Poj 2796　单调栈

关于单调栈的性质，和单调队列基本相同，只不过单调栈只使用数组的尾部，　类似于栈。

Accepted Code:

 1 /*************************************************************************
 2     > File Name: 2796.cpp
 3     > Author: Stomach_ache
 4     > Mail: [email protected]
 5     > Created Time: 2014年07月21日 星期一 22时45分56秒
 6     > Propose:
 7  ************************************************************************/
 8 #include <cmath>
 9 #include <string>
10 #include <cstdio>
11 #include <fstream>
12 #include <cstring>
13 #include <iostream>
14 #include <algorithm>
15 using namespace std;
16
17 typedef long long LL;
18 const int maxn = 100005;
19 int n;
20 // b[i] holds the smallest index where a[b[i]]...a[i] are not larger than a[i], and e[i] is similar, just extends to the right of a[i]
21 int a[maxn], b[maxn], e[maxn], st[maxn];
22 LL sum[maxn]; //prefix sum
23
24 int
25 main(void) {
26 #ifndef ONLINE_JUDGE
27       freopen("in.txt", "r", stdin);
28 #endif
29       while (~scanf("%d", &n)) {
30           for (int i = 1; i <= n; i++) scanf("%d", a+i);
31         for (int i = 1; i <= n; i++) b[i] = e[i] = i;
32         sum[0] = 0;
33         for (int i = 1; i <= n; i++) {
34             sum[i] = sum[i-1] + a[i];
35         }
36         // the top of no decreasing stack
37         int top = 0;
38         for (int i = 1; i <= n; i++) {
39               if (!top || a[i] > a[st[top-1]]) {
40                 // push an element to stack
41                   st[top++] = i;
42             } else {
43                   while (top > 0 && a[st[top-1]] > a[i]) {
44                       // update e[st[top-1]]
45                       e[st[top-1]] = i - 1;
46                     // update b[i]
47                     b[i] = b[st[top-1]];
48                     // pop the toppest element of stack
49                       top--;
50                 }
51                 if (!top || a[st[top-1]] != a[i]) {
52                       // push
53                       st[top++] = i;
54                 } else if(a[st[top-1] == a[i]]) {
55                       // update b[i]
56                       b[i] = b[st[top-1]];
57                 }
58             }
59         }
60         // update e[i] of those elements that still in the stack
61         for (int i = 0; i < top; i++) e[st[i]] = n;
62         // find the answer
63         LL ans = -1, bb, ee;
64         for (int i = 1; i <= n; i++) {
65               if (ans < a[i] * (sum[e[i]]-sum[b[i]-1])) {
66                   ans = a[i] * (sum[e[i]]-sum[b[i]-1]);
67                 bb = b[i];
68                 ee = e[i];
69             }
70         }
71         printf("%lld\n%lld %lld\n", ans, bb, ee);
72     }
73
74     return 0;
75 }

Poj 2796　单调栈