[AcWing 1076] 迷宫问题
作者:互联网
BFS + 记录方案
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#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1000 + 10;
#define x first
#define y second
int n;
int g[N][N];
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
pair<int,int> ne[N][N];
void bfs(int x, int y)
{
queue<pair<int,int>> q;
q.push({x, y});
memset(ne, -1, sizeof ne);
ne[x][y] = {1, 1};
while (q.size()) {
auto 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 < 1 || a > n || b < 1 || b > n || g[a][b])
continue;
if (ne[a][b].x != -1)
continue;
q.push({a, b});
ne[a][b] = t;
}
}
}
void solve()
{
cin >> n;
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= n; j ++)
cin >> g[i][j];
bfs(n, n);
pair<int,int> path(1, 1);
while (1) {
cout << path.x - 1 << ' ' << path.y - 1 << endl;
if (path.x == n && path.y == n)
break;
path = ne[path.x][path.y];
}
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
solve();
return 0;
}
- \(BFS\) 时从 \((n, n)\) 点往前搜,对于每个点 \(t\),和它能到的点 \(p\),让 \(ne[p.x][p.y] = t\)
标签:1076,void,迷宫,ne,long,bfs,int,define,AcWing 来源: https://www.cnblogs.com/wKingYu/p/16544795.html