POJ_2559_Largest Rectangle in a Histogram(栈)

Largest Rectangle in a Histogram

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17105 Accepted: 5531

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.

Input

The input contains several test cases. Each test case describes a histogram and starts with an integern, denoting the number of rectangles it is composed of. You may assume that1<=n<=100000. Then follow n integers h₁,...,h_n, where0<=h_i<=1000000000. These numbers denote the heights of the rectangles of the histogram in left-to-right order. The width of each rectangle is1. 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 34 1000 1000 1000 10000

Sample Output

84000

Hint

Huge input, scanf is recommended.

题意：柱状图是由一些宽度相等的长方形下端对齐后横向排列得到的图形。现在有由n个宽度为1，高度分别为h1，h2，……，hn的长方形从左到右依次排列组成的柱状图。问里面包含的长方形的最大面积是多少。

分析：如果我们能够预处理出以每个高度为高的长方形的左端点le[i]以及右端点ri[i]，那么就可以用O(n)的复杂度扫一遍就可以得到最大值。如何预处理出le[i]以及ri[i]呢？这里就要运用栈的知识点了。先考虑le[i]的情况，定义一个栈head，并且初始化为空。然后不断增加i的值，并且维护这个栈使得它按照下面得顺序存储用于推算后面的le值得元素：

设栈中元素从上到下的值为X(i)，则X(i) > X(i+1)且h(X(i)) > h(X(i+1))。

计算ri[i]的时候按照同样的方法求即可。

题目链接：http://poj.org/problem?id=2559

代码清单：

#include<set>#include<map>#include<cmath>#include<queue>#include<stack>#include<ctime>#include<cstdio>#include<string>#include<cctype>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;typedef unsigned int uint;typedef unsigned long long ull;const int maxn = 100000 + 5;int n;ll h[maxn];ll le[maxn],ri[maxn];stack<int>head;stack<int>tail;void init(){    while(!head.empty()) head.pop();    while(!tail.empty()) tail.pop();    memset(le,0,sizeof(le));    memset(ri,0,sizeof(ri));}void input(){    for(int i=1;i<=n;i++)        scanf("%I64d",&h[i]);}ll work(){    for(int i=1;i<=n;i++){        while(!head.empty()&&h[head.top()]>=h[i]) head.pop();        if(head.empty()) le[i]=1;        else le[i]=head.top()+1;        head.push(i);    }    for(int i=n;i>=1;i--){        while(!tail.empty()&&h[tail.top()]>=h[i]) tail.pop();        if(tail.empty()) ri[i]=n;        else ri[i]=tail.top()-1;        tail.push(i);    }    ll res=-1;    for(int i=1;i<=n;i++){        res=max(res,h[i]*(ri[i]-le[i]+1));    }    return res;}void solve(){    printf("%I64d\n",work());}int main(){    while(scanf("%d",&n)!=EOF&&n){        init();        input();        solve();    }return 0;}