磁磚樣式 藍橋真題

老套路 每次遞歸 都找最左上角的空白位置 這個地方是必須要填滿的 如果放不下合適的瓷磚直接返回即可 具有很強的剪枝效果

遞歸時的參數爲已鋪的面積 若已鋪滿 則判重後累加答案即可

 

#include <bits/stdc++.h>
using namespace std;
const int maxn=20;

map <int,bool> mp;
int book[maxn][maxn];
int n=3,m=10,sum=30;

void getpos(int &x,int &y)
{
    int i,j;
    for(i=1;i<=n;i++){
        for(j=1;j<=m;j++){
            if(book[i][j]==-1){
                x=i,y=j;
                return;
            }
        }
    }
}

bool judgeII(int x,int y)
{
    if(1<x&&1<y){
        if(book[x-1][y-1]==book[x-1][y]&&book[x-1][y]==book[x][y]&&book[x][y]==book[x][y-1]) return 0;
        else return 1;
    }
    else return 1;
}

bool judgeI(int x,int y)
{
    return judgeII(x,y)&&judgeII(x,y+1)&&judgeII(x+1,y)&&judgeII(x+1,y+1);
}

int dfs(int area)
{
    int res,i,j,x,y,val,cnt;
    if(area==sum){
        val=0,cnt=0;
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                val+=book[i][j]*(1<<cnt);
                cnt++;
            }
        }
        if(!mp[val]){
            mp[val]=1;
            return 1;
        }
        else return 0;
    }
    getpos(x,y);
    //printf("*%d %d*\n",x,y);
    res=0;
    if(x<n&&book[x+1][y]==-1){
        book[x][y]=book[x+1][y]=0;
        if(judgeI(x,y)&&judgeI(x+1,y)){
            res+=dfs(area+2);
        }
        book[x][y]=book[x+1][y]=1;
        if(judgeI(x,y)&&judgeI(x+1,y)){
            res+=dfs(area+2);
        }
        book[x][y]=book[x+1][y]=-1;
    }
    if(y<m&&book[x][y+1]==-1){
        book[x][y]=book[x][y+1]=0;
        if(judgeI(x,y)&&judgeI(x,y+1)){
            res+=dfs(area+2);
        }
        book[x][y]=book[x][y+1]=1;
        if(judgeI(x,y)&&judgeI(x,y+1)){
            res+=dfs(area+2);
        }
        book[x][y]=book[x][y+1]=-1;
    }
    return res;
}

int main()
{
    memset(book,-1,sizeof(book));
    printf("%d\n",dfs(0));
    return 0;
}