UVA 11573 - Ocean Currents
题目链接
题意：给定一个海面。数字分别代表海流方向，顺着海流不用费能量，逆海流要费1点能量，每次询问给一个起点一个终点，问起点到终点耗费的最小能量
思路：广搜。队列用优先队列，每次取能量最低的点出来进行状态的转移
代码：
#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;
}