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Problem Description
在N*N的方格棋盘放置了N个皇后，使得它们不相互攻击（即任意2个皇后不允许处在同一排，同一列，也不允许处在与棋盘边框成45角的斜线上。
你的任务是，对于给定的N，求出有多少种合法的放置方法。 |
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Input
共有若干行，每行一个正整数N≤10，表示棋盘和皇后的数量；如果N=0，表示结束。
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Output
共有若干行，每行一个正整数，表示对应输入行的皇后的不同放置数量。 |
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Sample Input
1 8 5 0 |
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Sample Output
1 92 10 |
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Code  View Code
1 #include <stdio.h> 2 #include <math.h> 3 int q[11]; //q[1] means the coordinate of queue is (1, q[1]) 4 int result[11];//to save the time, so record the previous time 5 int check(int k) //check if the kth queen is conflict with previous ones 6 { 7 int i; 8 i = 1; 9 while (i < k) 10 { 11 //k - i == abs(q[k] - q[i]) means the current queen is not 45° with other queens 12 if (q[i] == q[k] || k - i == abs(q[k] - q[i])) 13 return 0; 14 i++; 15 } 16 return 1; 17 } 18 //begin with the first queen, find the position for first queen 19 //and use DFS to go on to find next queen's position, if u can't find it, u can go back to change the previous queen's position 20 //untill there is no position for the fist queue to change 21 int Nqueens(int n) 22 { 23 int count, i, j, k; 24 count = 0; //record the count of rank 25 //begin with first queen and the zero position 26 k = 1; 27 q[k] = 0; 28 while (k > 0) 29 { 30 q[k]++; 31 //when q[k] <= n, u can go on to find q[k] satisfied the condition 32 while (q[k] <= n && check(k) == 0) 33 q[k]++; 34 //if q[k] <= n, which means u can find position for current queen 35 if (q[k] <= n) 36 { 37 //means u find the last queen, so u can record the counr 38 if (k == n) 39 count++; 40 //if it's not the last queen, u should go on to find the place of next queen 41 //and for next queen, u should begin with zero. 42 else 43 { 44 k++; 45 q[k] = 0; 46 } 47 } 48 else 49 //if u can't find the position for current queen, u should go back to modify the previous queen's position 50 k--; 51 } 52 return count; 53 } 54 int main() 55 { 56 int n; 57 while (scanf("%d", &n) != EOF && n) 58 { 59 if (result[n] == 0) 60 result[n] = Nqueens(n); 61 printf("%d\n", result[n]); 62 } 63 return 0; 64 }
View Code
1 #include <stdio.h> 2 #include <math.h> 3 #include <stdlib.h> 4 int q[11]; //q[1] means the coordinate of queue is (1, q[1]) 5 int result[11];//to save the time, so record the previous time 6 int n; 7 int check(int k) //check if the kth queen is conflict with previous ones 8 { 9 int i; 10 i = 1; 11 while (i < k) 12 { 13 //k - i == abs(q[k] - q[i]) means the current queen is not 45° with other queens 14 if (q[i] == q[k] || k - i == abs(q[k] - q[i])) 15 return 0; 16 i++; 17 } 18 return 1; 19 } 20 int count; //record the count of rank 21 int flag; 22 void DFS(int step) 23 { 24 int i, j, k; 25 if (step == n + 1) 26 count++; 27 else 28 { 29 for (i = 1; i <= n; i++) 30 { 31 q[step] = i; 32 if (check(step) == 0) 33 continue; 34 DFS(step + 1); 35 } 36 } 37 } 38 int main() 39 { 40 while (scanf("%d", &n) != EOF && n) 41 { 42 //memset(); 43 count = 0; 44 if (result[n] == 0) 45 { 46 DFS(1); 47 result[n] = count; 48 } 49 printf("%d\n", result[n]); 50 } 51 return 0; 52 }
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u can find that the first one use least time. |
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