Borg Maze
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12032 | Accepted: 3932 |
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.
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
题意：求输入图中'S'点到所有'A'点的最小距离。
题解：这个题出的很好，要把一幅图转换为另外一幅图，用BFS把每一个点到其余所有的点的距离全部求一遍，然后按照点的下标构造一副新的图。构造完后，用prim算法求最小生成树即可。
这个题的输入处理有点问题。。不能用getchar()，我是参考了别人的代码改成这样才AC
getchar()---------->char temp[51];
gets(temp);
#include <stdio.h> #include <algorithm> #include <string.h> #include <queue> using namespace std; const int N = 200; const int INF = 9999999999; char a[N][N]; bool vis[N][N]; int graph[N][N]; int dis[N][N]; int n,m,k; struct Point { int x,y; } p[N*2]; void input(int &k) { for(int i=0; i<n; i++){ gets(a[i]); for(int j=0; j<m; j++){ if(a[i][j]=='S'){ p[0].x = i; p[0].y = j; } if(a[i][j]=='A'){ p[k].x =i; p[k++].y =j; } } } for(int i=0;i<k;i++){ for(int j=0;j<k;j++) graph[i][j] = INF; } } void BFS(Point start,int start1){ int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}}; memset(vis,false,sizeof(vis)); memset(dis,0,sizeof(dis)); queue<Point> q; q.push(start); vis[start.x][start.y] = true; while(!q.empty()){ Point t = q.front(); q.pop(); for(int i=0;i<4;i++){ int x = t.x+dir[i][0]; int y = t.y+dir[i][1]; if(x<0||x>=n||y<0||y>=m||vis[x][y]||a[x][y]=='#') continue; dis[x][y]= dis[t.x][t.y]+1; vis[x][y] = true; Point p1; p1.x = x,p1.y = y; q.push(p1); } } for(int i=0;i<k;i++){ graph[start1][i] = dis[p[i].x][p[i].y]; } return; } bool vis1[N*2]; int low[N*2]; int prim(int n,int pos){ memset(vis1,false,sizeof(vis1)); memset(low,0,sizeof(low)); for(int i=0;i<n;i++){ low[i] = graph[pos][i]; } int cost = 0; vis1[pos] = true; low[pos] = 0; for(int i=1;i<n;i++){ int Min = INF; for(int j=0;j<n;j++){ if(!vis1[j]&&Min>low[j]){ pos = j; Min = low[j]; } } cost += Min; vis1[pos] = true; for(int j=0;j<n;j++){ if(!vis1[j]&&low[j]>graph[pos][j]) low[j] = graph[pos][j]; } } return cost; } int main() { int tcase; scanf("%d",&tcase); while(tcase--) { scanf("%d%d",&m,&n); char temp[51]; gets(temp); k=1; input(k); for(int i=0;i<k;i++) BFS(p[i],i); printf("%d ",prim(k,0)); } return 0; }