n /= 10;         if(!len) data[len++] = 0;        while(len--) putchar(data[len]+48);    }  int n;int a[N];int FMin[N][20];void Init(){    int i,j;    for(i=1;i<=n;i++)        FMin[i][0]=a[i];    for(i=1;(1<<i)<=n;i++){   //按区间长度递增顺序递推        for(j=1;j+(1<<i)-1<=n;j++){   //区间起点            FMin[j][i]=min(FMin[j][i-1],FMin[j+(1<<(i-1))][i-1]);        }    }}int Query(int l,int r){    int k=(int)(log(double(r-l+1))/log((double)2));    return min(FMin[l][k],FMin[r-(1<<k)+1][k]);}int main() {    while(rd(n), n){for(int i = 1; i <= n; i++) rd(a[i]);        Init();        ll ans = 0;        for(int i = 1; i <= n; i++)        {if(ans >= (ll)a[i]*n)continue;            int l = 1, r = i-1, L = i;            while(l<=r)            {                int mid = (l+r)>>1;                if(Query(mid, i-1) >= a[i])                    r = mid-1, L = min(L, mid);                else                    l = mid+1;            }            l = i+1, r = n; int R = i;            while(l <= r)            {                int mid = (l+r)>>1;                if(Query(i+1, mid) >= a[i])                    l = mid+1, R = max(R, mid);                else                    r = mid-1;            }            ans = max(ans, (ll)(R-L+1) * a[i]);        }        pt(ans);putchar('\n');    }    return 0;}

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