zoukankan      html  css  js  c++  java
  • ZOJ 2432-Greatest Common Increasing Subsequence

    You are given two sequences of integer numbers. Write a program to determine their common increasing subsequence of maximal possible length.
    Sequence S1, S2, ..., SN of length N is called an increasing subsequence of a sequence A1, A2, ..., AM of length M if there exist 1 <= i1 < i2 < ...< iN <= M such that Sj = Aij for all 1 <= j <= N, and Sj < Sj+1 for all 1 <= j < N.

    Input

    Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.

    Output

    On the first line of the output print L - the length of the greatest common increasing subsequence of both sequences. On the second line print the subsequence itself. If there are several possible answers, output any of them.

    This problem contains multiple test cases!

    The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

    The output format consists of N output blocks. There is a blank line between output blocks.

    Sample Input

    1

    5
    1 4 2 5 -12
    4
    -12 1 2 4

    Sample Output

    2
    1 4

    题目大意就是求公共最长上升自序列,并输出路径。

    之前写过怎么求公共上升子序列的长度,关于路径需要用二维数组来记录每一个点的前驱节点。由于还要记录层数,所以加上一个偏移量表示层数就可以了。

    #include<bits/stdc++.h>
    
    using namespace std;
    const int N = 1007;
    int n, m;
    int a[N], b[N], dp[N];
    int pre[N][N];
    
    void solve()
    {
        scanf("%d", &n);;
        for(int i=1; i<=n; ++ i)
            scanf("%d", &a[i]);
        scanf("%d", &m);
        for(int i=1; i<=m; ++ i)
            scanf("%d", &b[i]);
    
        memset(dp, 0, sizeof(dp));
        memset(pre, -1, sizeof(pre));
    
        for(int i=1; i<=n; ++ i)
        {
            int k = 0, x = 0;
            for(int j=1; j<=m; ++ j)
            {
                if(dp[j])
                    pre[i][j] = pre[i-1][j];
    
                if(a[i] == b[j] && dp[j] <= dp[k])
                    dp[j] = dp[k] + 1, pre[i][j] = x * 1000 + k;
    
                if(dp[j] > dp[k] && b[j] < a[i])
                    k = j, x = i;
            }
        }
    
        int ans = 0;
        for(int i=1; i<=m; ++ i)
        {
            if(dp[i] > dp[ans])
                ans = i;
        }
    
        printf("%d
    ", dp[ans]);
        stack<int> stk;
        for(int x = n, k = ans, t; pre[x][k] != -1; t = pre[x][k], x = t / 1000, k = t % 1000)
            stk.push(b[k]);
        while(stk.size())
        {
            printf("%d ", stk.top());
            stk.pop();
        }
        printf("
    ");
    }
    
    int main()
    {
        int t;
        scanf("%d", &t);
        for(int i=0; i<t; ++ i)
            solve();
        return 0;
    }
    
    

    写完这份代码感觉自己好傻逼啊,一个简单的错误调了半天bug。

    写这部分的时候

    for(int x = n, k = ans, t; pre[x][k] != -1; t = pre[x][k], x = t / 1000, k = t % 1000)
            stk.push(b[k]);
    

    之前是写成这样的

    for(int x = n, k = ans; pre[x][k] != -1; x = pre[x][k] / 1000, k = pre[x][k] % 1000)
            stk.push(b[k]);
    

    这样的一个错误调了老长时间。。

  • 相关阅读:
    python 企业微信告警
    容器启动报错listen unix /containerd-shim/moby/9a3b9086ece8fcd8746695836e3f057cc0313b3cdb722d76a5f571dfa428759e/shim.sock: bind: address already in use: unknown
    etcd查询k8s相关数据
    hadoop三大核心组件概念及原理
    使用nginx反代实现k8s apiserver高可用
    k8s useraccout账号创建及RDBA授权
    k8s之二进制部署
    git代码提交后jenkins的构建与持续部署
    dockerfile动态修改服务配置文件
    数据库操作语句DDL,DML,DCL
  • 原文地址:https://www.cnblogs.com/aiterator/p/6806207.html
Copyright © 2011-2022 走看看