POJ - 2823 Sliding Window （单调队列）
分别维护一个单调递减的队列和单调递增队列就可以了，但是这题莫名的坑，要用C++提交卡编译器的时间才能AC，真的是日了狗了，以后就用C++提交好了，不用别的了。
1 #include <iostream> 2 #include <queue> 3 #include <stack> 4 #include <cstdio> 5 #include <vector> 6 #include <map> 7 #include <set> 8 #include <bitset> 9 #include <algorithm> 10 #include <cmath> 11 #include <cstring> 12 #include <cstdlib> 13 #include <string> 14 #include <sstream> 15 #include <time.h> 16 #define x first 17 #define y second 18 #define pb push_back 19 #define mp make_pair 20 #define lson l,m,rt*2 21 #define rson m+1,r,rt*2+1 22 #define mt(A,B) memset(A,B,sizeof(A)) 23 using namespace std; 24 typedef long long LL; 25 //const double PI = acos(-1); 26 const int N=1e6+10; 27 const LL mod=1e9+7; 28 const int inf = 0x3f3f3f3f; 29 const LL INF=0x3f3f3f3f3f3f3f3fLL; 30 int a[N],MI[N],MA[N]; 31 struct node 32 { 33 int x,y; 34 }Q[N];//第一位是下标，第二位是数据 35 int main() 36 { 37 #ifdef Local 38 freopen("data.txt","r",stdin); 39 #endif 40 //ios::sync_with_stdio(false); 41 //cin.tie(0); 42 int n,k,w=0; 43 register int head=0,tail=0; 44 scanf("%d%d",&n,&k); 45 for(int i=0;i<n;i++)scanf("%d",&a[i]); 46 for(int i=0;i<n;i++) 47 { 48 while(tail-head>0&&Q[head+1].x<i-k+1)head++; 49 while(tail>head&&Q[tail].y>=a[i])tail--; 50 tail++; 51 Q[tail].y=a[i]; 52 Q[tail].x=i; 53 if(i>=k-1)MI[w++]=Q[head+1].y; 54 } 55 head=0;tail=0;w=0; 56 for(int i=0;i<n;i++) 57 { 58 while(tail-head>0&&Q[head+1].x<i-k+1)head++; 59 while(tail>head&&Q[tail].y<=a[i])tail--; 60 tail++; 61 Q[tail].y=a[i]; 62 Q[tail].x=i; 63 if(i>=k-1)MA[w++]=Q[head+1].y; 64 } 65 for(int i=0;i<w;i++) 66 { 67 if(i==w-1)printf("%d\n",MI[i]); 68 else printf("%d ",MI[i]); 69 } 70 for(int i=0;i<w;i++) 71 { 72 if(i==w-1)printf("%d\n",MA[i]); 73 else printf("%d ",MA[i]); 74 } 75 #ifdef Local 76 cerr << "time: " << (LL) clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl; 77 #endif 78 }
