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);
}
}