POJ2823 线段树OR单调队列
题目描述：
Sliding Window
| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 16559 | Accepted: 4745 | |
| Case Time Limit: 5000MS | ||
Description
An array of size
n ≤ 10
6 is given to you. There is a sliding window of size
k which is moving from the very left of the array to the very right. You can only see the
knumbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and k is 3.
The array is [1 3 -1 -3 5 3 6 7], and k is 3.
| Window position | Minimum value | Maximum value |
|---|---|---|
| [1 3 -1] -3 5 3 6 7 | -1 | 3 |
| 1 [3 -1 -3] 5 3 6 7 | -3 | 3 |
| 1 3 [-1 -3 5] 3 6 7 | -3 | 5 |
| 1 3 -1 [-3 5 3] 6 7 | -3 | 5 |
| 1 3 -1 -3 [5 3 6] 7 | 3 | 6 |
| 1 3 -1 -3 5 [3 6 7] | 3 | 7 |
Your task is to determine the maximum and minimum values in the sliding window at each position.
Input
The input consists of two lines. The first line contains two integers
n and
k which are the lengths of the array and the sliding window. There are
n integers in the second line.
Output
There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.
Sample Input
8 3 1 3 -1 -3 5 3 6 7
Sample Output
-1 -3 -3 -3 3 3 3 3 5 5 6 7
题意很简单，就是求区间的最大值和最小值。
我用线段树过的，直接改的以前的代码。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef struct{
int left,right;
int Max,Min;
} node;
int tcase,n;
int maxx,minn;
node tree[1000001*4];
int a1[1000001],a2[1000001];
const int MAX = 999999999;
const int MIN = -99999999;
void build (int node,int left,int right)
{
tree[node].left=left;
tree[node].right=right;
tree[node].Max=MIN;
tree[node].Min=MAX;
if(left==right) return ;
int mid;
mid=(left+right)/2;
build(node*2,left,mid);
build(node*2+1,mid+1,right);
}
void update(int node,int pos,int val)
{
tree[node].Max=max(tree[node].Max,val);
tree[node].Min=min(tree[node].Min,val);
if(tree[node].left==pos&&tree[node].right==pos)
return;
int mid=(tree[node].left+tree[node].right)/2;
if(pos<=mid)
update(node*2,pos,val);
else if(pos>mid)
update(node*2+1,pos,val);
return;
}
int query_max(int node,int left,int right)
{
if (tree[node].left == left && tree[node].right == right)
return tree[node].Max ;
if (tree[node].left > right || tree[node].right < left)
return 0;
int mid;
mid=(tree[node].left + tree[node].right) / 2 ;
if (right <= mid)
return query_max(node * 2, left, right) ;
else if (mid < left)
return query_max(node * 2 + 1, left, right) ;
else
return max(query_max(node * 2, left, mid) ,query_max(node * 2 + 1, mid + 1, right)) ;
}
int query_min(int node,int left,int right)
{
if (tree[node].left == left && tree[node].right == right)
return tree[node].Min ;
if (tree[node].left > right || tree[node].right < left)
return 0;
int mid;
mid=(tree[node].left + tree[node].right) / 2 ;
if (right <= mid)
return query_min(node * 2, left, right) ;
else if (mid < left)
return query_min(node * 2 + 1, left, right) ;
else
return min(query_min(node * 2, left, mid) ,query_min(node * 2 + 1, mid + 1, right)) ;
}
int main()
{
int i,j,x,y,k;
scanf("%d %d",&n,&k);
build(1,1,n);
for(i=1;i<=n;i++)
{
scanf("%d",&x);
update(1,i,x);
}
int num=0;
for(i=1;i+k-1<=n;i++)
{
maxx=query_max(1,i,i+k-1);
minn=query_min(1,i,i+k-1);
a1[num]=maxx;
a2[num++]=minn;
}
for(i=0;i<num-1;i++)
printf("%d ",a2[i]);
printf("%d\n",a2[num-1]);
for(i=0;i<num-1;i++)
printf("%d ",a1[i]);
printf("%d\n",a1[num-1]);
return 0;
}
不过上网查了下 别人的代码，发现这题可以不用UPDATE函数，因为不需要数据的维护。
这题目还可以用单调队列。。。
http://blog.pureisle.net/archives/477.html
这个文章不错。