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。
原文地址:https://www.cnblogs.com/liangrx06/p/5083882.html