NEUOJ 1117: Ready to declare（单调队列）

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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题目描写叙述

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<V[i]<10000). She will meet K(K<=N) boys at the same time, and when the first boy leave, the next boy join them. It is to say that she will meet
first K boys in the list at first, then the first boy leave ,the k+1-th boy join. So she will meet boys N-K+1 times.

You must want to join she and them when these boys are not handsome, so you need to calculate each meet's the least handsome value and the most.

here is a sample when N=8,K=3

boys                least     most

[5 3 2] 1 1 10 2 3 25

5 [3 2 1] 1 10 2 3 13

5 3 [2 1 1] 10 2 3 12

5 3 2 [1 1 10] 2 3 110

5 3 2 1 [1 10 2] 3 110

5 3 2 1 1 [10 2 3] 210

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.

例子输入

8 3
5 3 2 1 1 10 2 3

例子输出

2 1 1 1 1 2
5 3 2 10 10 10

提示

来源

2011.12

尽管正解是单调队列。但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<iostream>
#include<cstring>
#include<cstdio>
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"<<endl;
				ans2[t++]=get_max_min(1);
				deque();
				cnt--;
			}
		}
		for(int i=0;i<t;i++)printf("%d%c",ans1[i],i==t-1?'\n':' ');
		for(int i=0;i<t;i++)printf("%d%c",ans2[i],i==t-1?'\n':' ');
	}
	return 0;
}

补上单调队列版

#include <cstdio>
#include <iostream>
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;
}