zoukankan      html  css  js  c++  java
  • zoj 2432(最长递增上升子序列)

    算是LCIS的入门题吧, 这个dp方法以前没怎么看,现在好好地学习了下,dp真的是奥妙无穷...

    首先,dp的最开始是定义状态 dp[i][j] 表示A串的前i个,与B串的前j个,并以B[j]为结尾的LCIS 的长度.

    状态转移方程:

    if(A[i]==B[j]) dp[i][j]=max(dp[i-1][k])+1;  ( 1 <= k < j )

    else dp[i][j]=dp[i-1][j];

    然后选择循环顺序,就可以将算法的复杂度降为n*n.

    Greatest Common Increasing Subsequence

    Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

    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


    Source: Northeastern Europe 2003, Northern Subregion

    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    using namespace std;
    
    #define N 550
    
    struct node
    {
        int x,y;
    }path[N][N];
    
    int dp[N][N];
    int s[N],t[N];
    
    
    int main()
    {
        int t1;
        while(scanf("%d",&t1)!=EOF)
        {
            while(t1--)
            {
                memset(path,0,sizeof(path));
                int n,m;
                scanf("%d",&n);
                for(int i=1;i<=n;i++)
                    scanf("%d",&s[i]);
                scanf("%d",&m);
                for(int i=1;i<=m;i++)
                    scanf("%d",&t[i]);
                memset(dp,0,sizeof(dp));
                int mx=0;
                for(int i=1;i<=n;i++)
                {
                    mx=0;
                    int tx=0,ty=0;
                    for(int j=1;j<=m;j++)
                    {
                        dp[i][j] = dp[i-1][j];
                        path[i][j].x=i-1;
                        path[i][j].y=j;
                        if( s[i]>t[j] && mx<dp[i-1][j])
                        {
                            mx=dp[i-1][j];
                            tx=i-1;
                            ty=j;
                        }
                        if( s[i] == t[j] )
                        {
                            dp[i][j]=mx+1;
                            path[i][j].x=tx;
                            path[i][j].y=ty;
                        }
                    }
                }
                mx=-1;
                int id;
                for(int i=1;i<=m;i++) 
                    if(dp[n][i]>mx) 
                    {
                        mx=dp[n][i];
                        id=i;
                    }
                int save[N];
                int cnt=0;
                int tx,ty;
                tx=n; ty=id;
                while( dp[tx][ty] != 0 ) 
                {
                    int tmpx,tmpy;
                    tmpx=path[tx][ty].x;
                    tmpy=path[tx][ty].y;
                    if(dp[tx][ty]!=dp[tmpx][tmpy])
                    {
                        save[cnt++]=t[ty];
                    }
                    tx=tmpx; ty=tmpy;
                }
                printf("%d\n",mx);
                for(int i=cnt-1;i>=0;i--)
                    printf("%d ",save[i]);
                printf("\n");
            }
        }
        return 0;
    }
  • 相关阅读:
    Android 再次打开APP进入按Home键退出时的界面(thisTaskRoot)
    SQL server 安装成功到使用Sa SQL server验证登录等一系列问题
    Android JSON解析插件
    Android utils 工具类之MD5加密 MD5Utils
    Android 避免内存泄漏
    Android utils 之 日志工具类
    Android Studio 第一个Android项目
    Android Studio安装
    JDK 安装与环境配置配置——Android开发第一步
    U盘启动安装系统之旅----记录自己的第一次操作
  • 原文地址:https://www.cnblogs.com/chenhuan001/p/2982677.html
Copyright © 2011-2022 走看看