POJ3026——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.
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
Source
Svenskt M?sterskap i Programmering/Norgesmesterskapet 2001
这里的话，S和A其实没区别的，每个点都要到，然后总距离又要最小，那么就是MST了，但是需要计算点与点之间距离，而地图上有障碍物，所以可以bfs来计算，然后就是prim或者kruskal套下就行了。
kruskal：
这里的话，S和A其实没区别的，每个点都要到，然后总距离又要最小，那么就是MST了，但是需要计算点与点之间距离，而地图上有障碍物，所以可以bfs来计算，然后就是prim或者kruskal套下就行了。
kruskal：
#include<stdio.h>prim：
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<iterator>
using namespace std;
const int maxn=112;
bool vis2[maxn][maxn];
int dir[4][2]={1,0,-1,0,0,1,0,-1};
int x,y,n;
int num[maxn][maxn];
struct node
{
int from,to;
int weight;
}edge[maxn*maxn];
struct node2
{
int x,y;
int step;
}temp1,temp2;
int father[maxn];
char mat[maxn][maxn];
void unit(int n)
{
for(int i=0;i<=n;i++)
father[i]=i;
}
int find(int x)
{
if(x!=father[x])
father[x]=find(father[x]);
return father[x];
}
void merge(int aa,int bb)
{
if(aa!=bb)
father[aa]=bb;
}
int cmp(node a,node b)
{
return a.weight < b.weight;
}
int kruskal(int v_all)
{
sort(edge,edge+n,cmp);
unit(v_all);
int sum=0,m=0;
for(int i=0;i<n;i++)
{
int a=find(edge[i].from);
int b=find(edge[i].to);
if(a!=b)
{
merge(a,b);
sum+=edge[i].weight;
m++;
if(m==v_all-1)
break;
}
}
return sum;
}
bool is_legal(int x0,int y0)
{
if(x0<0 || y0<0 || x0>=y || y0>=x || mat[x0][y0]=='#')
return false;
return true;
}
void solve(int x,int y)
{
bool vis[maxn][maxn];
memset(vis,0,sizeof(vis));
queue<node2>qu;
while(!qu.empty())
qu.pop();
temp1.x=x;
temp1.y=y;
vis[x][y]=1;
temp1.step=0;
qu.push(temp1);
while(!qu.empty())
{
temp1=qu.front();
qu.pop();
for(int i=0;i<4;i++)
{
temp2=temp1;
temp2.x+=dir[i][0];
temp2.y+=dir[i][1];
temp2.step++;
if(!is_legal(temp2.x,temp2.y) || vis[temp2.x][temp2.y])
continue;
if(mat[temp2.x][temp2.y]=='A' || mat[temp2.x][temp2.y]=='S')
{
edge[n].from=num[x][y];
edge[n].to=num[temp2.x][temp2.y];
edge[n++].weight=temp2.step;
vis2[num[x][y]][num[temp2.x][temp2.y]]=1;
}
vis[temp2.x][temp2.y]=1;
qu.push(temp2);
}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
n=0;
memset(vis2,0,sizeof(vis2));
int cnt=0;
scanf("%d%d",&x,&y);
gets(mat[0]);
for(int i=0;i<y;i++)
gets(mat[i]);
for(int i=0;i<y;i++)
for(int j=0;j<x;j++)
if(mat[i][j]=='A' || mat[i][j]=='S')
num[i][j]=cnt++;
for(int i=0;i<y;i++)
for(int j=0;j<x;j++)
if(mat[i][j] == 'A' || mat[i][j] == 'S')
solve(i,j);
printf("%d\n",kruskal(cnt));
}
return 0;
}
#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<set> #include<vector> #include<iterator> using namespace std; const int maxn=105; int dir[4][2]={1,0,-1,0,0,1,0,-1}; int x,y; int num[maxn][maxn]; int dis[maxn][maxn]; struct node2 { int x,y; int step; }temp1,temp2; char mat[maxn][maxn]; int father[maxn]; int lowcost[maxn]; int closeset[maxn]; bool used[maxn]; bool is_legal(int x0,int y0) { if(x0<0 || y0<0 || x0>=y || y0>=x || mat[x0][y0]=='#') return false; return true; } int prim(int v_all) { int i,j,k; for(i=1;i<v_all;i++) { used[i]=0; lowcost[i]=dis[0][i]; // father[i]=-1; closeset[i]=0; } used[0]=1; int min,sum=0; for(i=1;i<=v_all-1;i++) { min=0x3f3f3f3f; j=0; for(k=0;k<v_all;k++) if(!used[k] &&lowcost[k] < min) { min=lowcost[k];//k这个点没进去，然后找他依附的点到他的距离 j=k; } used[j]=1; if(min==0x3f3f3f3f) return -1; sum+=min; father[j]=closeset[j]; for(k=1;k<v_all;k++) { if(!used[k] && lowcost[k]>dis[j][k]) { lowcost[k]=dis[j][k]; closeset[k]=j; } } } return sum; } void solve(int x,int y) { bool vis[maxn][maxn]; memset(vis,0,sizeof(vis)); queue<node2>qu; while(!qu.empty()) qu.pop(); temp1.x=x; temp1.y=y; vis[x][y]=1; temp1.step=0; qu.push(temp1); while(!qu.empty()) { temp1=qu.front(); qu.pop(); for(int i=0;i<4;i++) { temp2=temp1; temp2.x+=dir[i][0]; temp2.y+=dir[i][1]; temp2.step++; if(!is_legal(temp2.x,temp2.y) || vis[temp2.x][temp2.y]) continue; if(mat[temp2.x][temp2.y]=='A' || mat[temp2.x][temp2.y]=='S') { int a=num[x][y]; int b=num[temp2.x][temp2.y]; dis[a][b]=temp2.step; } vis[temp2.x][temp2.y]=1; qu.push(temp2); } } } int main() { int t; scanf("%d",&t); while(t--) { int cnt=0; scanf("%d%d",&x,&y); gets(mat[0]); for(int i=0;i<y;i++) gets(mat[i]); for(int i=0;i<y;i++) for(int j=0;j<x;j++) if(mat[i][j]=='A' || mat[i][j]=='S') num[i][j]=cnt++; for(int i=0;i<y;i++) for(int j=0;j<x;j++) if(mat[i][j] == 'A' || mat[i][j] == 'S') solve(i,j); printf("%d\n",prim(cnt)); } return 0; }