思路：dp[inx][x][y],表示用了前inx个指令后的最小费用。
对于一个指令，我们可以选择不走或者走，其他的我们可以添加四个方向的指令与使用过指令后的dp来比较。
1 #include <iostream>
2 #include <algorithm>
3 #include <cstdio>
4 #include <queue>
5 #include <cstring>
6
7 using namespace std;
8
9 const int N = 60;
10 const int INF = 1e9;
11 int mv_x[] = {1, -1, 0, 0};
12 int mv_y[] = {0, 0, -1, 1};
13 char mp[N][N];
14 int dp[N][N][N];
15 char str[N];
16 int n, m;
17
18 struct node
19 {
20 int inx, x, y;
21 }R, E;
22
23 bool inline check(int x, int y)
24 {
25 return x >= 0 && x < n && y >= 0 && y < m;
26 }
27
28 void bfs()
29 {
30 for(int i = 0; i < N; ++i)
31 for(int j = 0; j < N; ++j)
32 for(int k = 0; k < N; ++k)
33 dp[i][j][k] = INF;
34
35 int len = strlen(str);
36 queue<node > que;
37
38 dp[0][R.x][R.y] = 0;
39 que.push({0, R.x, R.y});
40
41 while(!que.empty()){
42 node now = que.front();
43 que.pop();
44
45 for(int p = 0; p < 4; ++p){
46 int dx = now.x + mv_x[p];
47 int dy = now.y + mv_y[p];
48
49 if(check(dx, dy) && mp[dx][dy] != '#'){
50 if(dp[now.inx][now.x][now.y] + 1 < dp[now.inx][dx][dy]){
51 dp[now.inx][dx][dy] = dp[now.inx][now.x][now.y] + 1;
52 que.push({now.inx, dx, dy});
53 }
54 }
55
56 if(now.inx < len){
57 //不走
58 if(dp[now.inx][now.x][now.y] + 1 < dp[now.inx + 1][now.x][now.y]){
59 dp[now.inx + 1][now.x][now.y] = dp[now.inx][now.x][now.y] + 1;
60 que.push({now.inx + 1, now.x, now.y});
61 }
62
63 //走
64 dx = now.x;
65 dy = now.y;
66 if(str[now.inx] == 'L') dy--;
67 if(str[now.inx] == 'R') dy++;
68 if(str[now.inx] == 'U') dx--;
69 if(str[now.inx] == 'D') dx++;
70 if(!check(dx,dy) || mp[dx][dy] == '#') dx = now.x, dy = now.y;
71 if(dp[now.inx][now.x][now.y] < dp[now.inx + 1][dx][dy]){
72 dp[now.inx + 1][dx][dy] = dp[now.inx][now.x][now.y];
73 que.push({now.inx + 1, dx, dy});
74 }
75 }
76 }
77 }
78 }
79
80 void solve()
81 {
82 scanf("%d%d", &n, &m);
83 for(int i = 0; i < n; ++i) scanf("%s", mp[i]);
84 scanf("%s", str);
85 //cout << "len = " << (strlen(str)) << endl;
86
87 for(int i = 0; i < n; ++i){
88 for(int j = 0; j < m; ++j){
89 if(mp[i][j] == 'R'){
90 R.x = i; R.y = j;
91 }
92 if(mp[i][j] == 'E'){
93 E.x = i; E.y = j;
94 }
95 }
96 }
97
98 bfs();
99
100 int ans = INF;
101 for(int i = 0; i < N; ++i){
102 ans = min(ans, dp[i][E.x][E.y]);
103 }
104 //printf("ans = %d\n", ans);
105 printf("%d\n", ans);
106 }
107
108
109 int main()
110 {
111
112 solve();
113
114 return 0;
115 }
来源：oschina
链接：https://my.oschina.net/u/4312837/blog/4322188