2013长沙现场赛 K . Pocket Cube

最新推荐文章于 2018-06-01 15:36:46 发布

Lanifer

最新推荐文章于 2018-06-01 15:36:46 发布

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分类专栏： ZOJ

本文链接：https://blog.youkuaiyun.com/lg_csu/article/details/16948751

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本文分享了一次比赛中代码调试的经历，通过对错误代码的分析，总结了代码风格的重要性，并提供了优化后的代码示例，展示了清晰易读的代码如何提高调试效率。

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题目链接

通过一次比赛，发现自己还是有很大差距。。

这题确实很简单，但现场时我没调出来，泪奔。。。

说明我得代码能力还是很弱的。现场时敲出来的代码风格非常不好，及难调试。

错误代码: (看上去很乱，难以检查，与别人代码对照后发现是表打错了。。)

/*
一份正确的表：

const int E[6][24]={
   {0,21,2,23,4,5,6,1,9,15,10,11,12,3,8,14,16,7,18,13,20,17,22,19},
   {0,7,2,13,4,5,6,17,14,8,10,11,12,19,15,9,16,21,18,23,20,1,22,3},
   {2,0,3,1,6,7,8,9,23,22,10,11,12,13,14,15,16,17,18,19,20,21,5,4},
   {1,3,0,2,23,22,4,5,6,7,10,11,12,13,14,15,16,17,18,19,20,21,9,8},
   {0,1,11,5,4,16,12,6,2,9,10,17,13,7,3,15,14,8,18,19,20,21,22,23},
   {0,1,8,14,4,3,7,13,17,9,10,2,6,12,16,15,5,11,18,19,20,21,22,23}
};
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;

const int E[6][24] = {  // 有错误
   {0,7,2,13,4,5,6,17,14,8,10,11,12,19,15,9,16,21,18,23,20,1,22,3},
   {0,21,2,23,4,5,6,1,8,9,10,11,12,3,14,15,16,7,18,13,20,17,22,19},
   {0,1,5,2,4,11,12,6,3,9,10,16,13,7,8,15,17,14,18,19,20,21,22,23},
   {0,1,3,8,4,2,7,13,14,9,10,5,6,12,17,15,11,16,18,19,20,21,22,23},
   {0,1,2,3,4,5,6,7,8,9,20,10,11,12,13,14,18,16,19,17,21,15,22,23},
   {0,1,2,3,4,5,6,7,8,9,11,12,13,14,15,21,17,19,16,18,10,20,22,23}
};

int N;
int sum;

struct STA{
    int c[24];
};

STA change(const STA &sta,int t){
    STA ret;
    for(int i=0;i<24;i++){
        ret.c[i] = sta.c[E[t][i]];
    }
    return ret;
}

int faces(const STA &s){
    int ans = 0;
    if(s.c[1] ==s.c[0]  && s.c[2]==s.c[0]   && s.c[3] ==s.c[0] )  ans++;
    if(s.c[7] ==s.c[6]  && s.c[12]==s.c[6]  && s.c[13]==s.c[6] )  ans++;
    if(s.c[5] ==s.c[4]  && s.c[10]==s.c[4]  && s.c[11]==s.c[4] )  ans++;
    if(s.c[9] ==s.c[8]  && s.c[14]==s.c[8]  && s.c[15]==s.c[8] )  ans++;
    if(s.c[17]==s.c[16] && s.c[18]==s.c[16] && s.c[19]==s.c[16])  ans++;
    if(s.c[21]==s.c[20] && s.c[22]==s.c[20] && s.c[23]==s.c[20])  ans++;
    return ans;
}

void DFS(STA s,int d){
    sum = max(sum , faces(s));
    for(int t=0;t<6;t++){
        STA tmp = change(s,t);
        if(d<N) DFS(tmp , d+1);
    }
}


int main()
{
    while(scanf("%d",&N)!=EOF){
        STA beg;
        for(int i=0;i<24;i++) scanf("%d",&beg.c[i]);
        sum = 0;
        DFS(beg ,0);
        printf("%d\n",sum);
    }
    return 0;
}

今天重写了一份：（更显清晰，便于查错）AC

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;
typedef int STA[24];

const int F[6][4] = {
    {0,1,2,3},
    {4,5,10,11},
    {6,7,12,13},
    {8,9,14,15},
    {16,17,18,19},
    {20,21,22,23}
};
const int R1[3][4] = {
    {6,7,13,12},
    {16,17,19,18},
    {8,9,15,14}
};
const int R2[3][8]= {
    {2,3,8,14,17,16,11,5},
    {12,13,14,15,21,20,10,11},
    {3,1,23,21,19,17,13,7}
};

int N;
int sum;
STA state[10];

int num_face(const STA &s){
    int ans = 0;
    for(int i=0;i<6;i++){
        bool ok = true;
        for(int j = 1;j<4;j++)
            if(s[F[i][j]] != s[F[i][0]]) { ok = false; break;}
        if(ok) ans++;
    }
    return ans;
}
void trans(const STA &a,STA &b,int t){
    memcpy(b,a,sizeof(STA));
    int r = t/2;
    if(t&1){
        for(int i=0;i<4;i++) b[R1[r][i]] = a[R1[r][(i+1)%4]];
        for(int i=0;i<8;i++) b[R2[r][i]] = a[R2[r][(i+2)%8]];
    }
    else{
        for(int i=0;i<4;i++) b[R1[r][i]] = a[R1[r][(i+3)%4]];
        for(int i=0;i<8;i++) b[R2[r][i]] = a[R2[r][(i+6)%8]];
    }
}

void DFS(int d){
    sum = max(sum , num_face(state[d]));
    if(d>=N) return;
    for(int i=0;i<6;i++){
        trans(state[d] , state[d+1] , i);
        DFS(d+1);
    }
}
int main()
{
    while(scanf("%d",&N)!=EOF){
        sum = 0;
        for(int i=0;i<24;i++) scanf("%d",&state[0][i]);
        DFS(0);
        printf("%d\n",sum);
    }
    return 0;
}