hdu1045

Posted 2020-08-27 王坤1993

#include<iostream>
using namespace std;
int count = 0, n = 0;
//判断该ch[x][y]是否可以放置
bool isOk(char **ch, int x, int y){
int i;
//向上检索
for (i = x - 1; i >=0; --i){
if(ch[i][y]==‘0‘){
return false;
}
//碰到墙
if(ch[i][y] == ‘X‘){
break;
}
}
//向左检索
for (i = y - 1; i >=0; --i){
if(ch[x][i]==‘0‘){
return false;
}
//碰到墙
if(ch[x][i] == ‘X‘){
break;
}
}
return true;
}
void search(char **ch, int k, int step){
int x, y;
if(k == n * n){//到达最末
//是否大于之前的count
if(step > count){
count = step;
return;
}
}else {
x = k / n;//行数
y = k % n;//列数
if(ch[x][y] == ‘.‘ && isOk(ch, x, y)){
ch[x][y] = ‘0‘;
search(ch, k+1, step+1);//进入k+1步的搜索
//关键理解下面两句！回溯！
ch[x][y] = ‘.‘;//重新赋值为‘.‘，为了下一轮的搜索
search(ch, k+1, step);
}else {
//ch[x][y]不为‘.‘，进入k+1步
search(ch, k+1, step);
}
}
return;
}
int main(){
while(cin>>n && n){
count = 0;
char **ch = new char* [n];
for(int i = 0; i < n; ++i){
ch[i] = new char[n];
}
for(int j = 0; j < n; ++j){
for(int k = 0; k < n; ++k){
cin>>ch[j][k];
}
}
search(ch, 0, 0);
cout<<count<<endl;
}
return 0 ;
}