zoukankan      html  css  js  c++  java
  • HDU 4679 String

    String

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
    Total Submission(s): 695    Accepted Submission(s): 254

    Problem Description
    Given 3 strings A, B, C, find the longest string D which satisfy the following rules:
    a) D is the subsequence of A
    b) D is the subsequence of B
    c) C is the substring of D
    Substring here means a consecutive subsequnce.
    You need to output the length of D.
     
    Input
    The first line of the input contains an integer T(T = 20) which means the number of test cases.
    For each test case, the first line only contains string A, the second line only contains string B, and the third only contains string C.
    The length of each string will not exceed 1000, and string C should always be the subsequence of string A and string B.
    All the letters in each string are in lowercase.
     
    Output
    For each test case, output Case #a: b. Here a means the number of case, and b means the length of D.
     
    Sample Input
    2 aaaaa aaaa aa abcdef acebdf cf
     
    Sample Output
    Case #1: 4 Case #2: 3
    Hint
    For test one, D is "aaaa", and for test two, D is "acf".
     
    Source
     
    >Recommend
    zhuyuanchen520
       
      
        祭奠一下这个题
    #include <iostream>
    #include <string>
    #include <cstring>
    #include <cstdio>
    #define N 1100
    using namespace std;
    string s1,s2,s3,s4,s5;
    char temp[N];
    int num1[N][N],num2[N][N];
    struct num
    {
        int sta,end;
    } a[N],b[N];
    int main()
    {
        //freopen("data.in","r",stdin);
        void get(int (*p)[N],string ch1,string ch2);
        int t,tem=1;
        scanf("%d",&t);
        while(t--)
        {
            cin>>s1>>s2>>s3;
            get(num1,s1,s2);
            int l1 = s1.size();
            for(int i=0; i<=l1-1; i++)
            {
                temp[l1-1-i] = s1[i];
            }
            temp[l1] = '';
            s4 = temp;
            int l2 = s2.size();
            for(int i=0; i<=l2-1; i++)
            {
                temp[l2-1-i] = s2[i];
            }
            temp[l2] = '';
            s5 = temp;
            get(num2,s4,s5);
            int l3 = s3.size(),Top1=0,Top2=0;
            for(int i=0; i<=l1-1; i++)
            {
                if(s1[i]==s3[0])
                {
                    int x = 0;
                    int sta = i;
                    int end = -1;
                    for(int j=i; j<=l1-1&&x<=l3-1; j++)
                    {
                        if(s1[j]==s3[x])
                        {
                            x++;
                        }
                        if(x==l3)
                        {
                            end = j;
                        }
                    }
                    if(end==-1)
                    {
                        continue;
                    }
                    a[Top1].sta = sta;
                    a[Top1++].end = end;
                }
            }
            for(int i=0; i<=l2-1; i++)
            {
                if(s2[i]==s3[0])
                {
                    int x = 0;
                    int sta = i;
                    int end = -1;
                    for(int j=i; j<=l2-1&&x<=l3-1; j++)
                    {
                        if(s2[j]==s3[x])
                        {
                            x++;
                        }
                        if(x==l3)
                        {
                            end = j;
                        }
                    }
                    if(end==-1)
                    {
                        continue;
                    }
                    b[Top2].sta = sta;
                    b[Top2++].end = end;
                }
            }
            int res = l3,Max=0;
            for(int i=0; i<=Top1-1; i++)
            {
                for(int j=0; j<=Top2-1; j++)
                {
                    int x1 = a[i].sta;
                    int y1 = a[i].end;
                    int x2 = b[j].sta;
                    int y2 = b[j].end;
                    int k1 = 0;
                    if(x1>0&&x2>0)
                    {
                        k1 = num1[x1-1][x2-1];
                    }
                    int k2 = 0;
                    if(y1<l1-1&&y2<l2-1)
                    {
                        k2 = num2[l1-2-y1][l2-2-y2];
                    }
                    Max = max(Max,res+k2+k1);
                }
            }
            printf("Case #%d: %d
    ",tem++,Max);
        }
        return 0;
    }
    void get(int (*p)[N],string ch1,string ch2)
    {
        int l1 = ch1.size();
        int l2 = ch2.size();
        memset(p,0,sizeof(p));
        for(int i=0; i<=l1-1; i++)
        {
            for(int j=0; j<=l2-1; j++)
            {
                if(i==0&&j==0)
                {
                    if(ch1[i]==ch2[j])
                    {
                        p[i][j] = 1;
                    }
                }
                else if(i==0&&j!=0)
                {
                    if(ch1[i]==ch2[j])
                    {
                        p[i][j] = 1;
                    }
                    else
                    {
                        p[i][j] = p[i][j-1];
                    }
                }
                else if(i!=0&&j==0)
                {
                    if(ch1[i]==ch2[j])
                    {
                        p[i][j] = 1;
                    }
                    else
                    {
                        p[i][j] = p[i-1][j];
                    }
                }
                else
                {
                    if(ch1[i]==ch2[j])
                    {
                        p[i][j] = p[i-1][j-1]+1;
                    }
                    else
                    {
                        p[i][j] = max(p[i][j-1],p[i-1][j]);
                    }
                }
            }
        }
    }
    

  • 相关阅读:
    30天内自动登录
    本地保存cookie
    thymeleaf 基本表达式
    适配器(adapter)与fragment之间、fragment与activity之间的通信问题
    String/Stringbuilder/StringBuffer
    if else与switch for与foreach
    iOS App 签名的原理
    iOS category内部实现原理
    iOS中的事件的产生和传递
    RunLoop
  • 原文地址:https://www.cnblogs.com/riskyer/p/3263099.html
Copyright © 2011-2022 走看看