poj 3026 Borg Maze(bfs+prim)

Borg Maze

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

题目大意：输入x,y，表示y行x列。字符为空格时，表示可以移动，为‘#’时，表示墙壁，不可穿越，当为‘A’，‘S’时，表示节点，‘S’为起点。        求所有点连通时，权值最小为多少

  1 /*注意：G++递交AC，C++递交编译错误  Compile Error*/
  2 #include <iostream>
  3 #include <cstdio>
  4 #include <cstdlib>
  5 #include <cstring>
  6 #include <queue>
  7 using namespace std;
  8 const int MAX=150;
  9 const int INF=0xfffffff;
 10 char g[MAX][MAX];  //存放地图
 11 int egde[MAX][MAX]; //字母间的最短距离
 12 int vis[MAX][MAX],num[MAX][MAX];//vis广搜用的标记数组，num给字母编号
 13 int dis[MAX][MAX];
 14 int t,x,y,cnt,nc;
 15 int dirx[]= {0,0,1,-1}; //四个方向的坐标
 16 int diry[]= {1,-1,0,0};
 17 struct Pos
 18 {
 19     int x;
 20     int y;
 21 };
 22 void fun()  //给字母编号
 23 {
 24     nc=0;  // 字母个数
 25     memset(num,-1,sizeof(num));
 26     for(int i=0; i<y; i++)
 27     {
 28         for(int j=0; j<x; j++)
 29         {
 30             if(g[i][j]==‘A‘||g[i][j]==‘S‘)
 31                 num[i][j]=nc++;
 32             if(g[i][j]==‘S‘)  //起点的编号
 33                 cnt=nc-1;
 34         }
 35     }
 36 }
 37 void bfs(int cx,int cy)  // 广搜确定最短距离
 38 {
 39     queue<Pos>Q;
 40     memset(vis,0,sizeof(vis));
 41     vis[cx][cy]=1;
 42     dis[cx][cy]=0;
 43     Q.push((Pos)
 44     {
 45         cx,cy
 46     });
 47     while(!Q.empty())
 48     {
 49         Pos e=Q.front();
 50         Q.pop();
 51         int x=e.x,y=e.y;
 52         if(num[x][y]!=-1)  //为-1时，即（x,y）为字母
 53             egde[num[cx][cy]][num[x][y]]=dis[x][y];
 54         for(int d=0; d<4; d++)
 55         {
 56             int sx=x+dirx[d];
 57             int sy=y+diry[d];
 58             if(sx<0&&sy<0&&sx>=y&&sy>=x)
 59                 continue;
 60             if(vis[sx][sy]||g[sx][sy]==‘#‘)
 61                 continue;
 62             vis[sx][sy]=1;
 63             dis[sx][sy]=dis[x][y]+1;
 64             Q.push((Pos)
 65             {
 66                 sx,sy
 67             });
 68         }
 69     }
 70 }
 71 int prim(int start) //套prim模板
 72 {
 73     int i,j,k,Min,ans=0;
 74     int d[MAX];
 75     bool visit[MAX];
 76     memset(visit,false,sizeof(visit));
 77     for(i=0; i<nc; i++)
 78         d[i]=egde[start][i];
 79     visit[start]=1;
 80     d[start]=0;
 81     for(i=0; i<nc-1; i++)
 82     {
 83         Min=INF,k=-1;
 84         for(j=0; j<nc; j++)
 85         {
 86             if(!visit[j]&&d[j]<Min)
 87             {
 88                 Min=d[j];
 89                 k=j;
 90             }
 91         }
 92         ans+=Min;
 93         visit[k]=1;
 94         for(j=0; j<nc; j++)
 95         {
 96             if(!visit[j]&&d[j]>egde[k][j])
 97                 d[j]=egde[k][j];
 98         }
 99     }
100     return ans;
101 }
102 int main()
103 {
104     cin>>t;
105     char s[500];
106     while(t--)
107     {
108         scanf("%d%d",&x,&y);
109         gets(s);  //注意：或许有无数空格
110         for(int i=0; i<y; i++)
111             gets(g[i]);
112         fun();
113         for(int i=0; i<y; i++)
114         {
115             for(int j=0; j<x; j++)
116             {
117                 if(g[i][j]==‘A‘||g[i][j]==‘S‘)
118                     bfs(i,j);
119             }
120         }
121         printf("%d\n",prim(cnt));
122     }
123     return 0;
124 }