zoukankan      html  css  js  c++  java
  • HUD 1506 Largest Rectangle in a Histogram

    Largest Rectangle in a Histogram

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 19580    Accepted Submission(s): 5921


    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.
     
    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
     
    Source
     
    Recommend
    LL   |   We have carefully selected several similar problems for you:  1505 1069 1087 1058 1176 
     
    经典问题   广告印刷
    有n个数ai。从中选出一段区间[L,R],使得(R-L+1)*min{a_L,…,a_R}最大
     
    两个单调队列求每个点向两侧扩展到的最远距离
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    #define LL long long 
    const int maxn= 100005;
    #ifdef WIN32
    #define lld "I64d"
    #else
    #define lld "lld"
    #endif
    
    int a[maxn],b[maxn];
    int l[maxn],r[maxn],tmp[maxn],q[maxn];
    int n,m;
    void work(int c[] ,int d[]) {
        q[1]=c[1]; tmp[1]=1;
        int head=1,tail=1;
        for(int i=2;i<=n;++i) {
            while(head<=tail&&q[tail]>c[i]) d[tmp[tail--]]=i-1;
            q[++tail]=c[i];
            tmp[tail]=i;
           }
           while(head<=tail) d[tmp[head++]]=n;
    }
    int main() {
        while(scanf("%d",&n)&&n!=0) {
            LL ans=0;
            for(int i=1;i<=n;++i) 
                scanf("%d",a+i),b[n-i+1]=a[i];
            work(a,r);     
            work(b,l);
            for(int i=1;i<=n;++i) tmp[i]=l[i];
            for(int i=1;i<=n;++i) l[n-i+1]=n-tmp[i]+1;
            for(int i=1;i<=n;++i) ans=max(ans,1ll*a[i]*(r[i]-l[i]+1));
            printf("%lld
    ",ans);
        }
        return 0;
    }
     
  • 相关阅读:
    DirectX:在graph自己主动连线中增加自己定义filter(graph中遍历filter)
    C3P0数据库连接池使用
    POJ
    【jQuery】复选框的全选、反选,推断哪些复选框被选中
    BestCoder Round #75 King&#39;s Cake 模拟&amp;&amp;优化 || gcd
    《Javascript_Dom 编程艺术》(第2版)读书笔记
    POJ 2947-Widget Factory(高斯消元解同余方程式)
    MFC 小知识总结四
    迭代器和iter()函数
    hdu1595find the longest of the shortest 最短路
  • 原文地址:https://www.cnblogs.com/sssy/p/7689274.html
Copyright © 2011-2022 走看看