// [9/19/2014 Sjm]
/*
dis[j][k] := 从 j 点到 k 点的最少步数，由于They can travel in all 8 adjacent direction in one step.故而 dis[j][k] = max( abs(Xj - Xk), abs(Yj - Yk) )f[j][i] := 在 i 状态下，最后收集坚果 j 的最少步数n 代表坚果的数目。。阶段i：按递增顺序枚举状态值 (0 <= i <= 2^n - 1)
状态 ：枚举状态 i 中最后被收集的坚果 j （1 <= j <= n, i&(2^(j-1)) != 0）
决策 ：枚举状态 i 以外的坚果 k（1 <= k <= n, i&(2^(k-1)) == 0） ,判断在状态 i，最后被收集的坚果为j的情况下，再收集坚果 k ,是否为最优决策。即： f[k][i+2^(k-1)] = min(f[k][i+2^(k-1)], f[j][i] + dis[j][k])求最终解：
枚举 f[j][2^n - 1] + dis[j][0] (1 <= j <= n),
*/
1 #include <iostream>
2 #include <cstdlib>
3 #include <cstdio>
4 #include <algorithm>
5 #define INF 0x3f3f3f3f
6 using namespace std;
7 int n, dis[20][20], myX, myY, f[20][(1 << 20)];
8
9 struct myNode {
10 int x, y;
11 }node[20];
12
13 void Init(int i, char str[])
14 {
15 for (int j = 0; j < myY; ++j){
16 if ('#' == str[j]) {
17 node[++n].x = i;
18 node[n].y = j;
19 }
20 else if ('L' == str[j]) {
21 node[0].x = i;
22 node[0].y = j;
23 }
24 }
25 }
26
27 void getDis() {
28 for (int i = 0; i <= n; ++i) {
29 for (int j = 0; j <= n; ++j) {
30 dis[i][j] = max(abs(node[i].x - node[j].x), abs(node[i].y - node[j].y));
31 }
32 }
33 }
34
35 void Solve() {
36 int finalState = (1 << n) - 1;
37 for (int j = 1; j <= n; ++j) {
38 for (int i = 0; i <= finalState; ++i) {
39 f[j][i] = INF;
40 }
41 }
42 for (int j = 1; j <= n; ++j) {
43 f[j][1 << (j - 1)] = dis[0][j];
44 }
45 for (int i = 0; i < finalState; ++i) {
46 for (int j = 1; j <= n; ++j) {
47 if (i & (1 << (j - 1))) {
48 for (int k = 1; k <= n; ++k) {
49 if (!(i & (1 << (k - 1)))) {
50 f[k][i + (1 << (k - 1))] = min(f[k][i + (1 << (k - 1))], f[j][i] + dis[j][k]);
51 }
52 }
53 }
54 }
55 }
56 int ans = INF;
57 for (int j = 1; j <= n; ++j) {
58 ans = min(ans, f[j][finalState] + dis[j][0]);
59 }
60 printf("%d\n", ans);
61 }
62
63 int main()
64 {
65 while (~scanf("%d %d", &myX, &myY)) {
66 char str[25];
67 n = 0;
68 for (int i = 0; i < myX; ++i) {
69 scanf("%s", str);
70 Init(i, str);
71 }
72 if (0 == n) { // 注意无坚果的情况。。
73 printf("0\n");
74 continue;
75 }
76 getDis();
77 Solve();
78 }
79 return 0;
80 }
)
