大家好，又见面了，我是全栈君。
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7998 | Accepted: 2675 |
Description
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
Output
Sample Input
2 6 5 ##### #A#A## # # A# #S ## ##### 7 7 ##### #AAA### # A# # S ### # # #AAA### #####
Sample Output
8 11
最小生成树问题，主要是要求出点点距离，即字母间边权。这里用BFS求出了边权，用prim算法求出了最小生成树路径。
AC代码例如以下：
<pre name="code" class="cpp">#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#define inf 100000
using namespace std;
char map[60][60];
int vis[60][60],v[60][60];
int n,m,tt,ans;
int dx[4]={-1,0,1,0};
int dy[4]={0,1,0,-1};
struct H
{
int h,z;
}a[10005];
int w[251][251],lc[251],vv[60][60];//W是边权记录，LCprim算法运用，VV记录第几个字母的坐标
struct HH
{
int h,z,st;
};
int bfs(int h,int z)
{
int i;
v[h][z]=1;
queue <HH> q;
HH a,b,c;
a.h=h;
a.z=z;
a.st=0;
q.push(a);
while(!q.empty())
{
b=q.front();
q.pop();
if(vv[b.h][b.z])//统计边权
{
w[vv[h][z]][vv[b.h][b.z]]=b.st;
w[vv[b.h][b.z]][vv[h][z]]=b.st;
//cout<<vv[h][z]<<" "<<vv[b.h][b.z]<<" "<<b.st<<endl;
}
for(i=0;i<4;i++)
{
c.h=b.h+dx[i];
c.z=b.z+dy[i];
if(c.h>=0&&c.h<n&&c.z>=0&&c.z<m&&map[c.h][c.z]!='#'&&!v[c.h][c.z])
{
v[c.h][c.z]=1;
c.st=b.st+1;
q.push(c);
}
}
}
}
void prim()
{
int i,j;
int m,id;
for(i=1;i<tt;i++)
{
lc[i]=w[1][i];
}
lc[1]=0;
for(j=1;j<tt-1;j++)
{
m=inf;
for(i=1;i<tt;i++)
if(lc[i]!=0&&lc[i]<m)
{m=lc[i];id=i;}
ans+=m;
lc[id]=0;
for(i=1;i<tt;i++)
{
if(lc[i]!=0&&lc[i]>w[id][i])
lc[i]=w[id][i];
}
}
}
int main()
{
int i,j,l;
int t;
char c[10];
scanf("%d",&t);
gets(c);
while(t--)
{
tt=1;
memset(vis,0,sizeof vis );
cin>>m>>n;
gets(c);
for(i=0;i<n;i++)
gets(map[i]);
//for(i=0;i<n;i++)
//cout<<map[i]<<endl;
int bj=0;
memset(vv,0,sizeof vv);
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
if(map[i][j]=='A'||map[i][j]=='S')
{
vv[i][j]=tt++;//记录字母序号及个数
}
}
for(i=1;i<tt;i++)
for(j=1;j<=i;j++)
if(i==j)
w[i][j]=0;
else {
w[i][j]=inf;
w[j][i]=inf;
}
for(i=0;i<n;i++)
for(j=0;j<m;j++)
{
if((map[i][j]=='A'||map[i][j]=='S')&&!vis[i][j])
{
vis[i][j]=1;
memset(v,0,sizeof v);
bfs(i,j);//求出这个字母到各个字母产生的边权
}
}
ans=0;
prim();//prim算法的运用
cout<<ans<<endl;
}
return 0;
}
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