hdu 5495 LCS (bestcoder #58 1002)

LCS

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Problem Description

You are given two sequence {a1,a2,...,an}
and {b1,b2,...,bn}.
Both sequences are permutation of {1,2,...,n}.
You are going to find another permutation {p1,p2,...,pn}
such that the length of LCS (longest common subsequence) of {ap1,ap2,...,apn}
and {bp1,bp2,...,bpn}
is maximum.

Input

There are multiple test cases. The first line of input contains an integer T,
indicating the number of test cases. For each test case:

The first line contains an integer n(1≤n≤105) -
the length of the permutation. The second line contains n integers a1,a2,...,an.
The third line contains nintegers b1,b2,...,bn.

The sum of n in
the test cases will not exceed 2×106.

Output

For each test case, output the maximum length of LCS.

Sample Input

2
3
1 2 3
3 2 1
6
1 5 3 2 6 4
3 6 2 4 5 1

Sample Output

2
4

出题人的解题思路：

Problem 1. LCS

题目中给出的是两个排列, 于是我们我们可以先把排列分成若干个环, 显然环与环之间是独立的. 事实上对于一个长度为l
(l > 1)l(l>1)的环,
我们总可以得到一个长度为l-1l−1的LCS,
于是这个题的答案就很明显了, 就是nn减去长度大于1
1的环的数目.

其实就是把数组a到b看成一条边，由于都是1～n，所以判断有多少个独立环，长度为一的时候特判，大于一的时候为l-1；

ps: 输入用scanf，用cin会超时；

代码：

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
struct node
{
int a,b;
}data[100005];
int cmp(const node &s1,const node &s2)
{
return s1.a<s2.a;
}
int main()
{
int vis[100005];
int t;
cin>>t;
while(t--)
{    int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&data[i].a);
for(int j=1;j<=n;j++)
scanf("%d",&data[j].b);
sort(data+1,data+n+1,cmp);
memset(vis,0,sizeof(vis));
int ans=0;
for(int i=1;i<=n;i++)
{
if(!vis[i])
{
int p=i,l=1;
while(data[p].b!=i)
{
l++;
p=data[p].b;
vis[p]=1;
}
vis[i]=1;
if(l==1) ans+=l;
else
ans+=(l-1);
}
}
cout<<ans<<endl;
}
return 0;
}