Largest Rectangle in a Histogram
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11137 Accepted Submission(s): 3047
Problem Description
A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles:
Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.
Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.
Input
The input contains several test cases. Each test case describes a histogram and starts with an integer n, denoting the number of rectangles it is composed of. You may assume that 1 <= n <= 100000. Then follow n integers h1, ..., hn, where 0 <= hi <= 1000000000. These numbers denote the heights of the rectangles of the histogram in left-to-right order. The width of each rectangle is 1. A zero follows the input for the last test case.
Output
For each test case output on a single line the area of the largest rectangle in the specified histogram. Remember that this rectangle must be aligned at the common base line.
Sample Input
7 2 1 4 5 1 3 3 4 1000 1000 1000 1000 0
Sample Output
8 4000
题意 求条形图中最大矩形的面积 输入给你条的个数 每个条的高度hi （以下等于也视为高）
只要知道第i个条左边连续多少个（a）比他高 右边连续多少个（b）比他高 那么以这个条为最大高度的面积就是hi*(a+b+1);
但是直接枚举每一个的话肯定会超时的 超时代码
#include
using namespace std;
const int N = 100005;
typedef long long ll;
ll h[N]; int n,wide[N];
int main()
{
while (scanf ("%d", &n), n)
{
for (int i = 1; i <= n; ++i)
scanf ("%I64d", &h[i]);
for (int i = 1; i <= n; ++i)
{
wide[i] = 1;
int k = i;
while (k > 1 && h[--k] >= h[i]) ++wide[i];
k = i;
while (k < n && h[++k] >= h[i]) ++wide[i];
}
ll ans = 0;
for (int i = 1; i <= n; ++i)
if (h[i]*wide[i] > ans) ans = h[i] * wide[i];
printf ("%I64d\n", ans);
}
return 0;
} 可以发现 当第i-1个比第i个高的时候 比第i-1个高的所有也一定比第i个高
于是可以用到动态规划的思想
令left[i]表示包括i在内比i高的连续序列中最左边一个的编号 right[i]为最右边一个的编号
那么有 当h[left[i]-1]>=h[i]]时 left[i]=left[left[i]-1] 从前往后可以递推出left[i]
同理 当h[right[i]+1]>=h[i]]时 right[i]=right[right[i]+1] 从后往前可递推出righ[i]
最后答案就等于 max((right[i]-left[i]+1)*h[i])了
#include
using namespace std;
const int N = 100005;
typedef long long ll;
ll h[N];
int n, left[N], right[N];
int main()
{
while (scanf ("%d", &n), n)
{
for (int i = 1; i <= n; ++i)
scanf ("%I64d", &h[i]), left[i] = right[i] = i;
h[0] = h[n + 1] = -1;
for (int i = 1; i <= n; ++i)
while (h[left[i] - 1] >= h[i])
left[i] = left[left[i] - 1];
for (int i = n; i >= 1; --i)
while (h[right[i] + 1] >= h[i])
right[i] = right[right[i] + 1];
ll ans = 0;
for (int i = 1; i <= n; ++i)
if (h[i] * (right[i] - left[i] + 1) > ans) ans = h[i] * ll (right[i] - left[i] + 1);
printf ("%I64d\n", ans);
}
return 0;
}