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  • 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]);
                    }
                }
            }
        }
    }
    

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  • 原文地址:https://www.cnblogs.com/riskyer/p/3263099.html
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