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[100010],sc[200010],q[200010],head,tail;
info sb[100010];
bool cmp(const info &x,const info &y){
    return x.num < y.num;
}
int bsearch(int lt,int rt,int val){
    int mid,pos = -1;
    while(lt <= rt){
        int mid = (lt+rt)>>1;
        if(val <= sb[mid].num){
            pos = mid;
            rt = mid-1;
        }else lt = mid+1;
    }
    return pos;
}
int binsearch(int lt,int rt,int val){
    while(lt <= rt){
        int mid = (lt+rt)>>1;
        if(q[mid] < val) lt = mid+1;
        else rt = mid-1;
    }
    return lt;
}
int main() {
    while(~scanf("%d %d",&n,&m)){
        head = tail = tot = 0;
        for(int i = 1; i <= n; i++) scanf("%d",sa+i);
        for(int i = 1; i <= m; i++){
            scanf("%d",&sb[i].num);
            sb[i].pos = i;
        }
        sort(sb+1,sb+m+1,cmp);
        for(int i = 1; i <= n; i++){
            int tmp = bsearch(1,m,sa[i]);
            while(tmp > 0 && sb[tmp].num == sa[i]) sc[tot++] = sb[tmp++].pos;
        }
        for(int i = 0; i < tot; i++){
            if(head == tail || q[head-1] < sc[i]){
                q[head++] = sc[i];
            }else{
                int tmp = binsearch(tail,head-1,sc[i]);
                q[tmp] = sc[i];
            }
        }
        printf("%d
",head-tail);
    }
    return 0;
}