hdu1506—Largest Rectangle in a Histogram（dp+单调栈）

题目链接：传送门

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18618    Accepted Submission(s): 5563

Problem 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 34 1000 1000 1000 10000

 

Sample Output

84000

题目大意：在给出的图中所能画出矩形的最大面积

解题思路：因为要求的是矩形的面积，则矩形的高只能是某个柱子的高h，现在要求的就是矩形的长度。假设第i个柱子左侧连续的柱子中第一个比该它高的柱子的序号为r，第i个柱子右侧连续的柱子中第一个比它高的柱子的序号为l，则长度L = l-r+1。矩形面积为s = h*L。然后用单调栈能在o(n)的时间内求出r和l。

不太懂单调栈的可以看下这篇博客：传送门

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <stack>#include <queue>#include <algorithm>#include <cmath>using namespace std;typedef long long ll;typedef pair<int,ll>PA;const int N = 100009;const int M = 30009;const int INF = 0x3fffffff;stack<PA>st;ll data[N],L[N],R[N];//L[i]:第i个数左侧连续的数中第一个比data[i]大的数//R[i]:第i个数右侧连续的数中第一个比data[i]大的数int main(){    int n;    while( ~scanf("%d",&n)&&n ){        for( int i = 1 ; i <= n ; ++i ){            scanf("%d",&data[i]);        }        while( !st.empty() ) st.pop();        L[1] = 1; st.push(PA(1,data[1]));        for( int i = 2 ; i <= n ; ++i ){            int flag = 0;            while( !st.empty() ){                PA s = st.top();                //找到左边连续的序列中第一个比data[i]大的数                if( s.second < data[i] ){                    L[i] = s.first+1;                    flag = 1;                    st.push(PA(i,data[i]));                    break;                }else{                    L[i] = L[s.first];                    st.pop();                }            }            if( !flag ) st.push(PA(i,data[i]));        }        while( !st.empty() ) st.pop();        R[n] = n; st.push(PA(n,data[n]));        for( int i = n-1 ; i >= 1 ; --i ){            int flag = 0;            while( !st.empty() ){                PA s = st.top();                //找到右边连续的序列中第一个比data[i]大的数                if( s.second < data[i] ){                    R[i] = s.first-1;                    flag = 1;                    st.push(PA(i,data[i]));                    break;                }else{                    R[i] = R[s.first];                    st.pop();                }            }            if( !flag ) st.push(PA(i,data[i]));        }        ll ans = 0;        for( int i = 1 ; i <= n ; ++i ){            ans = max( ans , (R[i]-L[i]+1)*data[i] );        }        printf("%lld\n",ans);    }    return 0;}