For sequences of integers a and b, if you can make the two sequences the same by deleting some elements in a and b, we call the remaining sequence "the common sub sequence". And we call the longest one the LCS.
Now you are given two sequences of integers a and b. You can arrange elements in a and b in any order. You are to calculate the max length of the LCS of each arrangement of a and b.
Input
Input will consist of multiple test cases. The first line of each case is two integers N(0 < N < 10000), M(0 < M < 10000) indicating the length of a and b. The second line is N 32-bit signed integers in a. The third line is M 32-bit signed integers in b.
Output
Each case one line. The max length of the LCS of each arrangement of a and b.
Sample Input
5 4 1 2 3 2 1 1 4 2 1
Sample Output
3
公共序列最长(乱序)
排序,然后找就行
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define N 100005
int a[N],b[N];
int main()
{
int aa,bb,i;
while(~scanf("%d%d",&aa,&bb))
{
for(i=0;i<aa;i++)
scanf("%d",&a[i]);
for(i=0;i<bb;i++)
scanf("%d",&b[i]);
sort(a,a+aa);
sort(b,b+bb);
int ans=0;
int j=0,temp=0;//temp是重点
for(i=0;i<aa;i++)
for(j=temp;j<bb;j++)
if(a[i]==b[j])
{
ans++;
temp=j+1;
break;
}
printf("%d\n",ans);
}
return 0;
}