zoukankan      html  css  js  c++  java
  • POJ2559——DP——Largest Rectangle in a Histogram

    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
    

    HINT

    Huge input, scanf is recommended.

    #include<cstdio>
    using namespace std;
    const int MAX = 1000000;
    int h[MAX],L[MAX],R[MAX];
    int stack[MAX];
    int n;
    void solve()
    {
      int n;
      int t = 0;
        for(int i = 0 ; i < n; ++i){
            while(t > 0 && h[stack[t-1]] >= h[i]) t--;
            L[i] = ( t == 0) ? 0: stack[t-1];
            stack[t++] = i;
        }
        t = 0;
        for(int i = n - 1 ; i >= 0 ;--i){
            while(t > 0 && h[stack[t-1]] >= h[i]) t--;
            R[i] = (t == 0) ? n : stack[t-1];
           stack[t++] = i;
        }
        long long max1 = 0;
       for(int i = 0 ; i < n ;i++){
          max1 = max(max1,(long long)h[i]*(R[i] - L[i] + 1));
       }
      printf("%lld
    ",max1);
    }
    int main()
    {
        while(~scanf("%d",&n)&&n){
            for(int i = 0 ; i < n; i++)
                scanf("%d",&h[i]);
            solve();
        }
        return 0;
    }
    View Code

    利用

    L[i] = (j <= i并且h[j-1] < h[i]的最大的j)

    R[i] = (j > i并且h[j] > h[i]的最小的j)

  • 相关阅读:
    cocostudio 使用教程
    anrdroid AVD启动不起来的问题。Waiting for HOME ('android.process.acore') to be launched
    Android SDK无法更新的问题解决办法
    Code(容斥,好题)
    莫比乌斯反演(转)
    随笔--新建查询
    11427
    uva11722
    uva11021
    How many integers can you find(容斥+dfs容斥)
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4418321.html
Copyright © 2011-2022 走看看