Largest Rectangle in a Histogram

 

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.

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 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.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

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我先提出两种做法，一种是dp统计向左向右扩展的最大距离

这个之前有个合唱队的

N位同学站成一排，音乐老师要请其中的(N-K)位同学出列，使得剩下的K位同学不交换位置就能排成合唱队形。 

合唱队形是指这样的一种队形：设K位同学从左到右依次编号为1, 2, …, K，他们的身高分别为T1, T2, …, TK，则他们的身高满足T1 < T2 < … < Ti , Ti > Ti+1 > … > TK (1 <= i <= K)。 

你的任务是，已知所有N位同学的身高，计算最少需要几位同学出列，可以使得剩下的同学排成合唱队形。 

Input输入的第一行是一个整数N（2 <= N <= 100），表示同学的总数。第一行有n个整数，用空格分隔，第i个整数Ti（130 <= Ti <= 230）是第i位同学的身高（厘米）。Output输出包括一行，这一行只包含一个整数，就是最少需要几位同学出列。Sample Input

8186 186 150 200 160 130 197 220

Sample Output

4

#include
     
      int main() {    int a[105],l[105],r[105],n;    scanf("%d",&n);    for(int i=1; i<=n; i++)        scanf("%d",&a[i]);    for(int i=1; i<=n; i++) {        l[i]=r[n-i+1]=1;        for(int j=1; j
      
       a[j]&&l[i]<=l[j])                l[i]=l[j]+1;            if(a[n-i+1]>a[n-j+1]&&r[n-i+1]<=r[n-j+1])                r[n-i+1]=r[n-j+1]+1;        }    }    int ans=0;    for(int i=1; i<=n; i++)        if(l[i]+r[i]>ans)ans=l[i]+r[i];    printf("%d\n",n-ans+1);    return 0;}
      
     

 

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这个直接这样也可以过的，无非就是乘了个高度和个数

单调栈水过

#include
     
      #include
      
       typedef __int64 LL;const int maxn = 100005;LL a[maxn],sta[maxn],left[maxn];int main(){    int N;    while (~scanf("%d",&N))    {        if(!N)break;        LL res = 0,tmp;        memset(sta,0,sizeof(sta));        memset(left,0,sizeof(left));        for (int i = 1; i <= N; i++)            scanf("%I64d",&a[i]);        a[++N] = -1;        int top = 0;        for (int i = 1; i <= N; i++)        {            if (!top||a[i]>a[sta[top-1]])            {                sta[top++] = i;                left[i] = i;                continue;            }            if (a[i] == a[sta[top-1]])            continue;            while (top > 0 && a[i] < a[sta[top-1]])            {                top--;                tmp = a[sta[top]]*((i-1)- (left[sta[top]]-1));                res = res