AcWing 1076. 迷宫问题（最短路模型）

题目链接

活动 - AcWing本课程系统讲解常用算法与数据结构的应用方式与技巧。https://www.acwing.com/problem/content/description/1078/

来源

《信息学奥赛一本通》, kuangbin专题 , POJ3984

代码

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

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 1010, M = N * N;

int n;
int g[N][N];
PII q[M];
PII pre[N][N];

void bfs(int sx, int sy)
{
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
    
    int hh = 0, tt = 0;
    q[0] = {sx, sy};
    
    memset(pre, -1, sizeof pre);
    pre[sx][sy] = {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 >= n) continue;
            if (g[a][b]) continue;
            if (pre[a][b].x != -1) continue;
            
            q[++tt] = {a, b};
            pre[a][b] = t;
        }
    }
}

int main()
{
    scanf("%d", &n);
    
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            scanf("%d", &g[i][j]);
    
    bfs(n - 1, n - 1);
    
    PII end(0, 0);
    
    while (true)
    {
        printf("%d %d\n", end.x, end.y);
        if (end.x == n - 1 && end.y == n - 1) break;
        end = pre[end.x][end.y];
    }
    
    return 0;
}

参考资料

1. AcWing算法提高课