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

题目传送门

 1 /*
 2     题意：宽度为1，高度不等，求最大矩形面积
 3     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
 4                   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
 5 */
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <algorithm>
 9 #include <stack>
10 #include <iostream>
11 using namespace std;
12
13 typedef long long ll;
14
15 const int MAXN = 1e5 + 10;
16 const int INF = 0x3f3f3f3f;
17 int a[MAXN], L[MAXN], R[MAXN];
18 int st[MAXN];
19
20 int main(void)        //POJ 2559 Largest Rectangle in a Histogram
21 {
22 //    freopen ("POJ_2559.in", "r", stdin);
23
24     int n;
25     while (scanf ("%d", &n) == 1)
26     {
27         if (n == 0)    break;
28         for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);
29         memset (st, 0, sizeof (st));
30
31         int p = 0;
32         for (int i=1; i<=n; ++i)
33         {
34             while (p >= 1 && a[st[p-1]] >= a[i])    p--;
35             L[i] = (p == 0) ? 0 : st[p-1];
36             st[p++] = i;
37         }
38
39         p = 0;
40         for (int i=n; i>=1; --i)
41         {
42             while (p >= 1 && a[st[p-1]] >= a[i])    p--;
43             R[i] = (p == 0) ? n + 1 : st[p-1];
44             st[p++] = i;
45         }
46
47         ll ans = 0;
48         for (int i=1; i<=n; ++i)
49         {
50             ans = max (ans, (ll) a[i] * (R[i] - L[i] - 1));
51         }
52         printf ("%I64d\n", ans);
53     }
54
55     return 0;
56 }