使用hash表加速尋找-POJ 3349

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

21 2 3 4 5 64 3 2 1 6 5

Sample Output

Twin snowflakes found.

題目大意：每片雪花有 6 個角長度的值，對於不同的雪花而言，這些值可能被反轉，並且開頭的長度未必是一致的

例如 1 2 3 4 5 6和 4 3 2 1 6 5，雖然他們表面上不同，但是經過反轉和位移以後，就是相同的雪花。 

在輸入資料中如果有相同的雪花，則輸出yes，否則輸出no。

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int HN = 9999; //HASH表的大小int n;struct Node {int index;Node * next;};int snow[100002][6]; //儲存資料Node nodes[100002]; //鏈表Node hashtable[HN]; //hash表，儲存鏈表頭//比較s1 和 s2 是否相同bool check(int s1, int s2) {bool flag;int temp;for (int j = 0; j < 6; j++) {flag = true;for (int step = 0; step < 6; step++) {if (snow[s1][(j + step) % 6] != snow[s2][step]) {flag = false;break;}}if (flag)return true;}for (int j = 0; j < 6; j++) {flag = true;for (int step = 0; step < 6; step++) {temp = j - step;if (temp < 0)temp += 6;if (snow[s1][temp] != snow[s2][step]) {flag = false;break;}}if (flag)return true;}return false;}int sum;bool solve() {for (int i = 1; i <= n; i++) {sum = 0;for (int j = 0; j < 6; j++) {scanf("%d", &snow[i][j]);sum += snow[i][j];}int t = sum % HN;//採用頭部插入法nodes[i].index = i;nodes[i].next = hashtable[t].next;hashtable[t].next = &nodes[i];}//在hash表中比較每個鏈表，一個鏈表即有衝突的節點集合for (int i = 0; i < HN; i++) {for (Node * n1 = hashtable[i].next;  n1 != NULL  ; n1 = n1->next)for (Node * n2 = n1->next; n2 != NULL ; n2 = n2->next)if (check(n1->index, n2->index))return true;}return false;}int main() {freopen("in.txt","r", stdin);cin >> n;memset(hashtable, 0, sizeof(hashtable));memset(nodes, 0, sizeof(nodes));if(solve())printf("Twin snowflakes found.\n");elseprintf("No two snowflakes are alike.\n");return 0;}