POJ 2559 Largest Rectangle in a Histogram 解题报告 单调栈

POJ 2559 Largest Rectangle in a Histogram 解题报告 单调栈

解题思路：两种方法，一种是dp，一种是单调栈。
1.dp。超时了，思路就是不断更新第i个矩形左右两边大于等于第i个矩形高度的边缘矩阵的下标。

#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<string.h>
#include<map>
#include<stack>
#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<fstream>
#include<sstream>
#include<set>
#pragma warning(disable:4996)
#define INF 0x3f3f3f3f
#define ll long long
#define PI acos(-1.0)
const int N = 100005;
const int maxn = 1e9;
using namespace std;
int n, k, t;
ll h[100005],l[100005],r[100005];
int main()
{
	while (scanf("%d", &n) != EOF && n)
	{
		for (int i = 1; i <= n; i++)
		{
			scanf("%lld", &h[i]);
			l[i] = i;
			r[i] = i;
		}
		for (int i = 1; i <= n; i++)
		{
			while (h[l[i] - 1] >= h[i])
			{
				l[i] = l[l[i] - 1];
			}
		}
		for (int i = n; i >= 1; i--)
		{
			while (h[r[i] + 1] >= h[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) * h[i]);
		}
		printf("%lld\n", ans);
	}
}

2.单调栈。就是笛卡尔树一边建一边拆，或者说是简化。代码里加了注释。

#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<string.h>
#include<map>
#include<stack>
#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<fstream>
#include<sstream>
#include<set>
#pragma warning(disable:4996)
#define INF 0x3f3f3f3f
#define ll long long
#define PI acos(-1.0)
const int N = 100005;
const int maxn = 1e9;
using namespace std;
int n, k, t;
int a[N],b[N];
int main()
{
	while (scanf("%d", &n) != EOF && n)
	{
		int top = 0;
		ll ans = 0;
		int tem;
		a[0] = -1;
		for (int i = 1; i <= n + 1; i++)
		{
			if (i != n + 1)
				scanf("%d", &tem);
			else
				tem = 0;
			if (tem > a[top])
			{
				++top;
				a[top] = tem;
				b[top] = 1;//记录高度达到tem的连续矩形数量
			}
			else
			{
				ll cnt = 0;
				while (tem <= a[top])
				{
					ans = max(ans, (cnt + b[top]) * a[top]);//如果tem比某个节点小，更新ans后直接舍弃这个节点,因为新节点比它小，原节点用不到了
					cnt += b[top];//但是旧节点也要计数，因为比新节点高
					top--;
				}
				++top;
				a[top] = tem;
				b[top] = cnt + 1;
			}
		}
		printf("%lld\n", ans);
	}
}