标签:nbsp next can efi opp oppo ext 公式 lock
题目难点在于对两片雪花的比较,哈希可以加快搜索速度,防止超时,而对于如何逆时针和顺时针比较雪花是否相同便成为重点。
在这里给出两条公式:
设i为A、B的第i片叶子,j为B当前顺时针转过的格数
那么 A(i) ---> B( (i+j)%6 )
设i为A、B的第i片叶子,j为B当前逆时针转过的格数
那么 A(i) ---> B( (5-i-j+6)%6 )
#include <iostream> using namespace std; #define mod 999983 typedef struct Hashtable { int arm[6]; struct Hashtable * next; }Hashtable; int Hash(int a[]){ int key=0; for (int i = 0; i<6; i++) key += a[i] % mod; return key%mod; } Hashtable hash1[mod]; void init(){ for (int i = 0; i<mod; i++) hash1[i].next = NULL; } bool clock(int a[], int b[]){ for (int i = 0; i<6; i++){ bool flag = true; for (int j = 0; j<6; j++){ if (a[j] != b[(i + j) % 6]){ flag = false; break; } } if (flag) return true; } return false; } bool opposite(int a[], int b[]){ for (int i = 0; i<6; i++){ bool flag = true; for (int j = 0; j<6; j++){ if (a[j] != b[(5 - i - j + 6) % 6]){ flag = false; break; } } if (flag) return true; } return false; } int main(){ int n; scanf("%d", &n); init(); int snow[6]; bool a = false; bool b = false; for (int i = 0; i<n; i++){ for (int j = 0; j<6; j++) scanf("%d", &snow[j]); int key = Hash(snow); Hashtable * node=new Hashtable; if (hash1[key].next == NULL){ hash1[key].next = node; node->next = NULL; for (int x = 0; x<6; x++) node->arm[x] = snow[x]; } else{ node->next = hash1[key].next; hash1[key].next = node; for (int x = 0; x<6; x++) node->arm[x] = snow[x]; Hashtable * p = node; while (p->next != NULL){ p = p->next; a = clock(p->arm, snow); b = opposite(p->arm, snow); if (a || b){ printf("Twin snowflakes found.\n"); return 0; } } } } if (!a&&!b) printf("No two snowflakes are alike.\n"); return 0; }
标签:nbsp next can efi opp oppo ext 公式 lock
原文地址:http://www.cnblogs.com/lvcoding/p/7493306.html