NYIST 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

上传者
TC_胡仁东

 

解题：一种LCS转LCS的nlogn的算法。是严格上升的LCS。

 

首先是LCS，我们把a序列中的每个元素在b中出现的位置保存起来，再按照降序排列，排列后再代入a的每个对应元素，那就转化为了求这个新的序列的最长上升子序列了。如：a[] = {a, b, c,} b[] = {a,b,c,b,a,d},那么a中的a，b，c在b中出现的位置分别就是{0,4},{1,3},{2}。分别按降序排列后代入a序列就是{4,0,2,3,1},之所以要按照降序排列，目的就是为了让每个元素只取到一次。

接下来的问题就是要求最长升序子序列问题了，也就是求LIS。

 特殊情况下，会退化得很严重。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 struct info{

18     int num,pos;

19 };

20 int n,m,tot,sa[100010],sc[200010],q[200010],head,tail;

21 info sb[100010];

22 bool cmp(const info &x,const info &y){

23     return x.num < y.num;

24 }

25 int bsearch(int lt,int rt,int val){

26     int mid,pos = -1;

27     while(lt <= rt){

28         int mid = (lt+rt)>>1;

29         if(val <= sb[mid].num){

30             pos = mid;

31             rt = mid-1;

32         }else lt = mid+1;

33     }

34     return pos;

35 }

36 int binsearch(int lt,int rt,int val){

37     while(lt <= rt){

38         int mid = (lt+rt)>>1;

39         if(q[mid] < val) lt = mid+1;

40         else rt = mid-1;

41     }

42     return lt;

43 }

44 int main() {

45     while(~scanf("%d %d",&n,&m)){

46         head = tail = tot = 0;

47         for(int i = 1; i <= n; i++) scanf("%d",sa+i);

48         for(int i = 1; i <= m; i++){

49             scanf("%d",&sb[i].num);

50             sb[i].pos = i;

51         }

52         sort(sb+1,sb+m+1,cmp);

53         for(int i = 1; i <= n; i++){

54             int tmp = bsearch(1,m,sa[i]);

55             while(tmp > 0 && sb[tmp].num == sa[i]) sc[tot++] = sb[tmp++].pos;

56         }

57         for(int i = 0; i < tot; i++){

58             if(head == tail || q[head-1] < sc[i]){

59                 q[head++] = sc[i];

60             }else{

61                 int tmp = binsearch(tail,head-1,sc[i]);

62                 q[tmp] = sc[i];

63             }

64         }

65         printf("%d\n",head-tail);

66     }

67     return 0;

68 }

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