POJ 1321 棋盘问题(DFS)

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 传送门:POJ 1321 棋盘问题

回溯暴力搜索也不会超时。下次补一下状态压缩dp的做法...

#include <iostream>
#include <string.h>
using namespace std;
int n, k, ans = 0;
int vis[10][10];

int read()
{
    cin >> n >> k;
    if(n == -1 && k == -1) return 0;

    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < n; ++i)
    {
        for(int j = 0; j < n; ++j)
        {
            char c;
            cin >> c;
            if(c == '.')
                vis[i][j] = -1;
        }
    }
    return 1;
}

int check(int x, int y)
{
    if(vis[x][y]) return 0;
    for(int j = 0; j < n; ++j)
        if(vis[x][j] == 1)
            return 0;
    return 1;
}

void dfs(int cur, int r)
{
    if(!r)
    {
        ++ans;
        return;
    }
    if(cur >= n) return;
    for(int i = 0; i < n; ++i)
    {
        if(check(i, cur))
        {
            vis[i][cur] = 1;
            dfs(cur+1, r-1);
            vis[i][cur] = 0;
        }
    }
    dfs(cur+1, r);
}

void solve()
{
    ans = 0;
    dfs(0, k);
    cout << ans << '\n';
}

int main()
{
	while(read())
        solve();
	return 0;
}

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转载自blog.csdn.net/HNUCSEE_LJK/article/details/88597397