POJ 3349 Snowflake

#include <stdio.h>
#include <stdlib.h>
#define PRI 93961

struct SF{
    int e[6];
    SF *next;
    SF(){ e[0] = -1; next = NULL; }
};

SF snow[PRI];

bool exactsame(SF s1, SF s2){
    for (int i = 0; i < 6; i++){
        if (s1.e[i] != s2.e[i]) return false;
    }
    return true;
}

int mycmp(const void *a, const void *b){
    return *(int *)a - *(int *)b;
}

int main(){
    bool ok = false;
    int nos, key;
    SF *p, tmp;
    scanf("%d", &nos);
    for (int i = 0; i < nos; i++){
        for (int j = 0; j < 6; j++) scanf("%d", &tmp.e[j]);
        if (!ok){
            key = 0;
            qsort(tmp.e, 6, sizeof(tmp.e[0]), mycmp); 
            for (int j = 0; j < 6; j++){
                key += tmp.e[j];
                key %= PRI;
            }
            if (snow[key].e[0] == -1) snow[key] = tmp;
            else if (exactsame(snow[key], tmp)) ok = true;
            else if (snow[key].next == NULL){
                if (exactsame(snow[key], tmp)) ok = true;
                else{
                    SF *newsf = new SF;
                    for (int k = 0; k < 6; k++) newsf->e[k] = tmp.e[k];                   
                    snow[key].next = newsf;
                }
            }
            else{
                p = snow[key].next;
                if (exactsame(*p, tmp)) ok = true;
                while (p->next && !ok){
                    p = p->next;
                    if (exactsame(*p, tmp)) ok = true;
                }                    
                if (!ok){
                    SF *newsf = new SF;
                    for (int k = 0; k < 6; k++) newsf->e[k] = tmp.e[k]; 
                    p->next = newsf;
                }
            }
        }
    }
    if (ok) printf("Twin snowflakes found.\n");
    else printf("No two snowflakes are alike.\n");
    return 0;
}

【开博第一篇(*^__^*) ……~】虽然是水题但是调了一下午TUT。坑点在神马clockwise，counterclockwise都是骗人的。。六条边correspond的匹配就好了~

嗯哼。来说思路，是每个雪花按arm从小到大用qsort排序好，然后用哈希开地址法比较，若比较时有相同则标记ok = true，否则将该雪花存入相应地址。关键码用的是加和取mod。判断部分写得稍微冗杂了些。><