HDU 1506 Largest Rectangle in a Histogram(最大长方形) - c++语言程序开发技术文章_c++编程 - 红黑联盟


这就是前几篇已经提到的求最大长方形那道题。
解题思路:
如果每次都向前向后扫描,有时会重复扫描多次,导致超时。
我们可以用一个单调栈 (类似单调队列) 由低到高来存储它的高度,并用数组对每个高度记录一下它前面一共有多少个比它高的,可以看做它的左宽。
按顺序考虑每个高度h,如果h大于栈顶元素,则入栈,此时它大于左面全部的元素(下文会提到为什么可以这样认为),所以将它的宽度初始为1。
否则,将栈内元素出栈,直到满足上面的条件。出栈时,我们要将出栈元素对问题的影响全部考虑进行处理,才能保证做法的正确性。
对于每个高度,它的作用无非两个:1、以自己作高,向外扩展        2、以别人作高,自己被扩展
【当以自己作高完毕以后,它的高度就没有什么意义了,为了方便,我们可以把它的高度看做与之后栈顶元素相等。
所以之后的栈顶元素可以看做是大于左面全部的元素。】此举只为容易理解,如不懂请跳过。
由于我们数组中已经记录了某个高度的左宽,所以我们只需要考虑它能不能向右扩展,如果能,能扩展多少?
首先,对于第一个出栈的元素来说,它的右宽一定是0。
然而对于第二个,它的右边有刚才出栈的元素,而且刚才出栈元素的总宽中所涉及的元素一定可以被自己扩展,所以自己的右宽为刚才出栈元素的总宽。 www.2cto.com
同理可知,第三个出栈元素的右宽为第二个出栈元素的总宽。依次类推。
而当h大于栈顶元素时,h的左宽应该是上次出栈元素的总宽+1(自己),然后入栈。
最后时,将所有元素出栈,即可将所有情况考虑。
int q[N]={-1},w[N];     //w记录的是从这个点开始,之前有几个高度大于等于此高度. 
int main() 

    int n,h; 
    while(scanf("%d",&n),n) 
    { 
        int top = 0; 
        __int64 ans = 0; 
        for(int i=1;i<=n+1;i++) 
        { 
            if(i != n+1) 
                scanf("%d",&h); 
            else 
                h = 0; 
            if(h > q[top]) 
                q[++top] = h , w[top] = 1; 
            else 
            { 
                __int64 cnt = 0; 
                while(h <= q[top]) 
                { 
                    ans = max(ans ,(cnt+w[top])*q[top] ); 
                    cnt += w[top--]; 
                } 
                q[++top] = h; 
                w[top] = cnt+1; 
            } 
        } 
        printf("%I64d\n",ans); 
    } 
    return 0; 

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