FZU - 2107 Hua Rong Dao 回溯

题目大意：给出一个N*4的方格，要求你用2*2，2*1，1*2，1*1的格子将其填满，2*2的格子必须要有，问有多少中填满N*4方格的方法

解题思路：回溯

#include<cstdio>
#include<cstring>
int ans[5] = {0,0,18}; 
int vis[5][4],res;
bool flag;

bool judge(int cur) {
	if(!flag)
		return false;
	for(int i = 0; i < cur; i++)
		for(int j = 0; j < 4; j++)
			if(!vis[i][j])
				return false;
	return true;
}

void DFS(int x,int y,int cur) {

	if(judge(cur)) {
		res++;
		return ;
	}

	if(y == 4) {
		x++;
		y = 0;
	}

	for(int i = x; i < cur; i++)
		for(int j = y; j < 4; j++) {
			if(!flag && i + 1 < cur && j + 1 < 4 && !vis[i][j] && !vis[i][j+1] && !vis[i+1][j] && !vis[i+1][j+1]) {
				vis[i][j] = vis[i][j+1] = vis[i+1][j] = vis[i+1][j+1] = 1;
				flag = true;
				DFS(i,j+2,cur);
				flag = false;
				vis[i][j] = vis[i][j+1] = vis[i+1][j] = vis[i+1][j+1] = 0;
			}
			if(i + 1 < cur && !vis[i][j] && !vis[i+1][j]) {
				vis[i][j] = vis[i+1][j] = 1;
				DFS(i,j+1,cur);	
				vis[i][j] = vis[i+1][j]	= 0;
			}
			if(j + 1 < 4 && !vis[i][j] && !vis[i][j+1]) {
				vis[i][j] = vis[i][j+1] = 1;
				DFS(i,j+2,cur);
				vis[i][j] =	vis[i][j+1] = 0;
			}

			if(!vis[i][j]) {
				vis[i][j] = 1;
				DFS(i,j+1,cur);
				vis[i][j] = 0;
			}
			if(j == 3)
				y = 0;
		}
}

int main() {
	int test,temp;
	memset(vis,0,sizeof(vis));
	flag = false;
	res = 0;
	DFS(0,0,3);
	ans[3] = res;

	memset(vis,0,sizeof(vis));
	flag = false;
	res = 0;	
	DFS(0,0,4);
	ans[4] = res;

	scanf("%d",&test);
	while(test--) {
		scanf("%d",&temp);
		printf("%d\n",ans[temp]);	
	}
	return 0;
}