HDU 1506 largest rectangle in a histogram matrix series (1)

Largest rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)

Total submission (s): 7496 accepted submission (s): 2106 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
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. you may assume that 1 <=n <= 100000. then follow N integers H1 ,..., HN,
Where 0 & lt; = Hi & lt; = 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
Question meaning:
Give you n, and then tell you the height of the N segment, thus forming a Histogram
What is the area of the largest rectangle in your histogram?
Solution: for every height H [I], iteration to find its leftmost and rightmost for H [I], if it is big or equal to the left H [I-1], then his leftmost must be the leftmost of the I-1, and then the iteration continues PS: the iteration process is very fast
# Include <iostream> # include <cstdio> # define maxn 100005 using namespace STD; int n, m ;__ int64 H [maxn], le [maxn], RI [maxn]; __int64 ans; void left () // iteration to the left {int I, j; for (I = 1; I <= N; I ++) {le [I] = 1; while (H [I-le [I]> = H [I]) {le [I] + = Le [I-le [I] ;}} void right () // right iteration {int I, j; for (I = N; i> = 1; I --) {Ri [I] = 1; while (H [I + Ri [I]> = H [I]) {Ri [I] + = Ri [I + Ri [I] ;}} void solve () {int I; ans =-1; for (I = 1; I <= N; I ++) // update ans {If (le [I] + Ri [I]-1) * H [I]> ans) ans = (le [I] + Ri [I]-1) * H [I] ;}} int main () {int I, J; while (scanf ("% d", & N), n) {for (I = 1; I <= N; I ++) {scanf ("% i64d ", & H [I]);} H [0] = H [n + 1] =-1; // Add the boundary condition left (); Right (); solve (); printf ("% i64d \ n", ANS);} return 0 ;}