1117: Ready to declare
时间限制: 1 Sec 内存限制: 128 MB提交: 358 解决: 41
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题目描述
Finally, you find the most good-looking girl... You are going to write a letter to her. But you are not convident to be better than other boys. So you think you need a good chance... You have known that these days, she will meet N boys at all(N<=100000). The ith boy has a handsome value V[i](0输入
There are several test cases, each case contains two lines. The first line is two number ,N K. The second line has N number, the handsome value.输出
Each case output two lines.The first line is each meet's min handsome value, the second line is each meet's max handsome value.
样例输入
8 3 5 3 2 1 1 10 2 3
样例输出
2 1 1 1 1 2 5 3 2 10 10 10
提示
来源
2011.12
虽然正解是单调队列，可是我用的是两个栈实现的队列水过的，存下代码
/*************************************************************************
> File Name: Euler.cpp
> Author: acvcla
> QQ:
> Mail: acvcla@gmail.com
> Created Time: 2014年10月30日 星期四 11时19分11秒
************************************************************************/
#include
#include
#include
using namespace std; const int maxn=1e5+5; int stack1[maxn],stack2[maxn],max2[maxn],max1,min2[maxn],min1; int top1,top2; void init(){ top1=top2=0; max1=max2[top2]=0; min1=min2[top2]=maxn; } int ans1[maxn],ans2[maxn]; void deque(){ if(top2>0){ stack2[--top2]; return ; } while(top1>0){ stack2[top2]=stack1[--top1]; max2[top2]=max(top2>0?max2[top2-1]:0,stack2[top2]); min2[top2]=min(top2>0?min2[top2-1]:maxn,stack2[top2]); top2++; } max1=0; min1=maxn; --top2; } int get_max_min(int x){ if(x)return max(max1,top2>0?max2[top2-1]:0); return min(min1,top2>0?min2[top2-1]:maxn); } void push(int x){ max1=max(x,max1); min1=min(x,min1); stack1[top1++]=x; } int main() { int n,k,x; while(~scanf("%d%d",&n,&k)){ init(); int t=0,cnt=0; for(int i=1;i<=n;i++){ scanf("%d",&x); push(x); ++cnt; if(cnt==k) { ans1[t]=get_max_min(0); //cout<<"sldl"<
补上单调队列版
#include
#include
using namespace std; const int maxn =1e5 + 10; struct Queue { int value; int index; }; Queue min_que[maxn],max_que[maxn]; int n,k,num[maxn],front,rear; int delete_rear_inc(int f,int r,int d) { int mid; while(f<=r){ mid=(f+r)/2; if(min_que[mid].value==d) return mid; if(min_que[mid].value>d) r=mid-1; else f=mid+1; } return f; } int delete_rear_dec(int f,int r,int d) { int mid; while(f<=r){ mid=(f+r)/2; if(max_que[mid].value==d) return mid; if(max_que[mid].value>d) f=mid+1; else r=mid-1; } return f; } int main() { while(~scanf("%d%d",&n,&k)){ for(int i=1;i<=n;i++)scanf("%d",&num[i]); min_que[1].value=num[1]; front=rear=min_que[1].index=1; for(int i=2;i<=k;i++) { rear=delete_rear_inc(front,rear,num[i]); min_que[rear].value=num[i]; min_que[rear].index=i; } printf("%d",min_que[1].value); for(int i=k+1;i<=n;i++) { rear=delete_rear_inc(front,rear,num[i]); min_que[rear].value=num[i]; min_que[rear].index=i; if(i-min_que[front].index>=k) front++; printf(" %d",min_que[front].value); if(i==n)puts(""); } max_que[1].value=num[1]; front=rear=max_que[1].index=1; for(int i=2;i<=k;i++) { rear=delete_rear_dec(front,rear,num[i]); max_que[rear].value=num[i]; max_que[rear].index=i; } printf("%d",max_que[1].value); for(int i=k+1;i<=n;i++) { rear=delete_rear_dec(front,rear,num[i]); max_que[rear].value=num[i]; max_que[rear].index=i; if(i-max_que[front].index>=k) front++; printf(" %d",max_que[front].value); if(i==n)puts(""); } } return 0; }