poj 2823 Sliding Window(单调队列模板)
2016-11-14 20:43
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还有这题输出用printf G++交会T,C++就能过.....而用cout两个都能过..好坑....
代码:
#include<cstdio>
using namespace std;
const int maxn = 1e6+5;
int n, k, ansMin[maxn], ansMax[maxn], queMax[maxn], queMin[maxn], val[maxn];
int main(void)
{
while(~scanf("%d%d", &n, &k))
{
int headMin = 0, tailMin = 0, headMax = 0, tailMax = 0;
for(int i = 0; i < n; i++)
{
while(headMin < tailMin && queMin[headMin] <= i-k) headMin++;
while(headMax < tailMax && queMax[headMax] <= i-k) headMax++;
scanf("%d", &val[i]);
while(headMin < tailMin && val[queMin[tailMin-1]] >= val[i]) tailMin--; tailMin++;
while(headMax < tailMax && val[queMax[tailMax-1]] <= val[i]) tailMax--; tailMax++;
queMin[tailMin-1] = queMax[tailMax-1] = i;
ansMin[i] = val[queMin[headMin]];
ansMax[i] = val[queMax[headMax]];
}
for(int i = k-1; i < n; i++) printf("%d%c", ansMin[i], i==n-1 ? '\n' : ' ');
for(int i = k-1; i < n; i++) printf("%d%c", ansMax[i], i==n-1 ? '\n' : ' ');
}
return 0;
}
Sliding Window
Description
An array of size n ≤ 106 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.
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
Sample Output
Source
POJ Monthly--2006.04.28, Ikki
还有这题输出用printf G++交会T,C++就能过.....而用cout两个都能过..好坑....
代码:
#include<cstdio>
using namespace std;
const int maxn = 1e6+5;
int n, k, ansMin[maxn], ansMax[maxn], queMax[maxn], queMin[maxn], val[maxn];
int main(void)
{
while(~scanf("%d%d", &n, &k))
{
int headMin = 0, tailMin = 0, headMax = 0, tailMax = 0;
for(int i = 0; i < n; i++)
{
while(headMin < tailMin && queMin[headMin] <= i-k) headMin++;
while(headMax < tailMax && queMax[headMax] <= i-k) headMax++;
scanf("%d", &val[i]);
while(headMin < tailMin && val[queMin[tailMin-1]] >= val[i]) tailMin--; tailMin++;
while(headMax < tailMax && val[queMax[tailMax-1]] <= val[i]) tailMax--; tailMax++;
queMin[tailMin-1] = queMax[tailMax-1] = i;
ansMin[i] = val[queMin[headMin]];
ansMax[i] = val[queMax[headMax]];
}
for(int i = k-1; i < n; i++) printf("%d%c", ansMin[i], i==n-1 ? '\n' : ' ');
for(int i = k-1; i < n; i++) printf("%d%c", ansMax[i], i==n-1 ? '\n' : ' ');
}
return 0;
}
Sliding Window
Time Limit: 12000MS | Memory Limit: 65536K | |
Total Submissions: 56208 | Accepted: 16159 | |
Case Time Limit: 5000MS |
An array of size n ≤ 106 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.
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 |
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
Source
POJ Monthly--2006.04.28, Ikki
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