Largest Rectangle in a Histogram
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 15831 Accepted: 5121 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 nintegers 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 0Sample Output
8 4000Hint
Huge input, scanf is recommended.Source
这个题目看第一眼想到用dp来做，但是刚开始dp了一阵，没有啥思路，后来多设置两个数组就行了，还有学长给说了一种比较好的方法，用单调栈来做，其实就是优化dp。
dp的思路就是第 i 个柱子，它能形成的最大面积就是以它为高，尽量向两边扩展为宽的的面积，关键就是怎么往两边扩展，这时候就可以想到只要左边的柱子高度大于它的高度就行，所以向左找到第一个小于它的高度，同样向右找到第一个小于他的高度，左右所有的加起来乘以它的高度就是它的面积。所以遍历所有的柱子就可以了。问题就是如果一个一个的向左或者向右遍历时间复杂度是O(n2)， 所以这时候用到dp，就是如果它比左边的柱子要小的话，直接跳到左边柱子的left边界位置，然后，继续比较是否小于边界，如果小于继续迭代。反之就是答案。具体代码如下：
代码一(dp)：
/************************************************************************* > File Name: poj_2559.cpp > Author: Howe_Young > Mail: 1013410795@qq.com > Created Time: 2015年04月08日 星期三 09时00分34秒 ************************************************************************/ #include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> #include <cstdio> #define INF 99999999999999 using namespace std; typedef long long LL; const int N = 100005; LL h[N], l[N], r[N]; LL Max(LL a, LL b) { return a > b ? a : b; } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); int n; while (~scanf("%d", &n) && n) { memset(l, 0, sizeof(l)); memset(r, 0, sizeof(r)); for (int i = 1; i <= n; i++) scanf("%lld", &h[i]); l[1] = 0; h[0] = -1; r[n] = n + 1; h[n + 1] = -1; for (int i = 2; i <= n; i++)//找左边元素 { if (h[i] < h[i - 1]) { int tmp = l[i - 1]; while (h[i] <= h[tmp])//动态规划方法找，如果不用这中方法，普通的tmp--找的话会超时 tmp = l[tmp]; l[i] = tmp; } else if (h[i] == h[i - 1]) l[i] = l[i - 1]; else l[i] = i - 1; } for (int i = n - 1; i > 0; i--)//找右边 { if (h[i] < h[i + 1]) { int tmp = r[i + 1]; while (h[i] <= h[tmp]) tmp = r[tmp]; r[i] = tmp; } else if (h[i] == h[i + 1]) r[i] = r[i + 1]; else r[i] = i + 1; } LL ans = -1; for (int i = 1; i <= n; i++) { h[i] = (r[i] - l[i] - 1) * h[i]; ans = Max(ans, h[i]); } printf("%lld ", ans); } return 0; }
单调栈的思路是将这些柱子分别一个一个的判断，如果大于前面的那个那么前面比他大的就是0， 所以直接压栈，如果小于的话，弹栈，知道弹出小于它的为止，等于它，弹出来一个，压进去一个。
代码二(单调栈)：
/************************************************************************* > File Name: poj_2559_stack.cpp > Author: Howe_Young > Mail: 1013410795@qq.com > Created Time: 2015年04月08日 星期三 09时48分09秒 ************************************************************************/ #include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> #include <cstdio> #include <stack> using namespace std; typedef long long LL; const int N = 100005; LL h[N], r[N], l[N]; LL Max(LL a, LL b) { return a > b ? a : b; } int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int n; while (~scanf("%d", &n) && n) { memset(l, 0, sizeof(l)); memset(r, 0, sizeof(r)); for (int i = 1; i <= n; i++) scanf("%lld", &h[i]); stack<int> S;//找出左边的元素比他大的或者等于它的个数，单调栈 S.push(0);//把额外的一个点压进去，防止栈弹空 h[0] = h[n + 1] = -1; for (int i = 1; i <= n; i++) { if (h[i] < h[i - 1])//如果后者比前者小 { int cnt = 0; while (h[S.top()] >= h[i]) { l[i] += l[S.top()] + 1; S.pop(); } S.push(i); } else if (h[i] == h[i - 1]) { S.pop(); S.push(i); l[i] = l[i - 1] + 1; } else { l[i] = 0; S.push(i); } } stack<int> S2;//找处右边大于等于它的个数 S2.push(n + 1); for (int i = n; i > 0; i--) { if (h[i] < h[i + 1]) { int cnt = 0; while (h[S2.top()] >= h[i]) { r[i] += r[S2.top()] + 1; S2.pop(); } S2.push(i); } else if (h[i] == h[i + 1]) { S2.pop(); S2.push(i); r[i] = r[i + 1] + 1; } else { S2.push(i); r[i] = 0; } } LL ans = -1; for (int i = 1; i <= n; i++) { h[i] *= (l[i] + r[i] + 1); ans = Max(ans, h[i]); } printf("%lld ", ans); } return 0; }
过了一段时间又做了一遍。感觉代码比上面两个要好一点
代码三(单调栈)：
/************************************************************************* > File Name: largest.cpp > Author: Howe_Young > Mail: 1013410795@qq.com > Created Time: 2015年09月10日 星期四 18时54分01秒 ************************************************************************/ #include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> using namespace std; typedef long long ll; const int maxn = 100100; int a[maxn]; int L[maxn], R[maxn]; int stack[maxn];//单调递增栈 int main() { int n; while (~scanf("%d", &n) && n) { for (int i = 1; i <= n; i++) scanf("%d", &a[i]); a[0] = a[n + 1] = -1;//添加两个端点位置 int top = 0; stack[++top] = 0; for (int i = 1; i <= n; i++)//求出左边做远能扩展到的位置 { if (a[i] > a[i - 1]) { L[i] = i; stack[++top] = i; } else { while (a[stack[top]] >= a[i]) top--; L[i] = L[stack[top + 1]]; stack[++top] = i; } } top = 0; stack[++top] = n + 1; for (int i = n; i >= 1; i--)//右边 { if (a[i] > a[i + 1]) { R[i] = i; stack[++top] = i; } else { while (a[stack[top]] >= a[i]) top--; R[i] = R[stack[top + 1]]; stack[++top] = i; } } long long ans = 0; for (int i = 1; i <= n; i++) ans = max(ans, (long long)a[i] * (R[i] - L[i] + 1)); printf("%lld ", ans); } return 0; }