動態規劃-Largest Rectangle in a Histogram- HOJ 1506

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7296    Accepted Submission(s): 2037


Problem DescriptionA 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. 

InputThe 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. 

OutputFor 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 34 1000 1000 1000 10000

 

Sample Output

84000

 

還有一個類似的題。

動態規劃-面積最大的全1子矩陣

關鍵點都是遍曆兩次，找到左界限和右界限。

#include <stdio.h>typedef long long int64;int64 n;int64 arr[100002];int64 left[100002], right[100002];int main() {//freopen("input.txt", "r", stdin);while (scanf("%I64d",&n), n) {for (int i = 1; i <= n; i++) {scanf("%I64d",&arr[i]);}arr[0]=-1;arr[n+1]=-1;int64 tmp;for (int i = 1; i <= n; i++) { //求左界限tmp = i;while(arr[i] <= arr[tmp-1]) {tmp = left[tmp-1];}left[i] = tmp;}int64 ans = 0,max;for(int i=n; i>=1; i--) { //求右界限，同時求最大值tmp=i;//while(arr[i] <= arr[tmp+1]) {tmp = right[tmp+1];}right[i] = tmp;max = (tmp-left[i] + 1) * arr[i];if(max > ans)ans = max;}printf("%I64d\n",ans);}return 0;}