終於寫夠一百道題啦，怎麼也沒想過他是這樣來臨的，在一個陽光明媚的冬日，寫完第九十九題之後還是非常激動的，不知道第一百題做什麼好，ACM羣裏不知怎麼的就討論起了八皇后問題（好像還是因爲我發了一個非常奇怪的問題求解答）我猶豫了三分鐘之後，算了，乾脆寫個八皇后問題吧，這題很熟悉，但是現在卻已經都忘記了題目是什麼樣子的，又翻了出來。。。。
In chess it is possible to place eight queens on the board so that no one queen can be taken by any
other. Write a program that will determine all such possible arrangements for eight queens given the
initial position of one of the queens.
Do not attempt to write a program which evaluates every possible 8 configuration of 8 queens placed
on the board. This would require 88
evaluations and would bring the system to its knees. There will
be a reasonable run time constraint placed on your program.
Input
The first line of the input contains the number of datasets,
and it’s followed by a blank line.
Each dataset contains a pair of positive integers separated
by a single space. The numbers represent the square on which
one of the eight queens must be positioned. A valid square
will be represented; it will not be necessary to validate the
input.
To standardize our notation, assume that the upper leftmost corner of the board is position (1,1). Rows run horizontally and the top row is row 1. Columns are vertical and
column 1 is the left-most column. Any reference to a square
is by row then column; thus square (4,6) means row 4, column
6.
Each dataset is separated by a blank line.
Output
Output for each dataset will consist of a one-line-per-solution representation.
Each solution will be sequentially numbered 1 . . . N. Each solution will consist of 8 numbers. Each
of the 8 numbers will be the ROW coordinate for that solution. The column coordinate will be indicated
by the order in which the 8 numbers are printed. That is, the first number represents the ROW in
which the queen is positioned in column 1; the second number represents the ROW in which the queen
is positioned in column 2, and so on.
Notes:
The sample input below produces 4 solutions. The full 8×8 representation of each solution is shown
below.
DO NOT SUBMIT THE BOARD MATRICES AS PART OF YOUR SOLUTION!
SOLUTION 1 SOLUTION 2 SOLUTION 3 SOLUTION 4
1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0
0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0
Submit only the one-line, 8 digit representation of each solution as described earlier. Solution #1
below indicates that there is a queen at Row 1, Column 1; Row 5, Column 2; Row 8, Column 3; Row
6, Column 4; Row 3,Column 5; … Row 4, Column 8.
Include the two lines of column headings as shown below in the sample output and print the
solutions in lexicographical order.
Sample Input
1
1 1
Sample Output
SOLN COLUMN
1 2 3 4 5 6 7 8
1 1 5 8 6 3 7 2 4
2 1 6 8 3 7 4 2 5
3 1 7 4 6 8 2 5 3
4 1 7 5 8 2 4 6 3
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int r,c,col[10],Ans;
bool vis[3][30];
void DFS(int cur) {
if(cur == 9) {
Ans ++;
printf("%2d ",Ans);
for(int i=1; i<=8; i++) printf(" %d",col[i]);
printf("\n");
return ;
}
if(cur == c) { DFS(cur + 1); return; }
for(int i=1; i<=8; i++) {
if(!vis[0][i] && !vis[1][cur + i] && !vis[2][cur - i + 8]) {
col[cur] = i;
vis[0][i] = vis[1][cur + i] = vis[2][cur - i + 8] = true;
DFS(cur + 1);
vis[0][i] = vis[1][cur + i] = vis[2][cur - i + 8]=false;
}
}
}
int main(int argc,char* agrv[]) {
int T; scanf("%d",&T);
while(T--) {
scanf("%d%d",&r,&c);
Ans = 0;
memset(vis,0,sizeof(vis));
col[c] = r;
vis[0][r] = vis[1][r + c] = vis[2][c - r + 8] = true;
printf("SOLN COLUMN\n # 1 2 3 4 5 6 7 8\n\n");
DFS(1);
if(T) printf("\n");
}
return 0;
}