POJ

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;
    }
}