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。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 struct info{
     int num,pos;
 };
 int n,m,tot,sa[],sc[],q[],head,tail;
 info sb[];
 bool cmp(const info &x,const info &y){
     return x.num < y.num;
 }
 int bsearch(int lt,int rt,int val){
     int mid,pos = -;
     while(lt <= rt){
         int mid = (lt+rt)>>;
         if(val <= sb[mid].num){
             pos = mid;
             rt = mid-;
         }else lt = mid+;
     }
     return pos;
 }
 int binsearch(int lt,int rt,int val){
     while(lt <= rt){
         int mid = (lt+rt)>>;
         if(q[mid] < val) lt = mid+;
         else rt = mid-;
     }
     return lt;
 }
 int main() {
     while(~scanf("%d %d",&n,&m)){
         head = tail = tot = ;
         for(int i = ; i <= n; i++) scanf("%d",sa+i);
         for(int i = ; i <= m; i++){
             scanf("%d",&sb[i].num);
             sb[i].pos = i;
         }
         sort(sb+,sb+m+,cmp);
         for(int i = ; i <= n; i++){
             int tmp = bsearch(,m,sa[i]);
             while(tmp >  && sb[tmp].num == sa[i]) sc[tot++] = sb[tmp++].pos;
         }
         for(int i = ; i < tot; i++){
             if(head == tail || q[head-] < sc[i]){
                 q[head++] = sc[i];
             }else{
                 int tmp = binsearch(tail,head-,sc[i]);
                 q[tmp] = sc[i];
             }
         }
         printf("%d\n",head-tail);
     }
     return ;
 }