HUST 1630 largest rectangle（思维题）

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本文介绍了一种寻找直方图中最大矩形面积的有效算法。通过枚举每个柱子作为最低点，计算其左右两侧高于等于该点的柱子数量，并利用前一次计算结果加速，实现了接近O(n)的时间复杂度。

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largest rectangle

Time Limit: 1 Sec Memory Limit: 128 MB
Submissions: 11 Solved: 4

Description

Given n positive integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height =[2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

Input

　　The first line of the input file contains an integer T (T<=100) specifying the number of test cases.
Each test case begins with a line containing an integer N (1<=N<=100000),The following N-1 lines each contain a integer A[i](1<=A[i]<=10000).

Output

For each query, output a single line containing largest area.

Sample Input

Sample Output

10
30

HINT

Source

The 8th(2013) ACM Programming Contest of HUST

直接枚举最低点，分别枚举最低点的位置，也就是以每一个柱子为最低点进行枚举，算左边有多少高于等于它的，右边同理

计算的时候利用上一次的结果进行跳转，接近O（n）

#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn = 110000;
int rec[maxn],n;
int l[maxn],r[maxn];
int main(){
    int i,j,k,t,now,ans;
    scanf("%d",&t);
    while(t--){
        ans=-1;
        scanf("%d",&n);
        for(i=1;i<=n;i++) scanf("%d",&rec[i]);
        memset(l,0,sizeof(l));
        memset(r,0,sizeof(r));
        for(i=2;i<=n;i++){
            now=i-1;
            while(now>0 && rec[i]<=rec[now]){
                l[i]+=l[now]+1;
                now-=l[now];
                now--;
            }
        }
        for(i=n-1;i>0;i--){
            now=i+1;
            while(now<=n && rec[i]<=rec[now]){
                r[i]+=r[now]+1;
                now+=r[now];
                now++;
            }
        }
        for(i=1;i<=n;i++)
        ans=max(ans,(l[i]+r[i]+1)*rec[i]);
        printf("%d\n",ans);
    }
    return 0;
}