poj 2559 Largest Rectangle in a Histogram - 單調棧
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 19782 | Accepted: 6393 |
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:
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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
Hint
Huge input, scanf is recommended.
/*/ 大二這學期開學真是忙爆了。作為一個班干部，天哪。。 好久沒有刷題了，想起以前聽了島娘的一節課，單調棧，好像有點似懂非懂，于是抽時間看了一下，總算是搞通了。。。 將讀入的數據一個個壓棧，獎數據與棧頂進行比較大小，如果這個數比棧頂小，就計算以站頂為高的最大矩形的大小，彈出棧頂。如果這個數比棧頂大，直接就壓進棧。這樣一系列操作之后，就會發現棧里面剩下一個遞增數列，用一個pair來保存此時棧高度和前面達到這個高度的個數，然后按照前面的思想去計算每一個高度最大矩形面積是多大。
AC代碼： /*/
#include "stdio.h"
#include "string.h"
#include "stack"
#include "algorithm"
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MX = 1e5 + 5;
int main() {
LL n,h;
while(~scanf("%lld",&n)) {
if(n==0)break;
stack<PII> Q;
while(!Q.empty()) {
Q.pop();
}
LL ans=0;
for(LL i=0; i<n; i++) {
scanf("%lld",&h);
LL Now_Big_W=0;
while(!Q.empty() && Q.top().first >= h ) {
LL H = Q.top().first;
LL W = Q.top().second;
Q.pop();
Now_Big_W+=W;
ans=max(ans,H*Now_Big_W);
}
Q.push(PII(h,Now_Big_W+1));
}
LL The_number_W=0;
while(!Q.empty()){
The_number_W+=Q.top().second;
ans=max(ans,Q.top().first*The_number_W);
Q.pop();
}
printf("%lld\n",ans);
}
return 0;
}
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