洛谷例14-1四阶数独-枚举序列方法

写出来了很开心。

321——上代码！

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct rect{
	int r[4][4];
	rect(){
		memset(r,0,sizeof(r));
	}
	bool column(){
		int judge[5]={0};
		for(int j=0;j<4;j++){
		    for(int i=0;i<4;i++){		    	
		    	judge[ r[i][j] ]++;
			}
			for(int t=1;t<=4;t++){
				if (judge[t]==0 || judge[t]>1) return 0;
			}
			memset(judge,0,sizeof(judge));
	    }
		return 1;
	}
	
	bool block(){
		
		for(int j=0;j<4;j+=2){
		    for(int i=0;i<4;i+=2){
			    int judge[5]={0};		    	
		    	judge[ r[i][j] ]++;
		    	judge[ r[i][j+1] ]++;
		    	judge[ r[i+1][j] ]++;
		    	judge[ r[i+1][j+1] ]++;
		    	for(int t=1;t<=4;t++){
				    if (judge[t]==0 || judge[t]>1) return 0;
			    }			    
		    }
  	    }
		return 1;
	}
	
};
int main()
{
	
	int a[4]={1,2,3,4},b[24][4]={{1,2,3,4}},i=1,ans=0;
	while(next_permutation(a,a+4)){
	    memcpy(b[i],a,sizeof(a));
	    i++;
	}
	for(int i=0;i<24;i++)
		for(int j=i+1;j<24;j++)
		    for(int m=j+1;m<24;m++)
			    for(int n=m+1;n<24;n++){
			    	rect rec;
			    	memcpy(rec.r[0],b[i],sizeof(b[i]));
			    	memcpy(rec.r[1],b[j],sizeof(b[j]));
			    	memcpy(rec.r[2],b[m],sizeof(b[m]));
			    	memcpy(rec.r[3],b[n],sizeof(b[n]));
			    	//列 
			    	if(!rec.column()) continue;
					//块
					if(!rec.block()) continue;
					ans++;
				} 
	cout<<ans;
}