【题目链接】:

【题意】

Chinese

【题解】

bfs题;

根据bfs的性质;

第一次到达的点肯定是转弯次数最少的;

每次往一个方向走到头就好了;

搞个数组判判重.

这里在往一个方向走的时候;

如果途中遇到了终点;

也算能到达终点;

其他的就没什么坑点了;

【Number Of WA】

3

【完整代码】

#include 
     
      using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long long#define rep1(i,a,b) for (int i = a;i <= b;i++)#define rep2(i,a,b) for (int i = a;i >= b;i--)#define mp make_pair#define pb push_back#define fi first#define se second#define ms(x,y) memset(x,y,sizeof x)typedef pair
      
        pii;typedef pair
       
         pll;const int dx[9] = {
    0,1,-1,0,0,-1,-1,1,1};const int dy[9] = {
    0,0,0,-1,1,-1,1,-1,1};const double pi = acos(-1.0);const int N = 510;struct node{    int x,y,fx;};int f[N][N][5],n,m,sx,sy,tx,ty;bool bo[N][N];char s[N];queue 
        
          dl;int bfs(int x,int y){    rep1(i,1,500)        rep1(j,1,500)            rep1(k,1,4)                f[i][j][k] = -2;    rep1(i,1,4)    {        f[x][y][i] = -1;        int tq = x,tw = y;        if (bo[tq+dx[i]][tw+dy[i]])        {            while (bo[tq+dx[i]][tw+dy[i]])            {                tq+=dx[i],tw+=dy[i];                if (tq==tx && tw==ty) return 0;            }            f[tq][tw][i] = 0;            dl.push(node{tq,tw,i});        }    }    while (!dl.empty())    {        node temp = dl.front();        int q = temp.x,w = temp.y,pre = temp.fx;        dl.pop();        rep1(i,1,4)        {            int tq = q,tw = w;            if (bo[tq+dx[i]][tw+dy[i]])            {                while (bo[tq+dx[i]][tw+dy[i]])                {                    tq+=dx[i],tw+=dy[i];                    if (tq==tx && tw==ty) return f[q][w][pre]+1;                }                if (f[tq][tw][i]==-2)                {                    f[tq][tw][i] = f[q][w][pre]+1;                    dl.push(node{tq,tw,i});                }            }        }    }    return -1;}int main(){    //freopen("F:\\rush.txt","r",stdin);    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use    //init??????    cin >> n >> m;    rep1(i,1,n)    {        cin >>(s+1);        rep1(j,1,m)        {            if (s[j]=='.')                bo[i][j]=true;            if (s[j]=='#')                bo[i][j]=false;            if (s[j]=='S')            {                bo[i][j] = true;                sx = i,sy = j;            }            if (s[j]=='T')            {                bo[i][j] = true;                tx = i,ty = j;            }        }    }    cout << bfs(sx,sy)<