【POJ 3349】 Snowflake Snow Snowflakes

【题目链接】

          http://poj.org/problem?id=3349

【算法】

           哈希

           若两片雪花相同，则它们六个角上的和一定相同，不妨令 H(A) = sigma(Ai) % P ,每次只要到哈希表里查是否出现相同雪花，即可

【代码】

       

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 100010
const int P = 1e5 + 7;

int i,j,n,tot;
int a[7],Head[P],nxt[MAXN],snow[MAXN][7];

inline int get()
{
        int i,ret = 0;
        for (i = 1; i <= 6; i++) ret = (ret + a[i]) % P;
        return ret;
}
inline bool Insert()
{
        int i;
        int Hash = get();
        bool same;
        for (i = Head[Hash]; i; i = nxt[i])
        {
                same = true;
                for (j = 1; j <= 6; j++) same &= (snow[i][j] == a[j]);
                if (same) return true;         
        }        
        tot++;
        nxt[tot] = Head[Hash];
        for (i = 1; i <= 6; i++) snow[tot][i] = a[i];
        Head[Hash] = tot;
        return false;
}

int main() 
{
        
        scanf("%d",&n);
        for (i = 1; i <= n; i++)
        {
                for (j = 1; j <= 6; j++) scanf("%d",&a[j]);        
                sort(a+1,a+7);
                if (Insert()) 
                {
                        printf("Twin snowflakes found.
");
                        return 0;
                }
        }
        printf("No two snowflakes are alike.
");
        
        return 0;
    
}