201312-5 I’m stuck

最开始想的是 起点先dfs跑一边 得到可以达到的点 然后在这些点里面继续跑dfs判断是否可以到达终点就行 爆肝1小时  可是只能拿到80分 

又想到完全没有必要直接正向跑一边dfs然后反向跑一遍 容斥判定一下就行 100分

80分代码

//#include 




const int MAX=55;
char a[MAX][MAX];
int b[MAX][MAX];
int c[MAX][MAX];
int vis[MAX][MAX];
int vis1[MAX][MAX];
int sx,sy;
int ex,ey;
int n,m;
int xd1[2]={1,-1};
int yd1[2]={0,0};
int xd2[2]={0,0};
int yd2[2]={1,-1};
struct node{
    int x,y;
};

void dfs(int x,int y,int o){
    queuequ;
    qu.push({x,y});
    while (!qu.empty()) {
        node now=qu.front();
        qu.pop();
        int x1=now.x;
        int y1=now.y;
        char p;
        p=a[x1][y1];
        vis[x1][y1]=1;
        if(p=='+'||p=='S'||p=='T'){
            for(int i=0;i<4;i++){
                int x2=x1+xd[i];
                int y2=y1+yd[i];
                p=a[x2][y2];
                if(x2<1||x2>n||y2<1||y2>m||p=='#'||vis[x2][y2])
                    continue;
//                cout<n||y2<1||y2>m||p=='#'||vis[x2][y2])
                    continue;
//                cout<<"-"<n||y2<1||y2>m||p=='#'||vis[x2][y2])
                    continue;
//                cout<n||y2<1||y2>m||p=='#'||vis[x2][y2])
                continue;
//            cout<qu;
    qu.push({x,y});
    while (!qu.empty()) {
        node now=qu.front();
        qu.pop();
        int x1=now.x;
        int y1=now.y;
        char p;
        p=a[x1][y1];
        vis1[x1][y1]=1;
        if(p=='+'||p=='S'||p=='T'){
            for(int i=0;i<4;i++){
                int x2=x1+xd[i];
                int y2=y1+yd[i];
                p=a[x2][y2];
                if(x2<1||x2>n||y2<1||y2>m||p=='#'||vis1[x2][y2])
                    continue;
//                cout<n||y2<1||y2>m||p=='#'||vis1[x2][y2])
                    continue;
//                cout<<"-"<n||y2<1||y2>m||p=='#'||vis1[x2][y2])
                    continue;
//                cout<n||y2<1||y2>m||p=='#'||vis1[x2][y2])
                continue;
//            cout<>n>>m;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            cin>>a[i][j];
            if(a[i][j]=='S'){
                sx=i;
                sy=j;
            }
            if(a[i][j]=='T'){
                ex=i;
                ey=j;
            }
        }
    }
//    cout< 

//#include 






const int N = 55;
int n, m;
char g[N][N];
int st1[N][N], st2[N][N]; //起点是否能到，能否能到终点

int dx[] = {-1, 0 ,1, 0};
int dy[] = {0, 1, 0, -1};

int check(int x, int y, int d) //(x,y) 能否朝 d 走
{
    char c = g[x][y];
    if(c == 'S' || c=='T' || c=='+') return 1;
    if(c == '|' && d%2==0) return 1;
    if(c == '-' && d%2==1) return 1;
    if(c == '.' && d==2) return 1;
    return 0;
}

void dfs1(int x, int y)
{
    if(st1[x][y]) return ;
    st1[x][y] = 1;
    for(int i=0; i<4; i++)
    {
        int a = x+dx[i], b = y+dy[i];
        if(a<0 || a>=n || b<0 || b>=m || g[a][b]=='#') continue;
        if(check(x, y, i)) dfs1(a, b);
    }
}

void dfs2(int x, int y)
{
    if(st2[x][y]) return ;
    st2[x][y] = 1;
    for(int i=0; i<4; i++) {
        int a = x+dx[i], b = y+dy[i];
        if(a<0 || a>=n || b<0 || b>=m || g[a][b]=='#') continue;
        if(check(a, b, i^2)) dfs2(a, b);
    }
}

int main()
{
    cin >> n >> m;
    for(int i=0; i> g[i];
    }

    int xe, ye;
    for(int i=0; i