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