ZOJ 3736 Pocket Cube（暴力，魔方）

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魔方的题，暴力睡过。1930ms，差点久挂了。
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <cmath>#include <cstring>#include <iostream>#include <queue>#include <map>#include <stack>#include <string>#include <algorithm>#include <vector>#include <list>#include <deque>#define LL long long#define DB double#define SI(a) scanf("%d",&a)#define SD(a) scanf("%lf",&a)#define SS(a) scanf("%s",a)#define PF printf#define MM(a,b) memset(a,b,sizeof(a))#define REP(i,a,b) for(int i=a;i<b;i++) #define REPD(i,a,b) for(int i=a;i>b;i--)#define INF 0x3f3f3f3f#define EPS 1e-8#define bug puts("bug")using namespace std;#define N 25int tun1[3][4] = {    2,3,1,0,    7,6,12,13,    8,9,15,14};int tun2[3][8] = {    4,5,6,7,8,9,23,22,    3,2,5,11,16,17,14,8,    23,21,19,17,13,7,3,1,};struct nod{    int re[N];    void in() {        REP(i,0,24) SI(re[i]);    }    int cc(int a,int b,int c,int d) {        if(re[a]==re[b]&&re[b]==re[c]&&re[c]==re[d]) return 1;        return 0;    }    int get() {        int ret =0;        if(cc(0,1,2,3)) ret++;        if(cc(4,5,11,10)) ret++;        if(cc(6,7,12,13)) ret++;        if(cc(8,9,14,15)) ret++;        if(cc(16,17,18,19)) ret++;        if(cc(20,21,22,23)) ret++;        return ret;    }    void oor(int k) {        int t;        t = re[tun1[k][3]];        REPD(i,3,0) re[tun1[k][i]] = re[tun1[k][i-1]];        re[tun1[k][0]] = t;        t = re[tun2[k][7]];        REPD(i,7,0) re[tun2[k][i]] = re[tun2[k][i-1]];        re[tun2[k][0]] = t;        t = re[tun2[k][7]];        REPD(i,7,0) re[tun2[k][i]] = re[tun2[k][i-1]];        re[tun2[k][0]] = t;    }    void l1(int k) {        oor(k);    }    void l2(int k) {        oor(k);oor(k);        oor(k);    }} ini,tmp;int pow(int k) {    int ret = 1;    while(k--) {        ret *= 6;    }    return ret;}int ans = 0;void solve(int c,int k) {    for(int i=0;i<c;i++) {        int t = k%6;k/=6;        if(t<3)           tmp.l1(t);        else           tmp.l2(t-3);     }    ans = max(ans,tmp.get());}int main(){    #ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    int n;    while(~SI(n)) {        ini.in();        ans =0;        for(int i=0;i<=n;i++) {            int k = pow(i);            for(int j=0;j<k;j++){                tmp = ini;                solve(i,j);            }        }        PF("%d\n",ans);    }    return 0;}