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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 210;
struct Node {
int x, y; // 坐标
int dis; // 距离
bool const operator<(const Node &b) const {
return dis > b.dis;
}
};
// 问:优先队列默认是大顶堆,如果想距离小的在前怎么办呢?
// 答:重载小于操作符。对比距离,谁的距离大谁就小。
char g[N][N]; // 地图
int sx, sy, tx, ty; // 起点坐标与终点坐标
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
int n, m; // 代表迷宫的高和宽
void bfs() {
priority_queue<Node> q; // 在priority_queue中优先队列默认是大顶堆
Node st{sx, sy, 0};
g[sx][sy] = '#';
q.push(st);
while (q.size()) {
Node u = q.top();
q.pop();
// 若已找到,则退出
if (u.x == tx && u.y == ty) {
cout << u.dis << endl;
return;
}
for (int i = 0; i < 4; i++) {
int x = u.x + dx[i];
int y = u.y + dy[i];
Node t = {x, y, 0};
if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] != '#') {
if (g[x][y] == 'X')
t.dis = u.dis + 2;
else
t.dis = u.dis + 1;
// 走过就覆盖掉
g[t.x][t.y] = '#';
// 新结点入队列
q.push(t);
}
}
}
// 没有结果,输出不可能到达
printf("impossible\n");
}
int main() {
while (cin >> n >> m) {
for (int i = 0; i < n; i++) cin >> g[i]; // 按字符串读入
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++) {
if (g[i][j] == 'S') sx = i, sy = j;
if (g[i][j] == 'T') tx = i, ty = j;
}
bfs();
}
return 0;
}