UVa1600 Patrol Robot (最短路)

链接：http://acm.hust.edu.cn/vjudge/problem/49764分析：多了一个不能连续穿过k个障碍物的限制条件，那么就在状态中增加一个表示已连续穿过cnt个障碍物的描述，vis也要增加一维变成vis[x][y][cnt]，毕竟vis用来记录当前状态是否曾经已经出现过。碰到障碍物且拿来扩展的状态cnt小于等于k就将cnt++，否则遇到空地就将cnt清零。

 1 #include <cstdio>
 2 #include <queue>
 3 #include <cstring>
 4 using namespace std;
 5
 6 const int maxn = 20 + 5;
 7
 8 int G[maxn][maxn], vis[maxn][maxn][maxn], d[maxn][maxn];
 9
10 struct Node {
11     int x, y, cnt;
12 };
13
14 const int dx[4] = {0, 0, 1, -1};
15 const int dy[4] = {1, -1, 0, 0};
16
17 int main() {
18     //freopen("slyar.in", "r", stdin);
19     //freopen("slyar.out", "w", stdout);
20     int T;
21     scanf("%d", &T);
22     while (T--) {
23         int m, n, k;
24         scanf("%d%d%d", &m, &n, &k);
25         memset(G, 0, sizeof(G));
26         for (int i = 1; i <= m; i++)
27             for (int j = 1; j <= n; j++)
28                 scanf("%d", &G[i][j]);
29         memset(vis, 0, sizeof(vis));
30         memset(d, -1, sizeof(d));
31         d[1][1] = 0; vis[1][1][0] = 1;
32         Node t; t.x = 1, t.y = 1, t.cnt = 0;
33         queue<Node> q;
34         q.push(t);
35         while (!q.empty()) {
36             Node u = q.front(); q.pop();
37             //  printf("x = %d, y = %d, cnt = %d\n", u.x, u.y, u.cnt);
38             if (u.x == m && u.y == n) break;
39             for (int i = 0; i < 4; i++) {
40                 int nx = u.x + dx[i], ny = u.y + dy[i];
41                 if (nx >= 1 && nx <= m && ny >= 1 && ny <= n) {
42                      if (G[nx][ny] == 1 && u.cnt + 1 > k) continue;
43                      Node v = u; v.x = nx, v.y = ny;
44                      if (G[nx][ny] == 1) v.cnt++; else v.cnt = 0;
45                      if (vis[nx][ny][v.cnt]) continue; else vis[nx][ny][v.cnt] = 1;
46                      d[nx][ny] = d[u.x][u.y] + 1;
47                      q.push(v);
48                 }
49             }
50         }
51         printf("%d\n", d[m][n]);
52     }
53     return 0;
54 }