Sliding Window
| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 32563 | Accepted: 9681 | |
| 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
k numbers 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.
| [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：
有N（1到10^6）个数，每K（1到N）个数一组。求出每组的最小值和最大值，最小值和最大值分别一行输出。
思路：
线段树基础题。线段树维护最大值最小值。
AC：
#include<stdio.h>
#define MAX 1000005
typedef struct
{
int l;
int r;
int min;
int max;
}node;
node no[MAX*5];
int num[MAX];
int fma,fmi;
void build(int from,int to,int i)
{
int mid=(from+to)/2;
no[i].l=from;
no[i].r=to;
if(from==to)
{
no[i].max=num[from];
no[i].min=num[from];
return;
}
build(from,mid,2*i);
build(mid+1,to,2*i+1);
no[i].max=(no[2*i].max>no[2*i+1].max?no[2*i].max:no[2*i+1].max);
no[i].min=(no[2*i].min<no[2*i+1].min?no[2*i].min:no[i*2+1].min);
}
void findmax(int from,int to,int i)
{
int mid=(no[i].l+no[i].r)/2;
if(from==no[i].l&&to==no[i].r)
{
fma=(fma>no[i].max?fma:no[i].max);
return;
}
if(from>=mid+1) findmax(from,to,2*i+1);
if(to<=mid) findmax(from,to,2*i);
if(from<=mid&&to>=mid+1)
{
findmax(from,mid,2*i);
findmax(mid+1,to,2*i+1);
}
}
void findmin(int from,int to,int i)
{
int mid=(no[i].l+no[i].r)/2;
if(from==no[i].l&&to==no[i].r)
{
fmi=(fmi>no[i].min?no[i].min:fmi);
return;
}
if(from>=mid+1) findmin(from,to,2*i+1);
if(to<=mid) findmin(from,to,2*i);
if(from<=mid&&to>=mid+1)
{
findmin(from,mid,2*i);
findmin(mid+1,to,2*i+1);
}
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&num[i]);
build(1,n,1);
for(int i=1;i<=n-m+1;i++)
{
fmi=num[i];
//WA了N遍，就是因为赋值这里
//一开始赋值fmi=MAX；
//因为fmin是int型的，存MAX不够存
findmin(i,i+m-1,1);
printf("%d",fmi);
i==n-m+1?printf("\n"):printf(" ");
}
for(int i=1;i<=n-m+1;i++)
{
fma=num[i];
//同理，这里也不可以赋值为fma=-MAX
findmax(i,i+m-1,1);
printf("%d",fma);
i==n-m+1?printf("\n"):printf(" ");
}
return 0;
}
总结：
1.细节的错误现在遇到得比较多，都是不容易发现的错误；
2.不是多输入和输出格式的错误，赋值的时候要注意要在数据范围内赋值。