Ultraviolet A 11573-ocean currents (BFS + priority queue)

Ultraviolet A 11573-ocean currents

Question Link

Given a sea surface, the number represents the current direction, and does not charge energy along the current. The reverse current consumes 1 point of energy. Each time you ask a start point, an endpoint is given, minimum energy consumption from the start point to the end point

Train of Thought: broad search, the queue uses a priority queue, each time the lowest energy point out for status transfer

Code:

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int d[8][2] = {{-1, 0}, {-1, 1}, {0, 1}, {1, 1}, {1, 0}, {1, -1}, {0, -1}, {-1, -1}};const int N = 1005;int n, m, vis[N][N];char g[N][N];struct Node {int x, y, val;Node() {}Node(int x, int y, int val) {this->x = x;this->y = y;this->val = val;}bool operator < (const Node& c) const {return val > c.val;}void read() {scanf("%d%d", &x, &y);}}s, e;int bfs() {priority_queue<Node> Q;s.val = 0;Q.push(s);memset(vis, -1, sizeof(vis));vis[s.x][s.y] = 0;while (!Q.empty()) {Node u = Q.top();if (u.x == e.x && u.y == e.y) return u.val;Q.pop();for (int i = 0; i < 8; i++) {int xx = u.x + d[i][0];int yy = u.y + d[i][1];int val = u.val;if (xx < 1 || xx > n || yy < 1 || yy > m) continue;if (i != g[u.x][u.y] - '0')val++;if (vis[xx][yy] == -1 || val < vis[xx][yy]) {vis[xx][yy] = val;Q.push(Node(xx, yy, val));}}}}int main() {while (~scanf("%d%d", &n, &m)) {for (int i = 1; i <= n; i++)scanf("%s", g[i] + 1);int q;scanf("%d", &q);while (q--) {s.read();e.read();printf("%d\n", bfs());}}return 0;}

Ultraviolet A 11573-ocean currents (BFS + priority queue)