4810. 【NOIP2016提高A组五校联考1】道路规划 (Standard IO)

Time Limits: 1000 ms  Memory Limits: 524288 KB  Detailed Limits  

Description

Input

Output

Sample Input

5
1 4 5 2 3
3 4 2 1 5

Sample Output

3

Data Constraint

Hint

Source / Author: 学军中学 road

题解：

显然最长下降子序列。

O(n log n)算法：

扫一遍a，对于新的一个a，二分一个恰比他小的替换掉，没有比他小的放在队尾。

#include<bits/stdc++.h>
#define N 100010
#define inf 2147483647
#define open(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
#define mem(a,b) memset(a,b,sizeof(a))
#define mcy(a,b) memcpy(a,b,sizeof(a))
using namespace std;

int n,i,j,len,ans;
int a[N],b[N],c[N],g[N];

int find(int x)
{
	int l=0,r=len,k=0;
	while(l<=r)
	{
		int mid=(l+r)/2;
		if(g[mid] < x)
		{
			k=mid;
			r = mid -1;
		} else l = mid +1;
	}
	return k;
}

void dp()
{
	g[0] = inf;
	for(i=1;i<=n;i++)
	{
		if(a[i] < g[len]) g[++len] = a[i];
		else
		{
			int  k = find(a[i]);
			g[k] = a[i];
		}
		ans = max(ans , len);
	}
	return ;
}

int main()
{
	open("road");
	scanf("%d",&n);
	for(i=1;i<=n;i++) scanf("%d",&a[i]);
	for(i=1;i<=n;i++) scanf("%d",&b[i]) , c[b[i]]=i;
	for(i=1;i<=n;i++) a[i] = c[a[i]];
	
	dp();
	
	printf("%d",ans);
	
	return 0;
}

标签：NOIP2016,4810,Limits,int,freopen,Input,sizeof,联考,define
来源： https://blog.csdn.net/Com_man_der/article/details/94572234