Place eight queens on an 8*8 chessboard, each queen cannot appear on its own row, column, and diagonal. There are a total of 92 placement situations.
Use r to denote rows and c[r] to denote columns to determine the condition. Because r is an increasing row, only rows without a queen will be found, so there is no need to judge the number of rows, and then the condition is reduced to three:
judge the column, judge the main pair Diagonal, judge the negative diagonal...
1.c[r]==c[j] j loops before r
2.c[r]-c[j]==rj;
3.c[j]-c[r]==rj;
Look for the function, exit at 8, make a boundary, and reach backtracking.
void search(int r){
if(r==8){
save();
return;
}
for(int i=0;i<8;i++){
c[r]=i;
bool ok=true;
for(int j=0;j<r;j++)
if(c[r]==c[j]||r-j==c[r]-c[j]||r-j==c[j]-c[r]){
ok=false;
break;
}
if(ok){
search(r+1);
ans=0;
}
}
}The save function, with the queen's coordinates, double loop adds the value to ans, and compares the maximum value at the last time.
void save(){
for(int i=0;i<8;i++){
for(int j=0;j<8;j++){
if(j==c[i]){
ans+=a[i][j];
}
}
}
Ans=max(Ans,ans);
}Main function, nothing to say
int main(){
int k;cin>>k;
while(k--){
for(int i=0;i<8;i++){
for(int j=0;j<8;j++){
cin>>a[i][j];
}
}
Ans=0;ans=0;
search(0);
cout<<setw(5)<<Ans<<endl;
}
return 0;
}