Codeforces 358E

本文介绍了一种基于深度优先搜索的迷宫路径寻找算法,并详细展示了如何通过递归方式遍历迷宫,寻找可能的最短路径。文章还讨论了算法实现过程中遇到的一些挑战,例如如何有效地标记已访问过的路径等。

真的不会做。。。
别人的代码没怎么看懂。。。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-8)
#define LL long long
#define ULL unsigned __int64
#define _LL __int64
#define INF 0x3f3f3f3f
#define Mod 20090717

using namespace std;

bool vis[1002][1002][4];

int num[1010][1010];

int dir[2][4] = {{-1,0,1,0},{0,-1,0,1}};

int total;

int ans;

void dfs(int x,int y,int d,int len)
{
    bool mark = false;
    for(int i = 0;i < 4; ++i)
    {
        if(0 == num[x+dir[0][i]][y+dir[1][i]] || vis[x][y][i] == true)
            continue;
        vis[x][y][i] = true,vis[x+dir[0][i]][y+dir[1][i]][(i+2)%4] = true;

        total -= 2;

        if(d != -1 && i != d)
            ans = __gcd(ans,len),dfs(x+dir[0][i],y+dir[1][i],i,1);
        else
            dfs(x+dir[0][i],y+dir[1][i],i,len+1);
         mark = true;
    }

    if(mark == false)
        ans = __gcd(ans,len);
}

int main()
{
    #ifdef DouBi
    freopen("in.cpp","r",stdin);
    #endif // DouBi
    int n,m,i,j,k;

    scanf("%d %d",&n,&m);

    memset(num,0,sizeof(num));

    for(i = 1;i <= n; ++i)
        for(j = 1;j <= m; ++j)
            scanf("%d",&num[i][j]);

    int odd = 0;

    ans = 0;
    total = 0;

    for(i = 1;i <= n; ++i)
        for(j = 1;j <= m; ++j)
        {
            if(num[i][j] == 0)
                continue;
            for(ans = 0,k = 0;k < 4; ++k)
                ans += num[i+dir[0][k]][j+dir[1][k]];
            total += ans;
            if(ans == 0)
                return puts("-1"),0;
            if(ans&1)
                odd++;
        }

    if(odd != 0 && odd != 2)
        return puts("-1"),0;

    memset(vis,false,sizeof(vis));

    ans = 0;

    for(i = 1;i <= n; ++i)
        for(j = 1;j <= m; ++j)
            if(num[i][j])
            {
                dfs(i,j,-1,0);
                goto V;
            }
    V:;

    if(total || ans <= 1)
        return puts("-1"),0;

    for(i = 2;i < ans; ++i)
        if(ans%i == 0)
            printf("%d ",i);
    printf("%d\n",ans);
    return 0;
}
### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值