題干：

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

題目大意：

這個題意是真難懂啊、、、

在一個迷宮中有很多外星人A，某個人在S點開始去抓捕，抓捕到外星人后這個人可以分身為任意個（也可以不分身），求把所有外星人抓完所用的最小距離，距離包括分身走的距離（即所有的距離都算上，比如：兩個分身同時走三步，那么走的距離就是6，而不是3），在迷宮之中只能向東南西北四個方向移動。

換種說法：

給你一個地圖，地圖中A代表外星人，S代表出發點，從S點出發，在起點S 或是 每遇到一個A點 便可不分裂或是分裂成多個。求把所有A都吃完需要多少步。注意不要以為當前點隨時都能分裂，而是只有在S或是遇到A才能分裂。

解題報告：

? 注意這題樣例 n m 和字符串之間可能有一行空行（而不一定是一個空格）真tm格式、、所以getchar會WA的。。

因為不難發現，既然這個題意的話，那么起點在哪并沒有怎么所謂了，因為S的所有特點和A的所有特點是完全相同的，所以可以把S點也看成是一個A點，那么題意轉化成：從一個A點出發，在所有A點相連的路徑中選擇一些路徑，使得走過的路最小。

那么解法很顯然了，假設有tot個S和A，那么用bfs構造出這tot個點之間的距離，然后求個MST就好，因為稠密圖而且是完全圖所以首選prim。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 200 + 5; const int INF = 0x3f3f3f3f; int n,m; char s[MAX][MAX]; int bk[MAX][MAX],cost[MAX][MAX],tot; bool vis[MAX][MAX]; int nx[4] = {0,1,0,-1}; int ny[4] = {1,0,-1,0}; struct Node {int x,y,t;Node(int x=0,int y=0,int t=0):x(x),y(y),t(t){} }; void bfs(int stx,int sty) {int id1 = bk[stx][sty];queue<Node> q;q.push(Node(stx,sty,0));memset(vis,0,sizeof vis);vis[stx][sty]=1;while(!q.empty()) {Node cur = q.front();q.pop();int id2 = bk[cur.x][cur.y];if(bk[cur.x][cur.y]) cost[id1][id2] = cur.t;for(int k = 0; k<4; k++) {int tx = cur.x+nx[k];int ty = cur.y+ny[k];if(tx<0||tx>n||ty<1||ty>m) continue;if(vis[tx][ty] || s[tx][ty] == '#') continue;vis[tx][ty] = 1;q.push(Node(tx,ty,cur.t+1));}} } int VIS[MAX*MAX],dis[MAX*MAX]; int prim() {int res = 0;for(int i = 1; i<=tot; i++) dis[i] = INF,VIS[i] = 0;dis[1] = 0;while(1) {int minv,minw=INF;for(int i = 1; i<=tot; i++) {if(VIS[i] == 0 && dis[i] < minw) minw = dis[i],minv = i;}if(minw == INF) break;VIS[minv] = 1;for(int i = 1; i<=tot; i++) {if(VIS[i] == 0 && dis[i] > cost[minv][i]) dis[i] = cost[minv][i];}res += dis[minv];}return res; } int main() {int t;cin>>t;while(t--) {tot=0;memset(bk,0,sizeof bk);memset(cost,0,sizeof cost);scanf("%d%d",&m,&n);string qq;getline(cin,qq);for(int i = 1; i<=n; i++) {cin.getline(s[i]+1,1005);for(int j = 1; j<=m; j++) {if(s[i][j] == 'S' || s[i][j] == 'A') bk[i][j] = ++tot;}}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(bk[i][j]) bfs(i,j);}}printf("%d\n",prim());} return 0 ; }

剛開始WA在數組就開了50*50這么大，導致cost數組也開了[50][50],但是其實應該[2501][2501]，但是其實這題說了最多100個外星人，所以可以開[105][105]就夠了。

總結

以上是生活随笔為你收集整理的【POJ - 3026】Borg Maze（bfs预处理 + 最小生成树，建图）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。