Checker Challenge（dfs深度搜索 打印八皇后）

Judge Info

Memory Limit: 65536KB
Case Time Limit: 3000MS
Time Limit: 3000MS
Judger: Number Only Judger

Description

Examine the 6x6 checkerboard below and note that the six checkers are arranged on the board so that one and only one is placed in each row and each column, and there is never more than one in any diagonal. (Diagonals run from southeast to northwest and southwest
to northeast and include all diagonals, not just the major two.)

Column
1   2   3   4   5   6
-------------------------
1 |   | O |   |   |   |   |
-------------------------
2 |   |   |   | O |   |   |
-------------------------
3 |   |   |   |   |   | O |
-------------------------
4 | O |   |   |   |   |   |
-------------------------
5 |   |   | O |   |   |   |
-------------------------
6 |   |   |   |   | O |   |
-------------------------

The solution shown above is described by the sequence 2 4 6 1 3 5, which gives the column positions of the checkers for each row from 1 to 6:

ROW	1	2	3	4	5	6
COLUMN	2	4	6	1	3	5

This is one solution to the checker challenge. Write a program that finds all unique solution sequences to the Checker Challenge (with ever growing values of N). Print the solutions using the column notation described above. Print the the first three solutions
in numerical order, as if the checker positions form the digits of a large number, and then a line with the total number of solutions.

Input

The first number

means how many test cases will followed. For every test case will be a single line that contains a single
integer N (6 <= N <= 13) that is the dimension of the N x N checkerboard.

Output

For each test cases, print the follow contains: The first three lines show the first three solutions found, presented as N numbers with a single space between them. The fourth line shows the total number of solutions found.

Sample Input

1
6

Sample Output

2 4 6 1 3 5
3 6 2 5 1 4
4 1 5 2 6 3
4

把八皇后的所有情况打印出来

#include<iostream>
using namespace std;

int plc[13];
int plc_p;
int map[13][13];
bool flag[13];
int n;
int ans;

void dfs()
{
if(plc_p==n)
{
ans++;
if(ans<4)
for(int i=0;i<n;i++)
{
if(i!=n-1) cout<<plc[i]+1<<" ";
else cout<<plc[i]+1<<endl;
}
}

for(int i=0;i<n;i++)
{
if((i!=0)&&flag[i-1]==0)
break;
if(flag[i]==0)
{
flag[i]=1;
for(int j=0;j<n;j++)
{
if(map[i][j]==0)
{
for(int k=0;k<n;k++)
map[k][j]++;
for(int k=0;k<n;k++)
map[i][k]++;
int ii=i,jj=j;
while(ii>-1&&jj>-1)
map[ii--][jj--]++;
ii=i,jj=j;
while(ii<n&&jj<n)
map[ii++][jj++]++;
ii=i,jj=j;
while(ii>-1&&jj<n)
map[ii--][jj++]++;
ii=i,jj=j;
while(ii<n&&jj>-1)
map[ii++][jj--]++;
map[i][j]=1;
plc[plc_p++]=j;
dfs();
plc_p--;
for(int k=0;k<n;k++)
map[k][j]--;
for(int k=0;k<n;k++)
map[i][k]--;
ii=i,jj=j;
while(ii>-1&&jj>-1)
map[ii--][jj--]--;
ii=i,jj=j;
while(ii<n&&jj<n)
map[ii++][jj++]--;
ii=i,jj=j;
while(ii>-1&&jj<n)
map[ii--][jj++]--;
ii=i,jj=j;
while(ii<n&&jj>-1)
map[ii++][jj--]--;
map[i][j]=0;
}
}
flag[i]=0;
}
}
}

int main()
{
int t;
cin>>t;
while(t--)
{
cin>>n;
for(int i=0;i<n;i++)
int plc_p=0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
map[i][j]=0;
for(int i=0;i<n;i++)
flag[i]=0;
ans=0;
dfs();
cout<<ans<<endl;
}
return 0;
}

dfs的基本过程：

dfs()

判明终止条件

for全部物件

标记

dfs()

恢复