【数据结构】第4章 队列

一、简答题

二、程序设计题

5. 设计一个函数，在一个\(N * N\)迷宫中找出从入口到出口的最短路径。入口在左上角，出口在右下角

分析：BFS，宽搜，需要用到队列

题目：844. 走迷宫

手写队列版：

#include <iostream>
#include <algorithm>
#include <cstring>

#define x first 
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 110, M = N * N;

int n, m;
int dist[N][N];
int g[N][N];
PII q[M];
int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

int bfs()
{
    int hh = 0, tt = 0; // 原本是-1,初始有一个元素，所以是0
    memset(dist, -1, sizeof dist);
    dist[0][0] = 0;
    q[0] = {0, 0};
    
    while (hh <= tt)
    {
        PII t = q[hh ++ ];
        for (int i = 0; i < 4; i ++ )
        {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (dist[a][b] != -1 || g[a][b] != 0) continue;
            dist[a][b] = dist[t.x][t.y] + 1;
            q[++ tt] = {a, b};
        }
    }
    return dist[n - 1][m - 1];
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < m; j ++ )
            scanf("%d", &g[i][j]);
    
    printf("%d\n", bfs());
    return 0;
}

STL队列版:

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>

#define x first 
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 110, M = N * N;

int n, m;
int dist[N][N];
int g[N][N];
int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};

int bfs()
{
    queue<PII> q;
    memset(dist, -1, sizeof dist);
    dist[0][0] = 0;
    q.push({0, 0});
    
    while (q.size())
    {
        PII t = q.front();
        q.pop();
        
        for (int i = 0; i < 4; i ++ )
        {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (dist[a][b] != -1 || g[a][b] != 0) continue;
            dist[a][b] = dist[t.x][t.y] + 1;
            q.push({a, b});
        }
    }
    return dist[n - 1][m - 1];
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < m; j ++ )
            scanf("%d", &g[i][j]);
    
    printf("%d\n", bfs());
    return 0;
}