• hdu-5904 LCIS(水题)


    题目链接:

    LCIS

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 458    Accepted Submission(s): 212


    Problem Description
    Alex has two sequences a1,a2,...,an and b1,b2,...,bm. He wants find a longest common subsequence that consists of consecutive values in increasing order.
     
    Input
    There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

    The first line contains two integers n and m (1n,m100000) -- the length of two sequences. The second line contains n integers: a1,a2,...,an (1ai106). The third line contains n integers: b1,b2,...,bm (1bi106).

    There are at most 1000 test cases and the sum of n and m does not exceed 2×106.
     
    Output
    For each test case, output the length of longest common subsequence that consists of consecutive values in increasing order.
     
    Sample Input
    3
    3 3
    1 2 3
    3 2 1
    10 5
    1 23 2 32 4 3 4 5 6 1
    1 2 3 4 5
    1 1
    2
    1
     
    Sample Output
    1
    5
    0
     
    题意:
     
    求这两个序列的最长公共子序列且是一段连续的值得最长长度;
     
    思路:
    两个序列A,B,fa[i]表示A序列中以i为结尾的最长的子序列的最长长度,fb[i]也是一样,然后取max(min(fa[i],fb[i]));
     
    AC代码:
     
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <bits/stdc++.h>
    #include <stack>
    #include <map>
      
    using namespace std;
      
    #define For(i,j,n) for(int i=j;i<=n;i++)
    #define mst(ss,b) memset(ss,b,sizeof(ss));
    #define lson o<<1
    #define rson o<<1|1
    typedef  long long LL;
      
    template<class T> void read(T&num) {
        char CH; bool F=false;
        for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
        for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
        F && (num=-num);
    }
    int stk[70], tp;
    template<class T> inline void print(T p) {
        if(!p) { puts("0"); return; }
        while(p) stk[++ tp] = p%10, p/=10;
        while(tp) putchar(stk[tp--] + '0');
        putchar('
    ');
    }
      
    const LL mod=1e9+7;
    const double PI=acos(-1.0);
    const LL inf=1e18;
    const int N=1e6+10;
    const int maxn=1e5+10;
    const double eps=1e-12;
    
    int n,m,a[maxn],b[maxn],fa[N],fb[N];
    
    int main()
    {
        int t;
        read(t);
        while(t--)
        {
            read(n);read(m);
            For(i,1,n)read(a[i]);
            For(i,1,m)read(b[i]);
            For(i,1,n)fa[a[i]-1]=fa[a[i]]=0,fb[a[i]-1]=fb[a[i]]=0;
            For(i,1,m)fa[b[i]-1]=fa[b[i]]=0,fb[b[i]-1]=fb[b[i]]=0;
            For(i,1,n)fa[a[i]]=fa[a[i]-1]+1;
            For(i,1,m)fb[b[i]]=fb[b[i]-1]+1;
            int ans=0;
            for(int i=1;i<=n;i++)ans=max(ans,min(fa[a[i]],fb[a[i]]));
            printf("%d
    ",ans);
        }    
        return 0;
    }
    

      

  • 相关阅读:
    使用批处理脚本在win10系统启动Redis 5.0.10
    异常分析 JedisConnectionException: java.net.SocketTimeoutException: Read timed out
    Spring Boot基于redis分布式锁模拟直播秒杀场景
    管理的经验二
    第三方api接口
    接口测试总结
    测试框架的基本能力
    接口测试的价值
    面试的经验
    管理的经验
  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5906395.html
Copyright © 2020-2023  润新知