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  • 九度OJ 1149:子串计算 (计数、排序)

    时间限制:1 秒

    内存限制:32 兆

    特殊判题:

    提交:1026

    解决:571

    题目描述:

    给出一个01字符串(长度不超过100),求其每一个子串出现的次数。

    输入:

    输入包含多行,每行一个字符串。

    输出:

    对每个字符串,输出它所有出现次数在1次以上的子串和这个子串出现的次数,输出按字典序排序。

    样例输入:
    10101
    样例输出:
    0 2
    01 2
    1 3
    10 2
    101 2
    来源:
    2010年北京大学计算机研究生机试真题

    思路:

    计数然后排序。呃,我怎么会写了这么长的代码。。。


    代码:

    #include <stdio.h>
    #include <string.h>
     
    #define M 100
     
    int main(void)
    {
        int n, i, j, k, m;
        char s[M+1], son[M][M][M+1];
        char *pson[M][M], *ptmp;
        int c[M][M], count[M];;
     
        while (scanf("%s", s) != EOF)
        {
            //printf("s=%s
    ", s);
            for (i=0; i<M; i++)
            {
                for (j=0; j<M; j++)
                    c[i][j] = 0;
            }
     
            n = strlen(s);
            for (i=1; i<=n; i++)
            {
                for (j=0; j<=n-i; j++)
                {
                    strncpy(son[i-1][j], s+j, i);
                    son[i-1][j][i] = '';
                }
                //for (j=0; j<=n-i; j++)
                //{
                //  printf("i=%d, j=%d, s=%s
    ", i, j, son[i-1][j]);
                //}
                count[i-1] = 0;
                for (j=0; j<=n-i; j++)
                {
                    ptmp = son[i-1][j];
                    if (j==0)
                    {
                        //printf("i=%d, j=%d, count[i-1]=%d, ptmp=%s
    ", i, j, count[i-1], ptmp);
                        pson[i-1][j] = ptmp;
                        count[i-1] ++;
                        c[i-1][j] ++;
                    }
                    else
                    {
                        //printf("i=%d, j=%d, count[i-1]=%d, ptmp=%s
    ", i, j, count[i-1], ptmp);
                        for (k=0; k<count[i-1]; k++)
                        {
                            /*
                            if (i==2 && k<2 )
                            {
                                printf("===
    ");
                                for (int jj=0; jj<count[i-1]; jj++)
                                {
                                    printf("i=%d, jj=%d, s=%s, c=%d
    ", i, jj, pson[i-1][jj], c[i-1][jj]);
                                }
                                printf("===
    ");
                            }
                            */
                            if (strcmp(ptmp, pson[i-1][k]) == 0)
                            {
                                //printf("strcmp==0, i=%d, j=%d, k=%d, ptmp=%s, pson[i-1][k]=%s
    ", i, j, k, 
    //ptmp, pson[i-1][k]);
                                c[i-1][k] ++;
                                break;
                            }
                            else if (strcmp(ptmp, pson[i-1][k]) < 0)
                            {
                                //printf("strcmp<0, i=%d, j=%d, k=%d, ptmp=%s, pson[i-1][k]=%s
    ", i, j, k, 
    //ptmp, pson[i-1][k]);
                                for (m=count[i-1]-1; m>=k; m--)
                                {
                                    pson[i-1][m+1] = pson[i-1][m];
                                    c[i-1][m+1] = c[i-1][m];
                                }
                                pson[i-1][k] = ptmp;
                                c[i-1][k] = 1;
                                count[i-1] ++;
                                break;
                            }
                            //printf("strcmp>0, i=%d, j=%d, k=%d, ptmp=%s, pson[i-1][k]=%s
    ", i, j, k, 
    //ptmp, pson[i-1][k]);
                            if (k == count[i-1]-1)
                            {
                                k ++;
                                pson[i-1][k] = ptmp;
                                c[i-1][k] = 1;
                                count[i-1] ++;
                                break;
                            }
                        }
                    }   
                    //printf("-----i=%d, j=%d, pson[i-1][count[i-1]-1]=%s
    ", i, j, pson[i-1][count[i-1]-1]);
                }           
                /*
                printf("===
    ");
                for (j=0; j<count[i-1]; j++)
                {
                    printf("i=%d, j=%d, s=%s, c=%d
    ", i, j, pson[i-1][j], c[i-1][j]);
                }
                printf("===
    ");
                */
            }
     
            int nk, nm;     
            int ctmp;       
            for (i=1; i<=n; i++)
            {                   
                for (j=0; j<count[i-1]; j++)
                {           
                    //printf("i=%d, j=%d
    ", i, j);
                    if (i==n && j==count[i-1]-1)
                        break;  
                    for (k=1; k<=n-i+1; k++)
                    {           
                        for (m=0; m<count[k-1]; m++) 
                        {           
                            //printf("i=%d, j=%d, k=%d, m=%d
    ", i, j, k, m);
                            if (k==n-i+1 && m>=count[k-1]-1-j)
                                break;
                            if (m==count[k-1]-1)
                            {   
                                nk = k+1;
                                nm = 0;
                            }
                            else
                            {   
                                nk = k;
                                nm = m+1; 
                            }   
                            if (strcmp(pson[k-1][m], pson[nk-1][nm]) > 0)
                            {
                                ptmp = pson[k-1][m];
                                pson[k-1][m] = pson[nk-1][nm];
                                pson[nk-1][nm] = ptmp;
                                ctmp = c[k-1][m];
                                c[k-1][m] = c[nk-1][nm];
                                c[nk-1][nm] = ctmp;
                            }
                        }
                        if (k==n-i+1 && m==count[k-1]-1-j)
                            break;
                    }
                }
                if (i==n && j==count[i-1]-1)
                    break;
            }
     
            for (i=1; i<=n; i++)
            {
                //printf("
    ");
                for (j=0; j<count[i-1]; j++)
                {
                    if (c[i-1][j] >= 2)
                        printf("%s %d
    ", pson[i-1][j], c[i-1][j]);
                }
                //printf("
    ");
            }
        }
     
        return 0;
    }
    /**************************************************************
        Problem: 1149
        User: liangrx06
        Language: C
        Result: Accepted
        Time:400 ms
        Memory:1952 kb
    ****************************************************************/


    编程算法爱好者。
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  • 原文地址:https://www.cnblogs.com/liangrx06/p/5083882.html
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