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
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