UVa 1619：Feel Good（单调栈）

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=845&page=show_problem&problem=4494

题意：给出一个长度为n(n≤100000)(n \le 100000)的正整数序列aia_i，求出一段连续子序列al,...,ara_l,...,a_r，使得(al+...+ar)∗min{al,...,ar}(a_l + ... +a_r)*min\{a_l,...,a_r\}尽量大。（本段摘自《算法竞赛入门经典（第2版）》）

分析：枚举每一个数，利用单调栈计算以它为最小值向左向右最远能够延伸到多远，每次更新答案即可。（注意：UVa这题貌似比较坑，题目中说如果有多组解，输出任意组即可，但是貌似不是的。具体输出看代码。）

代码：

#include <iostream>
#include <fstream>
#include <cstring>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>

using namespace std;

const int maxn = 1000000 + 5;

int n, L, R;
int l[maxn], r[maxn];
bool first;
long long ans, tmp;
long long a[maxn], sum[maxn];

int main()
{
while (~scanf("%d", &n))
{
if (first)
printf("\n");
else
first = true;
for (int i = 1; i <= n; ++i)
{
scanf("%lld", &a[i]);
sum[i] = sum[i - 1] + a[i];
}
a[0] = -1;
l[0] = 0;
stack< int > sta1;
sta1.push(0);
for (int i = 1; i <= n; ++i)
{
while (a[sta1.top()] >= a[i])
sta1.pop();
l[i] = sta1.top() + 1;
sta1.push(i);
}
a[n + 1] = -1;
r[n + 1] = n + 1;
stack< int > sta2;
sta2.push(n + 1);
for (int i = n; i >= 1; --i)
{
while (a[sta2.top()] >= a[i])
sta2.pop();
r[i] = sta2.top() - 1;
sta2.push(i);
}
L = 1;
R = 1;
ans = 0;
for (int i = 1; i <= n; ++i)
{
tmp = (sum[r[i]] - sum[l[i] - 1]) * a[i];
if (tmp > ans || (tmp == ans && R - L > r[i] - l[i]))
{
L = l[i];
R = r[i];
ans = tmp;
}
}
printf("%lld\n%d %d\n", ans, L, R);
}
return 0;
}