POJ 2796 Feel Good(单调栈)

题目链接：

POJ 2796 Feel Good

题意：

一个区间的权值是区间元素之和乘以区间最小的元素值。给一个长度为n的数组，求最大子区间权值。

数据范围:n≤106,data[i]∈[0,106]

分析：

考虑到每个数非负，只需要利用单调栈处理出每个数向左向右是最小值的最大区间就好了。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <climits>
using namespace std;
typedef long long ll;
const int MAX_N = 100010;

int n, top;
int data[MAX_N], sta[MAX_N], L[MAX_N], R[MAX_N];
ll sum[MAX_N];

int main()
{
while (~scanf("%d", &n)) {
sum[0] = 0;
for (int i = 1; i <= n; ++i) {
scanf("%d", &data[i]);
sum[i] = sum[i - 1] + data[i];
}
top = 0;
//单调非递减栈
data[n + 1] = -1;
for (int i = 1; i <= n + 1; ++i) {
while (top && data[sta[top]] > data[i]) {
R[sta[top]] = i - 1;
--top;
}
if(top == 0) L[i] = 1;
else L[i] = sta[top] + 1;
sta[++top] = i;
}
ll ans = -1;
int left, right;
for (int i = 1; i <= n; ++i) {
ll tmp = (sum[R[i]] - sum[L[i] - 1]) * data[i];
if (tmp > ans) {
ans = tmp;
left = L[i], right = R[i];
}
}
printf("%lld\n", ans);
printf("%d %d\n", left, right);
}
return 0;
}