hdu1324 Holedox Moving (bfs+状态压缩+A*)
2014-12-22 19:05
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题意:
一个贪吃蛇,地图上有障碍,问走到(1,1)的最少步数。思路:
对于状态的表示,由于蛇身的方块一定是一个连着一个的,所以可以只记录蛇头的坐标,然后用一个2位的二进制数表示接下来每个方块的方向。这样状态只有20*20*2^14种,不会超内存。
若当前贪吃蛇的状态为stat,接下来走的方向为pos,那么下一个状态就是:
pos = (pos + 2) % 4; (stat >> 2) + (pos << ((l << 1) - 4));
将当前的步数step与当前坐标与(1,1)的曼哈顿距离dist相加,每次取队列中step+dist最小的扩展(A*)。
完整代码:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <queue> using namespace std; struct Node{ int x, y, step, stat; Node() {} Node(int a, int b, int c, int d) : x(a), y(b), step(c), stat(d) {} bool operator < (const Node &n) const { int dist = x + y - 2; int ndist = n.x + n.y - 2; return step + dist > n.step + ndist; } }; int n, m, l, k; int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; int map[22][22]; bool vis[22][22][16388]; Node start; int direction[11]; bool check(int x, int y, Node cur) { if (x < 1 || x > n || y < 1 || y > m || map[x][y] == 1) return false; for (int i = l - 1; i >= 1; i--) { direction[i] = cur.stat & 3; cur.stat >>= 2; } for (int i = 1; i < l; i++) { cur.x += dir[direction[i]][0]; cur.y += dir[direction[i]][1]; if (cur.x == x && cur.y == y) return false; } return true; } int bfs() { if (start.x == 1 && start.y == 1) return 0; memset(vis, 0, sizeof(vis)); priority_queue<Node> q; q.push(start); vis[start.x][start.y][start.stat] = 1; while (!q.empty()) { Node cur = q.top(); q.pop(); for (int i = 0; i < 4; i++) { int nx = cur.x + dir[i][0]; int ny = cur.y + dir[i][1]; if (!check(nx, ny, cur)) continue; int pos = (i + 2) % 4; int stat = (cur.stat >> 2) + (pos << ((l << 1) - 4)); if (!vis[nx][ny][stat]) { vis[nx][ny][stat] = 1; if (nx == 1 && ny == 1) return cur.step + 1; q.push(Node(nx, ny, cur.step + 1, stat)); } } } return -1; } int main() { int cas = 1; while (~scanf("%d %d %d", &n, &m, &l) && n && m && l) { memset(map, 0, sizeof(map)); int x, y; scanf("%d %d", &x, &y); start = Node(x, y, 0, 0); int stat = 0; int prex = x, prey = y; int pos; for (int i = 1; i < l; i++) { scanf("%d %d", &x, &y); if (x == prex - 1) pos = 0; if (x == prex + 1) pos = 2; if (y == prey - 1) pos = 3; if (y == prey + 1) pos = 1; prex = x, prey = y; stat <<= 2; stat |= pos; } start.stat = stat; scanf("%d", &k); for (int i = 0; i < k; i++) { scanf("%d %d", &x, &y); map[x][y] = 1; } int ans = bfs(); printf("Case %d: %d\n", cas++, ans); } return 0; }
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