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 secondusing 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;
}