nyoj See LCS again 760 （LCS&&LIS） 好题

See LCS again

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

There are A, B two sequences, the number of elements in the sequence is n、m;

Each element in the sequence are different and less than 100000.

Calculate the length of the longest common subsequence of A and B.

输入

The input has multicases.Each test case consists of three lines;
The first line consist two integers n, m (1 < = n, m < = 100000);
The second line with n integers, expressed sequence A;
The third line with m integers, expressed sequence B;

输出

For each set of test cases, output the length of the longest common subsequence of A and B, in a single line.

样例输入

5 41 2 6 5 41 3 5 4

样例输出

3

//函数lower_bound()在first和last中的前闭后开区间进行二分查找，//返回大于或等于val的第一个元素位置。如果所有元素都小于val，//则返回last的位置，且last的位置是越界的！返回查找元素的第一个//可安插位置，也就是“元素值>=查找值”的第一个元素的位置

#include<stdio.h>#include<string.h>#include<algorithm>#define N 100010using namespace std;int a[N];int b[N];int dp[N];int main(){int n,m,i,j;int x;while(scanf("%d%d",&n,&m)!=EOF){memset(dp,0,sizeof(dp));for(i=1;i<=n;i++){scanf("%d",&x);dp[x]=i;}int r=0;for(i=1;i<=m;i++){scanf("%d",&x);if(dp[x])b[r++]=dp[x];}int p=0;dp[p++]=b[0];for(i=1;i<r;i++){if(dp[p-1]<b[i])dp[p++]=b[i];else{x=lower_bound(dp,dp+p,b[i])-dp;dp[x]=b[i];}}printf("%d\n",p);}return 0;}