USACO-Contact

来源：http://ace.delos.com/usacoprob2?a=5LTgWx8eTT9&S=contact

 

这题的想法是，把01串转换成二进制，用hash表存储。

为了区分0和00等类似的情况，将所有的子串的高位加个1，例如：

0就用10来存储，00用100存储。

hash表统计子串个数，然后排序输出就是了。

不过如果用string存储输出的数据，直接输出string，总是会多出一些不可见字符，只能一个一个字符输出了。

 

/*
ID:ay27272
PROG:contact
LANG:C++
*/

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
using namespace std;


char s[300005];
char temp[300000] = {0};
string out[1000];

#define NN 30000

struct data
{
    data(int count, int id): count(count), id(id) {}
    data() { count = id = 0; }
    int count, id;
    friend bool operator < (data a, data b)
    {
        return a.count>b.count;
    }
} sum[30000];

string change(int t)
{
    string p;
    int i;
    for (i=0; t>0; i++)
    {
        p = p + (char)(t%2 + '0');
        t /= 2;
    }
    p[i-1] = '\0';
    return p;
}



int main()
{
    freopen("contact.in","r",stdin);
    freopen("contact.out","w",stdout);
    int A, B, N;
    scanf("%d%d%d", &A, &B, &N);
    
    for (int i=0; i<NN; i++)
        sum[i] = data(0, i);

    int num, kk;
    memset(s, 0, sizeof(s));
    int ll=0;
    while (scanf("%s", temp) != EOF )
    {
        for (int i=0; temp[i]; i++)
            s[ll++] = temp[i];
        memset(temp, 0, sizeof(temp));
    }

    for (int i=0; s[i+A-1]; i++)
    {
        kk = 1; num = 0;
        for (int j=0; s[i+j] && j<A-1; j++)
        {
            num += kk*(s[i+j] - '0');
            kk *= 2;
        }
        for (int j=A-1; s[i+j] && j<B; j++)
        {
            num += kk*(s[i+j] - '0');
            kk *= 2;
            sum[num + kk].count++;
        }
    }

    sort(sum, sum+NN);


    int xx, ii;
    int i=0, count = 0;
    while (i<NN && count<N)
    if (sum[i].count)
    {
        count++;
        printf("%d\n", sum[i].count);
        xx = sum[i].count;
        ii = 0;
        while (sum[i].count == xx)
        {
            out[ii] = change(sum[i].id);
            i++;
            ii++;
        }
        for (int si=0; si<ii; si++)
            for (int sj=si+1; sj<ii; sj++)
                if (out[si].length() > out[sj].length()
                    || (out[si].length() == out[sj].length() && out[si] > out[sj]))
                {
                    string temp = out[si]; out[si] = out[sj]; out[sj] = temp;
                }
        for (int si=0; si<ii-1; si++)
        {
            for (int sj=0; sj<out[si].length()-1; sj++)
                printf("%c", out[si][sj]);
            if (si%6==5) printf("\n");
            else printf(" ");
        }
        for (int sj=0; sj<out[ii-1].length()-1; sj++)
            printf("%c", out[ii-1][sj]);
        printf("\n");
    }
    else i++;
    return 0;
}