| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 39324 | Accepted: 10298 |
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.
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.
思路
简单哈希题，用分离链接表解决冲突，另外用邻接表代替链表，为了将地址冲突降低到最小，这里取了10n（n = 100000）中最大的素数。输入雪花的信息，求出其key值，若在hash[]中还没出现，直接添加，否则说明此前出现过key值相同的雪花，因为key值相同的雪花的信息保存在一个链表中，所以遍历链表判断是否存在两片相同的雪花。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int mod = 999983;
const int maxn = 100005;
int tot = 0,head[mod+5],num[maxn][6],next[maxn];
int IndexHash(int id)
{
int hashval = 0;
for (int i = 0;i < 6;i++) hashval = (hashval%mod+num[id][i]%mod)%mod;
return hashval%mod;
}
bool cmp(int id1,int id2)
{
bool flag;
for (int i = 0;i < 6;i++) //顺时针比较雪花的各个角
{
flag = true;
for (int st = i,j = 0; j < 6;j++,st = (st+1) == 6?0:st+1)
{
if (num[id1][st] != num[id2][j])
{
flag = false;
break;
}
}
if (flag) return true;
}
for (int i = 0;i < 6;i++) //逆时针比较雪花的各个角
{
flag = true;
for (int st = i,j = 5;j >= 0;j--,st = (st+1)==6?0:(st+1))
{
if (num[id1][st] != num[id2][j])
{
flag = false;
break;
}
}
if (flag) return true;
}
return false;
}
void addnode(int hashval)
{
next[tot] = head[hashval];
head[hashval] = tot++;
}
bool solve(int id)
{
int hashval = IndexHash(id);
for (int i = head[hashval]; i != -1;i = next[i])
{
if (cmp(i,id)) return true;
}
addnode(hashval);
return false;
}
int main()
{
int N;
while (~scanf("%d",&N))
{
memset(head,-1,sizeof(head));
bool flag = false;
for (int i = 0;i < N;i++)
{
for (int j = 0;j < 6;j++) scanf("%d",&num[i][j]);
if (flag) continue;
flag = solve(i);
}
if (flag) printf("Twin snowflakes found.\n");
else printf("No two snowflakes are alike.\n");
}
return 0;
}