HDU-Largest Rectangle in a Histogram-1506 单调栈

Posted 2021-01-04 the-way-of-cas

连接：http://acm.hdu.edu.cn/showproblem.php?pid=1506

题意：给一个柱状图，每个小矩形宽为1，求这个柱状图最大子矩形的面积

思路：用单调栈求出每个小矩形所能组成最大矩形的左宽和有宽，分别存在l[i] , r[i]数组里，然后 ans = max( h[i] * ( r[i] - l[i] )

注意：int * int 结果可能超过2^32

代码：

#include <iostream>
#include <algorithm>
#define ll long long
#define maxn 100000+100
//#define scanf scanf_s

using namespace std;

int n;
int h[maxn];
int l[maxn], r[maxn];
int st[maxn];

void solve() {
    int t = 0;
    for (int i = 0; i < n; i++)
    {
        while (t > 0 && h[st[t - 1]] >= h[i]) t--;
        l[i] = t == 0 ? 0 : (st[t - 1] + 1);
        st[t++] = i;
    }
    t = 0;
    for (int i = n-1; i >= 0; i--)
    {
        while (t > 0 && h[st[t - 1]] >= h[i]) t--;
        r[i] = t == 0 ? n : st[t - 1];
        st[t++] = i;
    }
    ll res = 0;
    for (int i = 0; i < n; i++)
    {
        res = max(res, 1ll*h[i] * (r[i] - l[i]));
    }
    printf("%lld
", res);
}

int main() {
    while (scanf("%d", &n) && n) {
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &h[i]);
        }
        solve();
    }
    return 0;
}