POJ - 1376 Robot

来源：互联网发布：淘宝新手如何提高销量编辑：程序博客网时间：2024/05/15 23:52

题意：求在可以一秒沿着既定方向走1到3步和向左或右转90度的情况下，从起点到终点的最短时间

思路：坑的是这机器人还有体积，所以不能走到边界，然后就是单纯的BFS

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int MAXN = 110;struct node {int x,y;int time,pos;};int n,m,sx,sy,sdir,ex,ey,map[MAXN][MAXN];int vis[MAXN][MAXN][4];int dx[4] = {-1, 0, 1, 0};int dy[4] = {0, 1, 0, -1}; int getDir(char str[]) {if (!strcmp(str, "north"))return 0;if (!strcmp(str, "east"))return 1;if (!strcmp(str, "south"))return 2;return 3;}int check(int x, int y) {if (x < 1 || x >=n || y < 1 || y >= m)return 0;if (map[x][y] || map[x+1][y] || map[x][y+1] || map[x+1][y+1])return 0;return 1;}int bfs() {queue<node> q;node a;a.x = sx, a.y = sy, a.pos = sdir, a.time = 0;q.push(a);vis[sx][sy][sdir] = 1;while (!q.empty()) {node cur = q.front();q.pop();if (cur.x == ex && cur.y == ey) return cur.time;int nx = cur.x;int ny = cur.y;for (int i = 1; i < 4; i++) {nx += dx[cur.pos];ny += dy[cur.pos];if (!check(nx, ny))break;if (!vis[nx][ny][cur.pos]) {vis[nx][ny][cur.pos] = 1;node cnt;cnt.x = nx, cnt.y = ny;cnt.pos = cur.pos, cnt.time = cur.time+1;q.push(cnt);}}for (int i = 0; i < 4; i++) {if (max(cur.pos, i)-min(cur.pos, i) == 2)continue;if (vis[cur.x][cur.y][i])continue;vis[cur.x][cur.y][i] = 1;node cnt = cur;cnt.pos = i;cnt.time = cur.time+1;q.push(cnt);}}return -1;}int main() {while (scanf("%d%d", &n, &m) != EOF && n+m) {for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)scanf("%d", &map[i][j]);memset(vis, 0, sizeof(vis));scanf("%d%d%d%d", &sx, &sy, &ex, &ey);char dir[20];scanf("%s", dir);sdir = getDir(dir);if (sx == ex && sy == ey) {printf("0\n");continue;}if (!check(ex, ey)) {printf("-1\n");continue;}int ans = bfs();if (ans != -1)printf("%d\n", ans);else printf("-1\n");}return 0;}

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