LA 3401 Colored Cubes

题目：Colored Cubes

思路：枚举每一个正方体的状态，把所有正方体的每一个面都变成在这个面上出现最多的颜色。

注意：

1、不用枚举第一个正方体。

2、不要写错变量名。

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <cstring>
#include <map>
using namespace std;

#define maxn 4
#define m 6
#define maxch 24

const int chg[maxch+5][m+5]= {{0},{0,3,2,6,1,5,4},{0,3,1,2,5,6,4},{0,3,5,1,6,2,4},{0,3,6,5,2,1,4},{0,5,3,6,1,4,2},{0,6,3,2,5,4,1},{0,2,3,1,6,4,5},{0,1,3,5,2,4,6},{0,1,2,3,4,5,6},{0,5,1,3,4,6,2},{0,6,5,3,4,2,1},{0,2,6,3,4,1,5},{0,6,2,4,3,5,1},{0,2,1,4,3,6,5},{0,1,5,4,3,2,6},{0,5,6,4,3,1,2},{0,2,4,6,1,3,5},{0,1,4,2,5,3,6},{0,5,4,1,6,3,2},{0,6,4,5,2,3,1},{0,4,5,6,1,2,3},{0,4,6,2,5,1,3},{0,4,2,1,6,5,3},{0,4,1,5,2,6,3}};

int n;
map<string,int> mp;
int cnt;

int a[maxn+5][m+5];

int ans=0;

void init() {
	mp.clear();
	cnt=0;
	ans=(1<<30);
}

vector<int> st;

int find(){
	int b[maxn+5][m+5];
	memcpy(b[1],a[1],sizeof(a[1]));
	for(int i=2;i<=n;i++){
		for(int j=1;j<=m;j++){
			b[i][j]=a[i][chg[st[i-2]][j]];
		}
	}
	int sum=0;
	for(int j=1;j<=m;j++){
		int col[maxn*m+5]={0};
		for(int i=1;i<=n;i++){
			col[b[i][j]]++;
		}
		int s=0,w=0;
		for(int i=1;i<=cnt;i++){
			if(col[i]>s) s=col[i],w=i;
		}
		for(int i=1;i<=n;i++){
			if(b[i][j]!=w) sum++;
		}
	}
	return sum;
}

void dfs(int x){
	if(x==n) {
		ans=min(ans,find());
		return ;
	}
	for(int i=1;i<=maxch;i++){
		st.push_back(i);
		dfs(x+1);
		st.pop_back();
	}
}

int main() {
	while(~scanf("%d",&n)&&n) {
		init();
		for(int i=1; i<=n; i++) {
			for(int j=1; j<=m; j++) {
				string x;
				cin>>x;
				if(!mp.count(x)) mp[x]=++cnt;
				a[i][j]=mp[x];
			}
		}
		dfs(1);
		printf("%d\n",ans);
	}
	return 0;
}