题目链接
http://poj.org/problem?id=1321
思路
和N皇后问题类似 但是有一点不同的是 这个是只需要摆放K个棋子就可以了
所以 我们要做好 两个出口 并且要持续往下一层找
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climits> #include <ctime> #include <iostream> #include <algorithm> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> #include <limits> #define CLR(a) memset(a, 0, sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; typedef pair<string, int> psi; typedef pair<string, string> pss; const double PI = acos(-1); const double E = exp(1); const double eps = 1e-30; const int INF = 0x3f3f3f3f; const int maxn = 5e4 + 5; const int MOD = 1e9 + 7; string G[10]; int ans; int n, k, m; int v[10]; void dfs(int cur) { if (m == k) { ans++; return; } if (cur >= n) return; for (int i = 0; i < n; i++) { if (G[cur][i] == '#' && v[i] == 0) { v[i] = 1; m++; dfs(cur + 1); m--; v[i] = 0; } } dfs(cur + 1); } void init() { CLR(v); ans = 0; } int main() { while (scanf("%d%d", &n, &k)) { if (n == -1 && k == -1) break; for (int i = 0; i < n; i++) cin >> G[i]; init(); m = 0; dfs(0); cout << ans << endl; } }转载于:https://www.cnblogs.com/Dup4/p/9433140.html
