swust oj 191--迷宫逃离（打表搜索）

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <queue>
 5 using namespace std;
 6 
 7 #define inf 0x3f3f3f3f
 8 #define maxn  110
 9 struct node{
10     int x, y;
11 };
12 int n, m;
13 node p[4];
14 char mpt[maxn][maxn];
15 int vis[maxn][maxn][4];//vis分别代表每个人到每一个点需要消掉的障碍物
16 int dir[4][2] = { 1, 0, -1, 0, 0, 1, 0, -1 };
17 
18 void bfs(int index)
19 {
20     queue<node> q;
21     node now, next;
22     now.x = p[index].x;
23     now.y = p[index].y;
24     q.push(now);
25     vis[now.x][now.y][index] = 0;
26     while (!q.empty()){
27         now = q.front();
28         q.pop();
29         for (int i = 0; i < 4; i++){
30             next = now;
31             next.x += dir[i][0];
32             next.y += dir[i][1];
33             if (next.x<1 || next.x>n || next.y<1 || next.y>m) continue;
34             int tmp = vis[now.x][now.y][index];
35             if (mpt[next.x][next.y] == ‘#‘)tmp++;//遇到一个障碍物，那么此人到达这个点需要消掉的障碍物加1
36             if (tmp < vis[next.x][next.y][index]){
37                 //判断是否可以让需要消掉的障碍物更小
38                 vis[next.x][next.y][index] = tmp;
39                 q.push(next);
40             }
41         }
42     }
43 }
44 int main(){
45     while (~scanf("%d%d", &n, &m)){
46         memset(vis, inf, sizeof(vis));
47         for (int i = 1; i <= n; i++) scanf("%s", mpt[i] + 1);
48         for (int i = 1; i <= n; i++){
49             for (int j = 1; j <= m; j++){
50                 if (mpt[i][j] == ‘w‘) p[1].x = i, p[1].y = j;
51                 else if (mpt[i][j] == ‘W‘) p[2].x = i, p[2].y = j;
52                 else if (mpt[i][j] == ‘f‘) p[3].x = i, p[3].y = j;
53             }
54         }
55         //对每一个人计算出到每一个点需要消掉的最少障碍物
56         bfs(1);
57         bfs(2);
58         bfs(3);
59         int ans = inf;
60         for (int i = 1; i <= n; i++){
61             //如果三个人两两可以到达，那么他们可以集聚在一个点，下面枚举每一个这样的点
62             for (int j = 1; j <= m; j++){
63                 int tmp = 0;
64                 for (int k = 1; k <= 3; k++) tmp += vis[i][j][k];
65                 if (mpt[i][j] == ‘#‘) tmp -= 2;//如果在一个‘#‘集聚，那么就会多计算2个
66                 ans = ans < tmp ? ans : tmp;
67             }
68         }
69         printf("%d\n", ans);
70     }
71     return 0;
72 }