Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15013 Accepted Submission(s): 4357
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.
题解：让求最大矩形面积，宽为1，暴力超时
可以发现 当第i-1个比第i个高的时候 比第i-1个高的所有也一定比第i个高
于是可以用到动态规划的思想
令left[i]表示包括i在内比i高的连续序列中最左边一个的编号 right[i] 为最右边一个的编号
那么有 当 h[left[i]-1]>=h[i]]时 left[i]=left[left[i]-1] 从前往后可以递推出left[i]
同理 当 h[right[i]+1]>=h[i]]时 right[i]=right[right[i]+1] 从后往前可递推出righ[i]
最后答案就等于 max((right[i]-left[i]+1)*h[i]) 了；
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x,y) scanf("%lf%lf",&x,&y) #define P_ printf(" ") const int MAXN=100010; typedef long long LL; LL a[MAXN]; int l[MAXN],r[MAXN]; int main(){ int N; while(scanf("%d",&N),N){ for(int i=1;i<=N;i++)scanf("%lld",&a[i]),l[i]=i,r[i]=i; a[0]=a[N+1]=-1; for(int i=1;i<=N;i++){ while(a[l[i]-1]>=a[i]) l[i]=l[l[i]-1]; } for(int i=N;i>=1;i--){ while(a[r[i]+1]>=a[i]) r[i]=r[r[i]+1]; } LL ans=0; for(int i=1;i<=N;i++){ ans=max(ans,(r[i]-l[i]+1)*a[i]); } printf("%lld\n",ans); } return 0; }
Largest Rectangle in a Histogram（最大矩形面积，动态规划思想）
原文：http://www.cnblogs.com/handsomecui/p/5203681.html