• [USACO 6.5.3]Betsy's Tour


    题目大意

      从(1,1)格子开始不重复地经过N*N个格子(包括(1,1)和(N,1))到达(N,1),求路径方案数.

    题解

      搜索剪枝.

      剪枝:

        1.当把网格分成两部分时剪枝.

        2.走到角的附近时一定走角(N,1)除外.

    代码

    /*
    TASK:betsy
    LANG:C++
    */
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    
    using namespace std;
    
    const int d[2][4] = {{-1, 0, 1, 0},
                         {0, -1, 0, 1}};
    
    int n, f[10][10];
    bool vis[10][10];
    
    bool check(int x, int y)
    {
        if (!vis[x-1][y] && !vis[x+1][y] && vis[x][y-1] && vis[x][y+1]) return true;
        if (!vis[x][y-1] && !vis[x][y+1] && vis[x-1][y] && vis[x+1][y]) return true;
        return false;
    }
    
    long long dfs(int x, int y, int dep)
    {
        if (x == n && y == 1)
        {
            if (dep == n * n) return 1;
            return 0;
        }
        if (check(x, y)) return 0;
        int msc = 0, k;
        for (int i = 0; i < 4; ++i)
        {
            int tx = x + d[0][i], ty = y + d[1][i];
            if (vis[tx][ty])
            {
                f[tx][ty]--;
                if (f[tx][ty] == 1) msc++, k = i;
            }
        }
        vis[x][y] = false;
        long long res = 0;
        if (msc <= 1)
        {
            if (msc == 0)
                for (int i = 0; i < 4; ++i)
                {
                    int tx = x + d[0][i], ty = y + d[1][i];
                    if (vis[tx][ty]) res += dfs(tx, ty, dep + 1);
                }
            else res = dfs(x+d[0][k], y+d[1][k], dep+1);
        }
        for (int i = 0; i < 4; ++i)
        {
            int tx = x + d[0][i], ty = y + d[1][i];
            if (vis[tx][ty]) f[tx][ty]++;
        }
        vis[x][y] = true;
        return res;
    }
    
    int main()
    {
        freopen("betsy.in", "r", stdin);
        freopen("betsy.out", "w", stdout);
        scanf("%d", &n);
        memset(vis, true, sizeof(vis));
        for (int i = 0; i <= n; ++i)
            vis[0][i] = vis[i][0] = vis[n+1][i] = vis[i][n+1] = false;
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                f[i][j] = 4;
        for (int i = 1; i <= n; ++i) f[i][1] = f[1][i] = f[i][n] = f[n][i] = 3;
        f[n][1] = 10;
        f[1][1] = f[1][n] = f[n][n] = 2;
        cout << dfs(1, 1, 1) << endl;
        return 0;
    }
  • 相关阅读:
    UVa10036
    矩阵链乘法(动态规划)
    Codeforces 230A
    iOS 界面开发
    iOS 自动布局
    iOS 自动布局过程
    iOS 界面布局,设置约束
    iOS + UIWebView 实践
    iOS 参考 网络书籍
    iOS 框架 Nimbus
  • 原文地址:https://www.cnblogs.com/albert7xie/p/6343808.html
Copyright © 2020-2023  润新知