【链接】 我是链接,点我呀:)
【题意】
在这里输入题意
【题解】
先预处理出来一个正方形。
然后每次枚举新加的正方形左上角的坐标就可以。
注意覆盖的规则,控制一下就可以。
然后暴力判断是否相同。
暴力回溯即可(只用回溯一个正方形区域)
【代码】
/*
1.Shoud it use long long ?
2.Have you ever test several sample(at least therr) yourself?
3.Can you promise that the solution is right? At least,the main ideal
4.use the puts("") or putchar() or printf and such things?
5.init the used array or any value?
6.use error MAX_VALUE?
7.use scanf instead of cin/cout?
8.whatch out the detail input require
*/
/*
一定在这里写完思路再敲代码!!!
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 10;
const int M = 5;
char s[N+5][N+5],now[N+5][N+5];
char square[M+5][M+5];
void init(){
for (int i = 0;i <3;i++)
for (int j = 0;j<5;j++)
square[i][j] = ' ';
for (int i = 0;i < 2;i++) square[i+1][0] = square[i+1][4] = '|';
for (int j = 1;j < 5;j+=2) square[0][j]=square[2][j] = '_';
}
void Set(int x,int y){
for (int i = 0;i < 3;i++)
for (int j = 0;j < 5;j++){
if (i<=0 && now[i+x][j+y]!=' ' && square[i][j]==' ') continue;
now[i+x][j+y] = square[i][j];
}
for (int i = 1;i < 2;i++)
for (int j = 1;j<4;j++)
now[i+x][j+y] = ' ';
}
bool ok(){
for (int i = 0;i < 5;i++)
for (int j = 0;j < 9;j++)
if (now[i][j]!=s[i][j])
return false;
return true;
}
void out(){
for (int i = 0;i < 5;i++){
for (int j = 0;j < 9;j++)
cout <<now[i][j];
cout << endl;
}
cout << endl;
}
bool dfs(int dep){
if (dep > 1 && ok()) return true;
if (dep >= 7) return false;
int temp[M+5][M+5];
for (int i = 0;i < 3;i++){
for (int j = 0;j < 5;j+=2){
for (int k = 0;k <3;k++)
for (int l = 0;l < 5;l++)
temp[k][l] = now[k+i][l+j];
Set(i,j);
if (dfs(dep+1)) return true;
for (int k = 0;k <3;k++)
for (int l = 0;l < 5;l++)
now[k+i][l+j] = temp[k][l];
}
}
return false;
}
int main(){
#ifdef LOCAL_DEFINE
freopen("rush_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(0),cin.tie(0);
init();
int kase = 0;
while (1){
for (int i = 0;i < 5;i++){
cin.getline(s[i],15);
if (s[i][0]=='0') return 0;
}
for (int i = 0;i< 5;i++)
for (int j = 0;j < 9;j++)
now[i][j] = ' ';
if (dfs(1)){
cout <<"Case "<<++kase<<": Yes"<<endl;
}else{
cout <<"Case "<<++kase<<": No"<<endl;
}
}
return 0;
}