poj3007 Organize Your Train part II， 字符串hash

题意：

给定一个字符串，从任意位置把它切为两半，得到两条子串

定义 子串1为s1，子串2为s2，子串1的反串为s3，子串2的反串为s4

现在从s1 s2 s3 s4中任意取出两个串组合，问有多少种不同的组合方法

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#define for0(a,b) for(a=0;a<b;++a)
#define for1(a,b) for(a=1;a<=b;++a)
#define foru(i,a,b) for(i=a;i<=b;++i)
#define ford(i,a,b) for(i=a;i>=b;--i)
using namespace std;

typedef double db;
typedef long long ll;
const db eps = 1e-8;
const int inf = 1e9;
const int M = 1e4;
const int maxn = 80;

char s[maxn], p[maxn], q[maxn], rp[maxn], rq[maxn], str[maxn];
char val[M][maxn];
int cnt, n;

//BKDR Hash Function
int hash(char *v){
    unsigned int seed = 131;
    unsigned int hash = 0;
    for(int i=0; v[i]; ++i)
        hash = hash * seed + v[i];
    return (hash & 0x7FFFFFFF) % M;
}

void contact(char *a, char *b){
    int k = 0;
    for(int i=0; a[i]; ++i) str[k++] = a[i];
    for(int i=0; b[i]; ++i) str[k++] = b[i];
    str[k] = '\0';
}

void insert(char *a, char *b){
    contact(a, b);
    int k = hash(str);
    while( (strlen(val[k])>0) && strcmp(str,val[k])!=0){
        k = (k+1) % M;
    }
    if(strlen(val[k])== 0){
        strcpy(val[k], str);
        cnt++;
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.cpp", "r", stdin);
    freopen("out.cpp","w", stdout);
#endif // ONLINE_JUDGE
    int n, i, j;
    scanf("%d", &n);
    while(n--)
    {
        scanf("%s", s);
        for0(i,M) val[i][0]='\0';
        int len = strlen(s);
        int k;
        cnt = 0;
        for1(i,len-1) {
            k = 0;
            for0(j,i){
                p[j] = s[j];
                rp[k++] = s[i-j-1];
            }
            p[k] = rp[k] = '\0';
            k = 0;
            foru(j,i,len-1) {
                q[j-i] = s[j];
                rq[k] = s[len-k-1];
                ++k;
            }
            q[k] = rq[k] = '\0';
            insert(p, q); insert(q, p);
            insert(p, rq); insert(rq, p);
            insert(q, rp); insert(rp, q);
            insert(rp, rq); insert(rq, rp);
        }
        printf("%d\n", cnt);
    }
    return 0;
}