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  • [洛谷] P2802 回家

    dfs 标记 + 回溯

    //#pragma GCC optimize(2)
    #include <cstdio>
    #include <iostream>
    #include <cstdlib>
    #include <cmath>
    #include <cctype>
    #include <string>
    #include <cstring>
    #include <algorithm>
    #include <stack>
    #include <queue>
    #include <set>
    #include <map>
    #include <ctime>
    #include <vector>
    #include <fstream>
    #include <list>
    #include <iomanip>
    #include <numeric>
    using namespace std;
    typedef long long ll;
    
    const int MAXN = 1e2 + 10;
    
    int n, m, bx, by;
    
    int movel[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    
    int arr[MAXN][MAXN], ans = 0x3f3f3f3f;
    
    int used[MAXN][MAXN];
    
    void dfs(int x, int y, int hp, int step)
    {
        if(hp == 0 || arr[x][y] == 0 || used[x][y] == 1)
            return ;
    
        if(!(x >= 0 && x <= n && y >= 0 && y <= m))
            return ;
    
        if(arr[x][y] == 3)
        {
            ans = min(ans, step);
            return ;
        }
    
        if(arr[x][y] == 4)
            hp = 6;
        
        used[x][y] = 1;
        for(int i = 0; i < 4; i++)
        {
    
            dfs(x + movel[i][0], y + movel[i][1], hp - 1, step + 1);
        }
        used[x][y] = 0;
    }
    
    int main()
    {
        //ios::sync_with_stdio(false);
    
        //cin.tie(0);     cout.tie(0);
    
        cin>>n>>m;
    
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
            {
                cin>>arr[i][j];
                if(arr[i][j] == 2)
                    bx = i, by = j;
            }
        
        dfs(bx, by, 6, 0);
    
        ans >= 0x3f3f3f3f ? cout<<"-1"<<endl : cout<<ans<<endl;
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zeolim/p/12270436.html
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