UVa1352 - Colored Cubes

最新推荐文章于 2017-10-12 08:40:50 发布

原创最新推荐文章于 2017-10-12 08:40:50 发布 · 417 阅读

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本文介绍了一种通过枚举和动态规划算法优化骰子不同面的排列组合问题的方法，旨在寻找最优解，减少不必要的翻转次数。

这代码太好了

#include<bits/stdc++.h>
using namespace std;
/*
int left_[]={4,0,2,3,5,1};
int up[]={2,1,5,0,4,3};
void rot(int* T,int* p){
    int q[6];
    memcpy(q,p,sizeof(q));
    for(int i=0;i<6;i++) p[i]=T[ q[i] ];
}
void enumerate_permutations(){
    int p0[6]={0,1,2,3,4,5};
    printf("int dice24[24][6] = {\n");
    for(int i=0;i<6;i++){
        int p[6];
        memcpy(p,p0,sizeof(p0));
        if(i==0) rot(up,p);
        if(i==1) { rot(left_,p); rot(up,p); }
        if(i==3) { rot(up,p); rot(up,p); }
        if(i==4) { rot(left_,p); rot(left_,p); rot(left_,p); rot(up,p); }
        if(i==5) { rot(left_,p); rot(left_,p); rot(up,p); }
        for(int j=0;j<4;j++){
            printf("{%d, %d, %d, %d, %d, %d},\n",p[0],p[1],p[2],p[3],p[4],p[5]);
            rot(left_,p);
        }
    }
    printf("};\n");
}
int main()
{
    enumerate_permutations();
    return 0;
}*/
int dice24[24][6] = {
{2, 1, 5, 0, 4, 3},
{2, 0, 1, 4, 5, 3},
{2, 4, 0, 5, 1, 3},
{2, 5, 4, 1, 0, 3},
{4, 2, 5, 0, 3, 1},
{5, 2, 1, 4, 3, 0},
{1, 2, 0, 5, 3, 4},
{0, 2, 4, 1, 3, 5},
{0, 1, 2, 3, 4, 5},
{4, 0, 2, 3, 5, 1},
{5, 4, 2, 3, 1, 0},
{1, 5, 2, 3, 0, 4},
{5, 1, 3, 2, 4, 0},
{1, 0, 3, 2, 5, 4},
{0, 4, 3, 2, 1, 5},
{4, 5, 3, 2, 0, 1},
{1, 3, 5, 0, 2, 4},
{0, 3, 1, 4, 2, 5},
{4, 3, 0, 5, 2, 1},
{5, 3, 4, 1, 2, 0},
{3, 4, 5, 0, 1, 2},
{3, 5, 1, 4, 0, 2},
{3, 1, 0, 5, 4, 2},
{3, 0, 4, 1, 5, 2},
};
const int maxn = 4;
int n,dice[maxn][6], ans;

vector<string> names;
int ID(const char* name){
    string s(name);
    int n=names.size();
    for(int i=0;i<n;i++){
        if(names[i]==s) return i;
    }
    names.push_back(s);
    return n;
}
int r[maxn],color[maxn][6];

void check(){
    for(int i=0;i<n;i++)
        for(int j=0;j<6;j++) color[i][dice24[ r[i] ][j] ] = dice[i][j];

    int tot=0;
    for(int j=0;j<6;j++){
        int cnt[maxn*6];
        memset(cnt,0,sizeof(cnt));
        int maxface=0;
        for(int i=0;i<n;i++)
            maxface = max(maxface, ++cnt[color[i][j] ]);
        tot+= n-maxface;
    }
    ans = min(ans,tot);
}

void dfs(int d){
    if(d==n) check();
    else for(int i=0;i<24;i++){
        r[d]=i;
        dfs(d+1);
    }
}

int main()
{
    while(scanf("%d",&n)==1&&n){
        names.clear();
        for(int i=0;i<n;i++)
        for(int j=0;j<6;j++){
            char name[30];
            scanf("%s",name);
            dice[i][j]=ID(name);
        }
        ans = n*6;
        r[0]=0;
        dfs(1);
        printf("%d\n",ans);
    }
    return 0;
}