POJ 3349 Snowflake Snow Snowflakes (哈希表)

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题意:给出n组序列, 每组序列6个数, 判断是否存在两个序列同构(即某个序列从某个位置起顺时针读出来的数与另外一个序列1-6的数一一对应)

解法: 具体看代码, 这题解得比较菜。。3000+ms。。。

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<vector>using namespace std;const int maxn = 100010;bool flag=false;int num[maxn][8];vector<int>vec[maxn];void deal(int v, int sum){int i,j;if(vec[sum].size() == 0) vec[sum].push_back(v); //当前key值对应的vector数组没有存储else {//与之前输入的序列中key值与它相同的序列做比较int len = vec[sum].size();for(i=0; i<len; i++){int p = vec[sum][i];for(j=0; j<6; j++){ //顺时针if(num[p][j%6] != num[v][0])  continue;if(num[p][(j+1)%6] != num[v][1]) continue;if(num[p][(j+2)%6] != num[v][2]) continue;if(num[p][(j+3)%6] != num[v][3]) continue;if(num[p][(j+4)%6] != num[v][4]) continue;if(num[p][(j+5)%6] != num[v][5]) continue;flag=true;return ;}for(j=5; j>=0; j--){//逆时针if(num[p][j%6] != num[v][0]) continue;if(num[p][(j+5)%6] != num[v][1]) continue;if(num[p][(j+4)%6] != num[v][2]) continue;if(num[p][(j+3)%6] != num[v][3]) continue;if(num[p][(j+2)%6] != num[v][4]) continue;if(num[p][(j+1)%6] != num[v][5]) continue;flag=true;return ;}}if(!flag) vec[sum].push_back(v);}return ;}int main(){int i, j, n;memset(num, 0, sizeof(num));scanf("%d", &n);for(i=1;i<=n; i++){int sum=0;for(j=0; j<6; j++){scanf("%d", &num[i][j]);sum+=num[i][j];//将六个数字加起来作为key值}if(!flag) deal(i, sum%100007);}if(flag){printf("Twin snowflakes found.\n");} else {printf("No two snowflakes are alike.\n");}return 0;}

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