POJ 3349 Snowflake Snow Snowflakes

Time Limit: 4000MS Memory Limit: 65536K

Total Submissions: 24826 Accepted: 6474

Description

 

You may have heard that no two snowflakes are alike. Your task is to write a program to determine whether this is really true. Your program will read information about a collection of snowflakes, and search for a pair that may be identical. Each snowflake has six arms. For each snowflake, your program will be provided with a measurement of the length of each of the six arms. Any pair of snowflakes which have the same lengths of corresponding arms should be flagged by your program as possibly identical.

 

Input

 

The first line of input will contain a single integer n, 0 < n ≤ 100000, the number of snowflakes to follow. This will be followed by n lines, each describing a snowflake. Each snowflake will be described by a line containing six integers (each integer is at least 0 and less than 10000000), the lengths of the arms of the snow ake. The lengths of the arms will be given in order around the snowflake (either clockwise or counterclockwise), but they may begin with any of the six arms. For example, the same snowflake could be described as 1 2 3 4 5 6 or 4 3 2 1 6 5.

 

Output

 

If all of the snowflakes are distinct, your program should print the message:

No two snowflakes are alike.

If there is a pair of possibly identical snow akes, your program should print the message:

Twin snowflakes found.

 

Sample Input

 

2

1 2 3 4 5 6

4 3 2 1 6 5

Sample Output

 

Twin snowflakes found.

Source

 

CCC 2007

 这道题目用哈希做，以前没用过，在用的时候犯糊涂，用哈希可能会有冲突，一定要有处理冲突的方法，这道题目在旋转的时候不必从六个点都进行正反方向的旋转，这样会超时，只要记录一下最小值，从最小值进行顺时针和逆时针的旋转就好了。注意最小值不止一个啊，可能有多个同时是最小值

[plain]

#include <stdio.h>  

#include <string.h>  

struct num  

{  

    int pos;  

    int next;  

} a[1000000];  

int b[1000003],c[10],e[7],vex[200010][7];  

int maxlen=1000003,s;  

int check[1000003];  

int main()  

{  

    int i,j,n,m,t,sum,tag,x,y,min,k,u;  

    scanf("%d",&n);  

    tag=0;  

    s=1;  

    memset(b,-1,sizeof(b));  

    memset(check,0,sizeof(check));  

    for(i=1; i<=n; i++)  

    {  

        min=0x7fffffff;k=0;  

        for(j=1; j<=6; j++)  

        {  

            scanf("%d",&c[j]);  

            if(min>c[j])  

            {  

                min=c[j];k=0;  

                e[k++]=j;  

            }else if(min==c[j])  

            {  

                e[k++]=j;  

            }  

        }  

        if(!tag)  

        {  

            for(u=0;u<=k-1;u++)  

            {  

                j=e[u];  

                y=1;  

                sum=c[j]%maxlen;  

                vex[s][y++]=c[j];  

                for(x=j-1; x>=1; x--)  

                {  

                    vex[s][y++]=c[x];  

                    sum=(sum*10%maxlen+c[x]%maxlen)%maxlen;  

                }  

                for(x=6; x>=j+1; x--)  

                {  

                    vex[s][y++]=c[x];  

                    sum=(sum*10%maxlen+c[x]%maxlen)%maxlen;  

                }  

                if(!check[sum])  

                {  

                    check[sum]=1;  

                    a[s].pos=s;  

                    a[s].next=b[sum];  

                    b[sum]=s;  

                    s++;  

                }  

                else  

                {  

                    for(x=b[sum]; x!=-1; x=a[x].next)  

                    {  

                        for(y=1; y<=6; y++)  

                        {  

                            if(vex[s][y]!=vex[a[x].pos][y])  

                            {  

                                break;  

                            }  

                        }  

                        if(y==7)  

                        {  

                            break;  

                        }  

              &nbs

补充：软件开发 , C++ ,