C/C++教程

[AcWing 844] 走迷宫

本文主要是介绍[AcWing 844] 走迷宫,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

image


BFS 使用STL中的queue


点击查看代码
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
typedef pair<int, int> PII;
const int N = 100 + 10;
int n, m;
int g[N][N], d[N][N];
queue<PII> q;
int bfs()
{
    q.push({0, 0});
    memset(d, -1, sizeof(d));
    d[0][0] = 0;
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
    while (q.size()) {
        PII t = q.front();
        q.pop();
        for (int i = 0; i < 4; i ++) {
            int x = t.first + dx[i], y = t.second + dy[i];
            if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 0 && d[x][y] == -1) {
                d[x][y] = d[t.first][t.second] + 1;
                q.push({x, y});
            }
        }
    }
    return d[n - 1][m - 1];
}
int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i ++)
        for (int j = 0; j < m; j ++)
            cin >> g[i][j];
    cout << bfs() << endl;
    return 0;
}


BFS 数组模拟队列


点击查看代码
#include<iostream>
#include<cstring>
using namespace std;
typedef pair<int, int> PII;
const int N = 100 + 10;
int n, m;
int g[N][N], d[N][N];
PII q[N * N];
int bfs()
{
    int hh = 0, tt = 0;
    q[0] = {0, 0};
    memset(d, -1, sizeof(d));
    d[0][0] = 0;
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
    while (hh <= tt) {
        PII t = q[hh ++];
        for (int i = 0; i < 4; i ++) {
            int x = t.first + dx[i], y = t.second + dy[i];
            if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 0 && d[x][y] == -1) {
                d[x][y] = d[t.first][t.second] + 1;
                q[++ tt] = {x, y};
            }
        }
    }
    return d[n - 1][m - 1];
}
int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i ++)
        for (int j = 0; j < m; j ++)
            cin >> g[i][j];
    cout << bfs() << endl;
    return 0;
}

这篇关于[AcWing 844] 走迷宫的文章就介绍到这儿,希望我们推荐的文章对大家有所帮助,也希望大家多多支持为之网!