nyoj-130-相同的雪花(hash)

 题目链接

/*
    Name:NYOJ-130-相同的雪花 
    Copyright:
    Author:
    Date: 2018/4/14 15:13:39
    Description:
    将雪花各个分支上的值加起来，hash
*/
#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 1e5+10;
int snowflakes[MAXN][7];
vector<int> snow[5001];//hash
int Like(int x,int y)//判断两片雪花是否相同 
{
    for(int i=0;i<6;i++)
    {
        if((snowflakes[x][0]==snowflakes[y][i]&&snowflakes[x][1]==snowflakes[y][(i+1)%6]&&snowflakes[x][2]==snowflakes[y][(i+2)%6]&&snowflakes[x][3]==snowflakes[y][(i+3)%6]&&snowflakes[x][4]==snowflakes[y][(i+4)%6]&&snowflakes[x][5]==snowflakes[y][(i+5)%6])///顺时针方向判断两个雪花是否相同
         ||(snowflakes[x][0]==snowflakes[y][(i+5)%6]&&snowflakes[x][1]==snowflakes[y][(i+4)%6]&&snowflakes[x][2]==snowflakes[y][(i+3)%6]&&snowflakes[x][3]==snowflakes[y][(i+2)%6]&&snowflakes[x][4]==snowflakes[y][(i+1)%6]&&snowflakes[x][5]==snowflakes[y][i]))///逆时针方向判断两个雪花是否相同
            return 1;
    }
    return 0;
}
int main()
{
//    freopen("in.txt", "r", stdin);
    int t;
    cin>>t;
    while (t--) {
        memset(snow, 0, sizeof(snow));
        int n;
        cin>>n;
        for (int i=0; i<n; i++) {
            int sum = 0;
            for (int j=0; j<6; j++) {
                int tmp;
                cin>>tmp;
                snowflakes[i][j] = tmp;
                sum += tmp;
            }
            sum = sum % 5000;//防止越界
            snow[sum].push_back(i);
        }
        int flag = 0 ;
        for (int i=0; i<5000; i++) {
            if (flag) break;
            if (snow[i].size() < 2) continue;
            for (int j=0; j<snow[i].size()-1; j++) {
                for (int k=j+1; k<snow[i].size(); k++) {
                    if (Like(snow[i][j], snow[i][k])) {
                        flag = 1;
                        break;
                    }
                    
                }
            }
        }
        if(flag)printf("Twin snowflakes found.\n");  
            else printf("No two snowflakes are alike.\n");  
    }
    return 0;
}