<span>Largest Rectangle in a Histogram HDU - 1506 （单调栈）</span>
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 3 4 1000 1000 1000 1000 0
Sample Output
8 4000
首先考虑最大面积的矩形X的左右边界的性质：
设其左边界为L，右边界为R，则其高H = min{h[i] | L <= i <= R}
此时最大面积为 (R - L + 1) * H
若此时左边界的左边那个矩形的高度 h[L-1] >= H
则左边界可以向左拓展，则新的面积为：
(R - (L-1) + 1) * H > 原面积
则与原假设条件冲突
故左边界左边的那个矩形的高度 :h[L-1] < H
同理右边界右边的那个矩形的高度： h[R+1] < H
设H = h[i]
所以左边界L是满足h[j-1] < h[i]的最大的j，即从i点向左遍历的第一个高度比i小的点的右边一个点
而右边界R是满足 h[j+1] < h[i]的最小的j，即从i点向右遍历第一个高度比i小的点的左边一个点
所以我们可以利用单调栈的性质得到每个确定点，即确定高度的最大面积矩形的左右边界，然后枚举取最大即可。
取自原文：https://blog.csdn.net/wubaizhe/article/details/70136174
细节见我的AC代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define db(x) cout<<"== [ "<<x<<" ] =="<<endl; using namespace std; typedef long long ll; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;} inline void getInt(int* p); const int maxn=1000010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ int n; ll h[maxn]; int l[maxn]; int r[maxn]; stack<int> s; int main() { while(~scanf("%d",&n)) { if(n==0) { break; } repd(i,1,n) { scanf("%I64d",&h[i]); } // 1 2 5 7 // while(!s.empty()) { s.pop(); } repd(i,1,n) { while(s.size()&&h[s.top()]>=h[i]) { s.pop(); } if(s.empty()) { l[i]=1; }else { l[i]=s.top()+1; } s.push(i); } while(!s.empty()) { s.pop(); } // 1 2 5 7 for(int i=n;i>=1;i--) { while(s.size()&&h[s.top()]>=h[i]) { s.pop(); } if(s.empty()) { r[i]=n+1; }else { r[i]=s.top(); } s.push(i); } ll ans=0ll; repd(i,1,n) { ans=max(ans,1ll*h[i]*(r[i]-l[i])); } printf("%I64d\n",ans ); } return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ' ' || ch == '\n'); if (ch == '-') { *p = -(getchar() - '0'); while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 - ch + '0'; } } else { *p = ch - '0'; while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 + ch - '0'; } } }
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