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:

 /*************************************************************************
     > File Name: 2796.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年07月21日 星期一 22时45分56秒
     > Propose:
  ************************************************************************/
 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long LL;
 const int maxn = ;
 int n;
 // 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]
 int a[maxn], b[maxn], e[maxn], st[maxn];
 LL sum[maxn]; //prefix sum  

 int
 main(void) {
 #ifndef ONLINE_JUDGE
       freopen("in.txt", "r", stdin);
 #endif
       while (~scanf("%d", &n)) {
           for (int i = ; i <= n; i++) scanf("%d", a+i);
         for (int i = ; i <= n; i++) b[i] = e[i] = i;
         sum[] = ;
         for (int i = ; i <= n; i++) {
             sum[i] = sum[i-] + a[i];
         }
         // the top of no decreasing stack
         int top = ;
         for (int i = ; i <= n; i++) {
               if (!top || a[i] > a[st[top-]]) {
                 // push an element to stack
                   st[top++] = i;
             } else {
                   while (top >  && a[st[top-]] > a[i]) {
                       // update e[st[top-1]]
                       e[st[top-]] = i - ;
                     // update b[i]
                     b[i] = b[st[top-]];
                     // pop the toppest element of stack
                       top--;
                 }
                 if (!top || a[st[top-]] != a[i]) {
                       // push
                       st[top++] = i;
                 } else if(a[st[top-] == a[i]]) {
                       // update b[i]
                       b[i] = b[st[top-]];
                 }
             }
         }
         // update e[i] of those elements that still in the stack
         for (int i = ; i < top; i++) e[st[i]] = n;
         // find the answer
         LL ans = -, bb, ee;
         for (int i = ; i <= n; i++) {
               if (ans < a[i] * (sum[e[i]]-sum[b[i]-])) {
                   ans = a[i] * (sum[e[i]]-sum[b[i]-]);
                 bb = b[i];
                 ee = e[i];
             }
         }
         printf("%lld\n%lld %lld\n", ans, bb, ee);
     }

     return ;
 }