算法——广度优先搜索——跨步迷宫

原题链接

思路：找出最短路径，然后判断是否存在连续三个点是横纵坐标相等的，如果有就步数减1

但是有两个样例过不了

错误原因：在错误的测试案例中，最短路径可能有多条，而我刚好选了一条比较曲折的，不能一下子走两步

#include <bits/stdc++.h>

using namespace std;

int main() {

    int n, m;
    cin >> n >> m;
    vector<vector<int>> arr(n + 2, vector<int>(m + 2, -1));
    vector<vector<int>> state(n + 2, vector<int>(m + 2, 0));
    vector<vector<pair<int, int>>> lastNode(n + 2, vector<pair<int, int>>(m + 2, {0, 0}));

    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            int num;
            cin >> num;
            arr[i][j] = num;
        }
    }

    int step = 0;
    queue<pair<int, int>> p, c;
    bool isFind = false;
    p.emplace(1, 1);
    while (!p.empty()) {
        while (!p.empty()) {
            auto cur = p.front();
            if (cur.first == n && cur.second == m) {
                isFind = true;
                break;
            }
            p.pop();
            if (arr[cur.first - 1][cur.second] == 0 && state[cur.first - 1][cur.second] == 0) {
                c.emplace(cur.first - 1, cur.second);
                state[cur.first - 1][cur.second] = 1;
                lastNode[cur.first - 1][cur.second] = {cur.first, cur.second};
            }

            if (arr[cur.first + 1][cur.second] == 0 && state[cur.first + 1][cur.second] == 0) {
                c.emplace(cur.first + 1, cur.second);
                state[cur.first + 1][cur.second] = 1;
                lastNode[cur.first + 1][cur.second] = {cur.first, cur.second};
            }

            if (arr[cur.first][cur.second - 1] == 0 && state[cur.first][cur.second - 1] == 0) {
                c.emplace(cur.first, cur.second - 1);
                state[cur.first][cur.second - 1] = 1;
                lastNode[cur.first][cur.second - 1] = {cur.first, cur.second};
            }

            if (arr[cur.first][cur.second + 1] == 0 && state[cur.first][cur.second + 1] == 0) {
                c.emplace(cur.first, cur.second + 1);
                state[cur.first][cur.second + 1] = 1;
                lastNode[cur.first][cur.second + 1] = {cur.first, cur.second};
            }
        }
        if (isFind) break;
        step++;
        p = c;
        while (!c.empty()) c.pop();
    }
    vector<pair<int, int>> result;
    if (!isFind) {
        cout << -1 << endl;
        return 0;
    } else {
        int i = n, j = m;
        while (i != 1 || j != 1) {
            result.emplace_back(i, j);
            auto post = lastNode[i][j];
            i = post.first;
            j = post.second;
        }
    }
//    cout << "1 1" << endl;
//    for (int i = result.size() - 1; i >= 0; --i) {
//        auto cur = result[i];
//        cout << cur.first << " " << cur.second << endl;
//    }
    result.emplace_back(1, 1);
    int count = result.size() - 1; // 经过多少个点，减去1 就是单步走的最短路径
    for (int i = result.size() - 1; i >= 2; --i) {
        if (result[i].first == result[i - 1].first && result[i].first == result[i - 2].first) {
            count--;
            i = i - 1; // 循环结束会减一 所以这里只要减1 一共减2
        } else if (result[i].second == result[i - 1].second && result[i].second == result[i - 2].second) {
            count--;
            i = i - 1;
        }
    }
    cout << count << endl;
    return 0;
}