题目链接：http://poj.org/problem?id=3026
Svenskt Masterskap我程序员/ Norgesmesterskapet 2001
Description
The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the group consciousness of the Borg civilization. Each Borg individual is linked to the collective by a sophisticated subspace network that insures each member is given constant supervision and guidance.
Your task is to help the Borg (yes, really) by developing a program which helps the Borg to estimate the minimal cost of scanning a maze for the assimilation of aliens hiding in the maze, by moving in north, west, east, and south steps. The tricky thing is that the beginning of the search is conducted by a large group of over 100 individuals. Whenever an alien is assimilated, or at the beginning of the search, the group may split in two or more groups (but their consciousness is still collective.). The cost of searching a maze is definied as the total distance covered by all the groups involved in the search together. That is, if the original group walks five steps, then splits into two groups each walking three steps, the total distance is 11=5+3+3.
Input
On the first line of input there is one integer, N <= 50, giving the number of test cases in the input. Each test case starts with a line containg two integers x, y such that 1 <= x,y <= 50. After this, y lines follow, each which x characters. For each character, a space `` ‘‘ stands for an open space, a hash mark ``#‘‘ stands for an obstructing wall, the capital letter ``A‘‘ stand for an alien, and the capital letter ``S‘‘ stands for the start of the search. The perimeter of the maze is always closed, i.e., there is no way to get out from the coordinate of the ``S‘‘. At most 100 aliens are present in the maze, and everyone is reachable.
Output
For every test case, output one line containing the minimal cost of a succesful search of the maze leaving no aliens alive.
Sample Input 2 6 5 ##### #A#A## # # A# #S ## ##### 7 7 ##### #AAA### # A# # S ### # # #AAA### ##### Sample Output 8 11
分析：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stack>
6 #include<math.h>
7 #include<queue>
8 #include<map>
9 using namespace std;
10
11 #define INF 0x3f3f3f3f
12 #define N 1200
13
14 int n,m,maps[N][N],b[N][N],dist[N],vis[N][N],viss[N];
15 char s[N][N];
16 int dir[4][2]={ {0,1},{0,-1},{1,0},{-1,0} };
17
18 struct node
19 {
20 int x,y,step;
21 }a[N];
22
23 void Init()
24 {
25 int i,j;
26 for(i=0;i<m*n;i++)
27 dist[i]=INF;
28 for(j=0;j<n*m;j++)
29 maps[i][j]=(i==j)?0:INF;
30 }
31
32 void bfs(int ss,int x,int y)
33 {
34 int i;
35
36 memset(vis,0,sizeof(vis));
37 queue<node>Q;
38 node B,Next,Now;
39 B.x=x;
40 B.y=y;
41 B.step=0;
42 Q.push(B);
43 vis[x][y]=1;
44
45 while(Q.size())
46 {
47 Now=Q.front();
48 Q.pop();
49
50 if(s[Now.x][Now.y]>=‘A‘&&s[Now.x][Now.y]<=‘Z‘)
51 maps[ss][b[Now.x][Now.y]]=Now.step;
52
53 for(i=0;i<4;i++)
54 {
55 Next.x=Now.x+dir[i][0];
56 Next.y=Now.y+dir[i][1];
57 if(Next.x<m&&Next.x>=0&&Next.y<n&&Next.y>=0&&vis[Next.x][Next.y]==0&&s[Next.x][Next.y] != ‘#‘)
58 {
59 vis[Next.x][Next.y]=1;
60 Next.step=Now.step+1;
61 Q.push(Next);
62 }
63 }
64 }
65 }
66
67 int Prim(int e)
68 {
69 int i,j;
70 viss[1]=1;
71 for(int i=1;i<=e;i++)
72 dist[i]=maps[1][i];
73 int sum=0;
74 for(i=1;i<=e;i++)
75 {
76 int Min=INF,index=-1;
77
78 for(j=1;j<=e;j++)
79 if(!viss[j]&&Min>dist[j])
80 {
81 Min=dist[j];
82 index=j;
83 }
84 if(index==-1)break;
85 sum+=Min;
86 viss[index]=1;
87
88 for(j=1;j<=e;j++)
89 if(!viss[j]&&dist[j]>maps[index][j])
90 dist[j]=maps[index][j];
91 }
92
93 return sum;
94 }
95
96 int main()
97 {
98 int T,i,j;
99
100 scanf("%d", &T);
101
102 while(T--)
103 {
104 memset(a,0,sizeof(a));
105 memset(viss,0,sizeof(viss));
106
107 scanf("%d%d ", &n,&m);
108 Init();
109
110 int cnt=1;
111 for(i=0;i<m;i++)
112 {
113 gets(s[i]);
114 for(j=0;j<n;j++)
115 {
116 if(s[i][j]<=‘Z‘&&s[i][j]>=‘A‘)
117 b[i][j]=cnt++;///记录cnt号动点位置
118 }
119 }
120
121 for(i=0;i<m;i++)
122 {
123 for(j=0;j<n;j++)
124 {
125 if(s[i][j]<=‘Z‘&&s[i][j]>=‘A‘)
126 bfs(b[i][j],i,j);///搜索，记录各个点之间步数
127 }
128 }
129
130 int ans=Prim(cnt-1);///最小生成树求最终值
131 printf("%d\n", ans);
132 }
133 }