stack(数组模拟) POJ 2559 Largest Rectangle in a Histogram

 

题目传送门

/*
    题意：宽度为1，高度不等，求最大矩形面积
    stack(数组模拟)：对于每个a[i]有L[i]，R[i]坐标位置 表示a[L[i]] < a[i] < a[R[i]] 的极限情况
                  st[]里是严格单调递增，若不记录的话还要O(n)的去查找L，R，用栈的话降低复杂度
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <iostream>
using namespace std;

typedef long long ll;

const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int a[MAXN], L[MAXN], R[MAXN];
int st[MAXN];

int main(void)        //POJ 2559 Largest Rectangle in a Histogram
{
//    freopen ("POJ_2559.in", "r", stdin);

    int n;
    while (scanf ("%d", &n) == 1)
    {
        if (n == 0)    break;
        for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);
        memset (st, 0, sizeof (st));

        int p = 0;
        for (int i=1; i<=n; ++i)
        {
            while (p >= 1 && a[st[p-1]] >= a[i])    p--;
            L[i] = (p == 0) ? 0 : st[p-1];
            st[p++] = i;
        }

        p = 0;
        for (int i=n; i>=1; --i)
        {
            while (p >= 1 && a[st[p-1]] >= a[i])    p--;
            R[i] = (p == 0) ? n + 1 : st[p-1];
            st[p++] = i;
        }

        ll ans = 0;
        for (int i=1; i<=n; ++i)
        {
            ans = max (ans, (ll) a[i] * (R[i] - L[i] - 1));
        }
        printf ("%I64d\n", ans);
    }

    return 0;
}