hdoj 1506 Largest Rectangle in a Histogram 【单调栈】
2016-05-02 21:51
316 查看
题目链接:hdoj 1506 Largest Rectangle in a Histogram
Largest Rectangle in a Histogram
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15601 Accepted Submission(s): 4542
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 3
4 1000 1000 1000 1000
0
Sample Output
8
4000
题意:有n根柱子,要求你找到一个最大的矩形覆盖柱子(覆盖处不能为空),问你最大的矩形面积。
思路:用单调栈求出第i根柱子可以向左、向右延伸的最大长度。维护最大值就好了。
AC代码:
Largest Rectangle in a Histogram
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15601 Accepted Submission(s): 4542
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 3
4 1000 1000 1000 1000
0
Sample Output
8
4000
题意:有n根柱子,要求你找到一个最大的矩形覆盖柱子(覆盖处不能为空),问你最大的矩形面积。
思路:用单调栈求出第i根柱子可以向左、向右延伸的最大长度。维护最大值就好了。
AC代码:
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #define CLR(a, b) memset(a, (b), sizeof(a)) #define fi first #define se second using namespace std; typedef long long LL; typedef pair<int, int> pii; const int MAXN = 1e5 +10; const int INF = 0x3f3f3f3f; pii a[MAXN], Stack[MAXN]; int main() { int n; while(scanf("%d", &n), n) { LL ans = 0, top = 0; int l, r = 0; for(int i = 1; i <= n; i++) { scanf("%d", &a[i].fi); a[i].se = i; r = 0; while(top && Stack[top-1].fi >= a[i].fi) { if(r == 0) r = Stack[top-1].se; if(top == 1) { l = Stack[top-1].se; } else { l = Stack[top-1].se - Stack[top-2].se; } ans = max(ans, 1LL * (l + r - Stack[top-1].se) * Stack[top-1].fi); top--; } Stack[top++] = a[i]; } r = 0; while(top) { if(r == 0) r = Stack[top-1].se; if(top == 1) { l = Stack[top-1].se; } else { l = Stack[top-1].se - Stack[top-2].se; } ans = max(ans, 1LL * (l + r - Stack[top-1].se) * Stack[top-1].fi); top--; } printf("%lld\n", ans); } return 0; }
相关文章推荐
- malloc和free的实现原理
- Poj 3250 Bad Hair Day 【单调栈】
- elasticsearch自定义分析器
- 1.5linux安装redis
- 欢迎使用CSDN-markdown编辑器
- neural artistic style艺术风格图片生成测试
- google base库中的WaitableEvent
- 《构建之法》第六七章读后感
- Nginx 之四: Nginx服务器的rewrite、全局变量、重定向和防盗链相关功能
- 生成函数
- Linux系统资源的相关命令
- DLNA 笔记
- 电商大数据学习笔记:用户画像
- 常见布局修复方案—外边距叠加问题
- C++ 结构体指针的定义
- js prototype
- 【leetcode】226. Invert Binary Tree
- WPS中公式与文字无法居中对齐
- 百练+归并排序求逆序数+注意最后是按逆序数大小输出原来的序列啊!
- 编译器的工作过程