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

231 2 33 2 161 5 3 2 6 43 6 2 4 5 1

 

Sample Output

24

 

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$l(l>1)$的环, 我们总可以得到一个长度为$l-1$的LCS, 于是这个题的答案就很明显了, 就是$n$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;}