牛客寒假算法基础集训营4 - C - Applese走迷宫(BFS)

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题目链接：https://ac.nowcoder.com/acm/contest/330/C

思路：BFS多开一维标记状态，然后在搜索的时候单独判断一下遇到'@'的情况就好了(将切换状态和不变都扔入队列中)，然后按题意模拟即可。

AC代码：

#include <bits/stdc++.h>
using namespace std;
struct node
{
    int x, y, sta, step;
}S, T, Now, Net;
int dir[4][2] = {1,0,0,1,-1,0,0,-1};
bool vis[105][105][2];
string mp[105];
int n, m;
bool Check(int x, int y, int z)
{
    if(x < 0 || y < 0 || x >= n || y >= m)return false;
    if(mp[x][y] == '#') return false;
    if(vis[x][y][z] == true) return false;
    if((z == 0 && mp[x][y] == 'w') || (z == 1 && mp[x][y] == '~')) return false;
    return true;
}
int BFS()
{
    memset(vis,false,sizeof(vis));
    queue <node> Q;
    S.sta = 0, S.step = 0;
    Q.push(S);
    while(!Q.empty())
    {
        Now = Q.front();
        Q.pop();
        if(Now.x == T.x && Now.y == T.y) return Now.step;
        if(mp[Now.x][Now.y] == '@' && vis[Now.x][Now.y][Now.sta^1] == false)
        {
            Net.x = Now.x, Net.y = Now.y;
            Net.step = Now.step + 1;
            Net.sta = Now.sta ^ 1;
            vis[Net.x][Net.y][Net.sta] = true;
            Q.push(Net);
        }
        for(int i = 0; i < 4; i++)
        {
            Net.x = Now.x + dir[i][0];
            Net.y = Now.y + dir[i][1];
            if(Check(Net.x, Net.y, Now.sta))
            {
                Net.step = Now.step + 1;
                Net.sta = Now.sta;
                vis[Net.x][Net.y][Net.sta] = true;
                Q.push(Net);
            }
        }
    }
    return -1;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i = 0; i < n; i++)
        cin >> mp[i];
    for(int i = 0; i < n; i++)
    for(int j = 0; j < m; j++)
    {
        if(mp[i][j] == 'S')
            S.x = i, S.y = j;
        if(mp[i][j] == 'T')
            T.x = i, T.y = j;
    }
    int ans = BFS();
    printf("%d\n",ans);
}