nyoj 760 See LCS again

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 4
1 2 6 5 4
1 3 5 4

样例输出

3

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #include<iostream>
 using namespace std;
 int dp[],g[];
 int main()
 {
     int n,m;
     while(~scanf("%d%d",&n,&m))
     {
         memset(dp,,sizeof(dp));
         int x;
         for(int i = ; i <= n ; ++ i)
             scanf("%d",&x),dp[x]=i;
         int r =  ;
         for(int i =  ; i <= m ; ++ i)
         {
             scanf("%d",&x);
             if(dp[x])
             g[r++]=dp[x];
         }
         int p =  ;
         dp[p++] = g[];
         for(int i =  ; i < r ; ++ i)
         if(dp[p-] < g[i])
             dp[p++] = g[i];
         else
         {
             x  = lower_bound(dp,dp+p,g[i])-dp;
             dp[x] = g[i];
         }
         printf("%d\n",p);
     }
     return ;
 }

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