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

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 #include<iostream>
 5 using namespace std;
 6 int dp[100005],g[100005];
 7 int main()
 8 {
 9     int n,m;
10     while(~scanf("%d%d",&n,&m))
11     {
12         memset(dp,0,sizeof(dp));
13         int x;
14         for(int i = 1; i <= n ; ++ i)
15             scanf("%d",&x),dp[x]=i;
16         int r = 0 ;
17         for(int i = 1 ; i <= m ; ++ i)
18         {
19             scanf("%d",&x);
20             if(dp[x])
21             g[r++]=dp[x];
22         }
23         int p = 0 ;
24         dp[p++] = g[0];
25         for(int i = 1 ; i < r ; ++ i)
26         if(dp[p-1] < g[i])
27             dp[p++] = g[i];
28         else
29         {
30             x  = lower_bound(dp,dp+p,g[i])-dp;
31             dp[x] = g[i];
32         }
33         printf("%d
",p);
34     }
35     return 0;
36 }

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