搜索题锦——BFS
BFS是搜索中经常考察的一种搜索方式。
题目:
分析:
用个四维来写,差点没写死我。明明很简单的题目,从早上七点一直写到下午四点。
代码:
#include <bits/stdc++.h> #define rep(i, a, b) for (int i = a; i <= b; i ++ ) using namespace std; const int N = 25; char g1[N][N], g2[N][N]; bool st[N][N][N][N]; struct Node{ int x1, y1, x2, y2; }pre[N][N][N][N]; int dx[4] = {1, 0, 0, -1}, dy[8] = {0, -1, 1, 0, 0, 1, -1, 0}; char c[10] = {'D', 'L', 'R', 'U'}; void bfs(int x1, int y1, int x2, int y2) { queue<Node> q; q.push({x1, y1, x2, y2}); st[x1][y1][x2][y2] = true; while (q.size()) { Node t = q.front(); q.pop(); // printf("[%d, %d], [%d, %d]\n", t.x1, t.y1, t.x2, t.y2); if (t.x1 == 1 && t.y1 == 20 && t.x2 == 1 && t.y2 == 1) break; rep(i, 0, 3) { x1 = t.x1 + dx[i], y1 = t.y1 + dy[i], x2 = t.x2 + dx[i], y2 = t.y2 + dy[i + 4]; if (x1 <= 0 || x1 > 20 || y1 <= 0 || y1 > 20 || g1[x1][y1] == '#') x1 = t.x1, y1 = t.y1; if (x2 <= 0 || x2 > 20 || y2 <= 0 || y2 > 20 || g2[x2][y2] == '#') x2 = t.x2, y2 = t.y2; if (!st[x1][y1][x2][y2]) { // printf("[%d, %d], [%d, %d]\n", x1, y1, x2, y2); q.push({x1, y1, x2, y2}); st[x1][y1][x2][y2] = true; pre[x1][y1][x2][y2] = t; } } } #if 1 x1 = 1, y1 = 20, x2 = 1, y2 = 1; string ans; g1[x1][y1] = g2[x2][y2] = 'A'; // printf("[%d, %d], [%d, %d]\n", x1, y1, x2, y2); while (x1 != 20 || y1 != 20 || x2 != 20 || y2 != 1) { Node t = pre[x1][y1][x2][y2]; g1[t.x1][t.y1] = g2[t.x2][t.y2] = 'A'; // printf("now : [%d, %d], [%d, %d]\n", x1, y1, x2, y2); // printf("pre : [%d, %d], [%d, %d]\n", t.x1, t.y1, t.x2, t.y2); rep(i, 0, 3) { int X1 = t.x1 + dx[i], Y1 = t.y1 + dy[i]; int X2 = t.x2 + dx[i], Y2 = t.y2 + dy[i + 4]; // printf("%c direction: [%d, %d], [%d, %d]\n", c[i], X1, Y1, X2, Y2); if (X1 <= 0 || X1 > 20 || Y1 <= 0 || Y1 > 20 || g1[X1][Y1] == '#') X1 = t.x1, Y1 = t.y1; if (X2 <= 0 || X2 > 20 || Y2 <= 0 || Y2 > 20 || g2[X2][Y2] == '#') X2 = t.x2, Y2 = t.y2; if (X1 == x1 && Y1 == y1 && X2 == x2 && Y2 == y2) { ans += c[i]; break; } } x1 = t.x1, y1 = t.y1, x2 = t.x2, y2 = t.y2; } cout << ans.size() << endl; reverse(ans.begin(), ans.end()); cout << ans << endl; rep(i, 1, 20) { cout << g1[i] + 1 << " " << g2[i] + 1 << endl; } #endif } int main() { rep(i, 1, 20) { rep(j, 1, 20) cin >> g1[i][j]; rep(j, 1, 20) cin >> g2[i][j]; } bfs(20, 20, 20, 1); return 0; }