Borg Maze - poj 3026(BFS + Kruskal 算法)

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

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<queue>
  4 #include<algorithm>
  5 #include <iostream>
  6 using namespace std;
  7 int map[102][102];
  8 int vis[102][102];
  9 int parent[102];
 10 int row, col;
 11 int edge_num = 0;
 12 int point_num=0;
 13 struct edge {
 14     int u, v, w;
 15 } e[102 * 102];
 16 struct points {
 17     int u, v, d;
 18 } point;
 19 bool cmp(const edge e1,const edge e2){
 20     return e1.w<e2.w;
 21 }
 22 int Kruskal(){
 23     int mindis=0;
 24     for(int i=1;i<=point_num;i++){
 25         parent[i]=i;
 26     }
 27     sort(e,e+edge_num,cmp);
 28     int pnum=0;
 29     for(int i=0;i<edge_num&&pnum<point_num;i++){
 30         int k,g;
 31         int u=e[i].u;
 32         int v=e[i].v;
 33         for(k=parent[u];parent[k]!=k;k=parent[parent[k]]);
 34         for(g=parent[v];parent[g]!=g;g=parent[parent[g]]);
 35
 36         if(k!=g){
 37             parent[g]=k;
 38             mindis+=e[i].w;
 39             pnum++;
 40         }
 41         }
 42     return mindis;
 43 }
 44 void BFS(int x, int y) {
 45     queue<points> p;
 46     points s;
 47     s.u = x;
 48     s.v = y;
 49     s.d = 0;
 50     p.push(s);
 51     memset(vis, 0, 102 * 102 * sizeof(int));
 52     vis[x][y] = 1;
 53
 54     while (!p.empty()) {
 55         points tmp = p.front();
 56         p.pop();
 57         for (int i = -1; i < 2; i += 2) {
 58             points next;
 59             next.u = tmp.u + i;
 60             next.v = tmp.v;
 61             next.d = tmp.d + 1;
 62             if (next.u >= 0 && next.u < row) {
 63                 if (!vis[next.u][next.v]) {
 64                     int pos = map[next.u][next.v];
 65                     vis[next.u][next.v] = 1;
 66                     if (pos >= 0)
 67                         p.push(next);
 68                     if (pos >= 1) {
 69                         e[edge_num].u = map[x][y];
 70                         e[edge_num].v = pos;
 71                         e[edge_num].w = next.d;
 72                         edge_num++;
 73                     }
 74                 }
 75             }
 76         }
 77         for (int i = -1; i < 2; i += 2) {
 78             points next;
 79             next.u = tmp.u;
 80             next.v = tmp.v + i;
 81             next.d = tmp.d + 1;
 82             if (next.v >= 0 && next.v < col) {
 83                 if (!vis[next.u][next.v]) {
 84                     int pos = map[next.u][next.v];
 85                     vis[next.u][next.v] = 1;
 86                     if (pos >= 0)
 87                         p.push(next);
 88                     if (pos >= 1) {
 89                         e[edge_num].u = map[x][y];
 90                         e[edge_num].v = pos;
 91                         e[edge_num].w = next.d;
 92                         edge_num++;
 93                     }
 94                 }
 95             }
 96         }
 97
 98     }
 99 }
100 int main() {
101     int n;
102     scanf("%d", &n);
103     char tmp[102];
104     for (int i = 0; i < n; i++) {
105         point_num=0;
106         edge_num=0;
107         scanf("%d %d", &col, &row);
108         gets(tmp); //坑【一串空格】
109         for (int j = 0; j < row; j++) {
110             for (int k = 0; k < col; k++) {
111                 char c;
112                 scanf("%c", &c);
113                 if (c == ‘ ‘)
114                     map[j][k] = 0;
115                 else if (c == ‘#‘)
116                     map[j][k] = -1;
117                 else if (c == ‘\n‘)
118                     k--;
119                 else
120                     map[j][k] = ++point_num;
121             }
122
123         }
124         for (int j = 0; j < row; j++) {
125             for (int k = 0; k < col; k++) {
126                 if (map[j][k] > 0) {
127                     BFS(j, k);
128                 }
129             }
130         }
131         int mid=Kruskal();
132         printf("%d\n",mid);
133     }
134     return 0;
135 }