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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.
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.
8 3 1 3 -1 -3 5 3 6 7
-1 -3 -3 -3 3 3 3 3 5 5 6 7
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 const int maxn = 1000010; 6 int d[maxn],q[maxn],hd,tl,n,k; 7 bool getMin(int a,int b) { 8 return a <= b; 9 } 10 bool getMax(int a,int b) { 11 return a >= b; 12 } 13 void solve(bool (*f)(int,int)) { 14 hd = tl = 0; 15 bool flag = false; 16 for(int i = 0; i < n; ++i) { 17 while(hd < tl && f(d[i],q[tl-1])) --tl; 18 q[tl++] = i; 19 if(i + 1 >= k) { 20 while(hd < tl && i - q[hd] + 1 > k) ++hd; 21 if(flag) putchar(‘ ‘); 22 printf("%d",d[q[hd]]); 23 flag = true; 24 } 25 } 26 putchar(‘\n‘); 27 } 28 int main() { 29 while(~scanf("%d%d",&n,&k)) { 30 for(int i = 0; i < n; ++i) 31 scanf("%d",d+i); 32 solve(getMin); 33 solve(getMax); 34 } 35 return 0; 36 }
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原文地址：http://www.cnblogs.com/crackpotisback/p/4738009.html
