给定一个n*m大小的迷宫，其中*代表不可通过的墙壁，而.代表平地，S代表起点，T代表终点。
移动过程中，如果当前位置是(x,y)（下标从0开始），且每次只能前往上下左右四个位置的平地。
求从起点S到达终点T的最少步数。
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define SIS std::ios::sync_with_stdio(false) #define ll long long #define INF 0x3f3f3f3f const int mod = 1e9 + 7; const double esp = 1e-5; const double PI = 3.141592653589793238462643383279; using namespace std; const int N = 1e7 + 5; const int maxn = 1 << 20; ll powmod(ll a, ll b) { ll res = 1; a %= mod; while (b >= 0); for (; b; b >>= 1) { if (b & 1)res = res * a % mod; a = a * a % mod; }return res; } ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } /*void chafen(int l, int r, int k) {//差分函数 p[l] += k; p[r + 1] -= k; }*/ /*********************************************************/ bool vis[105][105]; char a[105][105]; int x[4] = { -1,0,1,0 }; int y[4] = { 0,1,0,-1 }; int u1, v1, u2, v2; int n, m; struct node { int x; int y; int step; }; bool test(int x, int y) { if (x < 0 || x >= n || y < 0 || y >= m) return false; if (a[x][y] == '*' || vis[x][y]) return false; return true; } int bfs() { queue q; node p;//起点 p.x = u1; p.y = v1; p.step = 0; q.push(p); vis[u1][v1] = true; while (!q.empty()) { node p1 = q.front(); q.pop(); if (p1.x == u2 && p1.y == v2) return p1.step; node t; for (int i = 0; i < 4; i++) { t.x = p1.x + x[i]; t.y = p1.y + y[i]; t.step = p1.step + 1; if (test(t.x,t.y)) { q.push(t); vis[t.x][t.y] = true; } } } return -1; } int main() { cin >> n >> m; memset(vis, false, sizeof(vis)); for (int i = 0; i < n; i++) { getchar(); for (int j = 0; j < m; j++) { cin >> a[i][j]; if (a[i][j] == 'S') { u1 = i; v1 = j; } if (a[i][j] == 'T') { u2 = i; v2 = j; } } } int a1, a2, a3, a4; cin >> a1 >> a2 >> a3 >> a4; cout << bfs() << endl; return 0; } /* 5 5 ..... .*.*. .*S*. .***. ...T* 2 2 4 3 */
