78 lines
2.2 KiB
Plaintext
78 lines
2.2 KiB
Plaintext
#include <iostream>
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#include <vector>
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#include <deque>
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using namespace std;
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// 方向数组：右上，右下，左下，左上 (对应格点的移动)
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const int dir[4][2] = {{-1, 1}, {1, 1}, {1, -1}, {-1, -1}};
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// 对应方向上，方格应该有的形状
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const char expect[4] = {'/', '\\', '/', '\\'};
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// 对应方向上，方格在 mat 中的坐标偏移
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const int dr[4] = {-1, 0, 0, -1};
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const int dc[4] = {0, 0, -1, -1};
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const int INF = 1e9;
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int main() {
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int n, m;
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if (!(cin >> n >> m)) return 0;
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// 奇偶性剪枝：如果终点坐标和为奇数，绝对无法到达
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if ((n + m) % 2 != 0) {
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cout << "NO SOLUTION" << endl;
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return 0;
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}
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vector<vector<char>> mat(n, vector<char>(m));
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// 修正：dist 必须是 int 类型，初始化为无穷大
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vector<vector<int>> dist(n + 1, vector<int>(m + 1, INF));
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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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cin >> mat[i][j];
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}
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}
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deque<pair<int, int>> dq;
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dist[0][0] = 0;
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dq.push_back({0, 0});
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while (!dq.empty()) {
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pair<int, int> curr = dq.front();
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dq.pop_front();
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int r = curr.first;
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int c = curr.second;
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for (int i = 0; i < 4; ++i) {
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int nr = r + dir[i][0];
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int nc = c + dir[i][1];
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// 检查格点是否越界
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if (nr >= 0 && nr <= n && nc >= 0 && nc <= m) {
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// 对应方格的坐标
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int tr = r + dr[i];
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int tc = c + dc[i];
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// 如果当前方向的字符与地图不符，权值为 1，否则为 0
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int weight = (mat[tr][tc] == expect[i] ? 0 : 1);
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if (dist[r][c] + weight < dist[nr][nc]) {
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dist[nr][nc] = dist[r][c] + weight;
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if (weight == 0) {
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dq.push_front({nr, nc}); // 权值为0，插到队头
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} else {
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dq.push_back({nr, nc}); // 权值为1，插到队尾
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}
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}
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}
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}
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}
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if (dist[n][m] == INF) cout << "NO SOLUTION" << endl;
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else cout << dist[n][m] << endl;
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return 0;
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}
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