2、題目大意：

給出n個矩形並排，給出他們的高度，求可以在此基礎上找出一個面積最大且底邊在基線上的矩形，輸出最大面積

一開始用的區間DP，樣例都過了，但發現dp[N][N]超記憶體了，看網上程式碼，居然是用結構體或兩個一維陣列表示的

3、題目：

Largest Rectangle in a Histogram

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 13118	Accepted: 4238

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,...,h_n, where 0<=h_i<=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.

Source

4、AC程式碼：

#include<stdio.h>
#include<math.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100005
struct node
{
    ll num;
    ll left;
    ll right;
} a[N];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        for(int i=1; i<=n; i++)
        {
            scanf("%lld",&a[i].num);
            a[i].left=i;
            a[i].right=i;
        }
        a[0].num=-1;
        a[n+1].num=-1;
        for(int i=1; i<=n; i++)
        {
            while(a[i].num<=a[a[i].left-1].num)
                a[i].left=a[a[i].left-1].left;
        }
        for(int i=n; i>=1; i--)
        {
            while(a[i].num<=a[a[i].right+1].num)
                a[i].right=a[a[i].right+1].right;
        }
        ll tmp=-1;
        for(int i=1; i<=n; i++)
        {
            if(a[i].num*(a[i].right-a[i].left+1)>tmp)
                tmp=a[i].num*(a[i].right-a[i].left+1);
        }
        printf("%lld\n",tmp);
    }
    return 0;
}

區間DP做的，超記憶體的程式碼：

#include<stdio.h>
#include<math.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 100005
#define INF 1000000005
int a[N];
int dp[22005][22005];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        break;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            dp[i][i]=a[i];
        }
        for(int i=2;i<=n;i++)
        {
            for(int j=1;j+i-1<=n;j++)
            {
                int s=j;
                int e=j+i-1;
                int minn=INF;
                for(int k=s;k<=e;k++)
                {
                    if(a[k]<minn)
                    minn=a[k];
                }
                dp[s][e]=max(dp[s][e],dp[s+1][e]);
                dp[s][e]=max(dp[s][e],(e-s+1)*minn);
                dp[s][e]=max(dp[s][e],dp[s][e-1]);
            }
        }
        printf("%d\n",dp[1][n]);
    }
    return 0;
}