See LCS again 最長遞增子序列到最長公共子序列的轉化

See LCS again

時間限制：1000 ms  |  內存限制：65535 KB

難度：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.

輸入

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;

輸出

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

樣例輸入

5 4
1 2 6 5 4
1 3 5 4

樣例輸出

3

由於數據量，所以把最長公共子序列轉化爲最長遞增子序列 而最長遞增子序列由nlogn算法

這就是這個題的經典之處

#include <cstdio>
#include <algorithm>
#include <climits>
#include <cstring>
using namespace std;
#define maxn 100000

int a[maxn + 10];
int b[maxn + 10];
int g[maxn + 10];
int n, m;


int main(void){
     int i;
     while(scanf("%d%d", &n, &m) == 2)
     {
          int t;
          memset(a, 0, sizeof(a));
          for (i = 1; i <= n; ++i)
          {
               scanf("%d", &t);
               a[t] = i;
          }
          int p = 0;
          for (i = 0; i < m; ++i)
          {
               scanf("%d", &t);
               if(a[t])
               {
                    b[p++] = a[t];
               }
          }
          int mmax = 0;
          for (i = 1; i <= p; ++i)
          {
               g[i] = 100000000;
          }
          for (i = 0; i < p; ++i)
          {
               int k = lower_bound(g + 1, g + p + 1, b[i]) - g;
               if(k > mmax)
                    mmax = k;
               g[k] = b[i];
          }
          printf("%d\n", mmax);
     }
     return 0;
}