POJ3026 Borg Maze(bfs求边+最小生成树)
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
题意:有x个点,其中有一个是起点,起点有不多于100个人,告诉你一张地图,地图上分布着这些点,起点用'S'表示,其余点用‘A’表示,可以走的路为‘ ’,不能走的为‘#’
求从起点出发到所有点所需的最小路径,可以在任意位置分出任意人向某点前进,保证人够,多人走同一路径距离只算一遍
题解:这题题意好难理解……理解后就好办了,题意可以转化为已知x个点的位置,用x-1条边将其连起来,使总边权最小,那不就是一道最小生成树吗?
因为有‘#’的存在,所以必须要把点与点之间的距离用bfs扫出来,之后写一遍最小生成树,然后就没问题了!
新年第一道题1A,真是愉快 代码如下:
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std; struct node
{
int from,to,dis;
}si[]; int vis[][],fa[],cnt[][],d[][],tmp=,sum=,ans=,dx[]={,,,-},dy[]={,,-,};
char c[][];
int ttt,n,m; void mem(int n)
{
for(int i=;i<=n;i++)
{
fa[i]=i;
}
} bool cmp(node x,node y)
{
return x.dis<y.dis;
} int find(int x)
{
if(fa[x]!=x)
{
fa[x]=find(fa[x]);
}
return fa[x];
} int union_(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx==fy)
{
return ;
}
fa[fy]=fx;
return ;
} void bfs(int x1,int y1)
{
memset(vis,,sizeof(vis));
memset(cnt,,sizeof(cnt));
queue< pair<int,int> > q;
q.push(make_pair(x1,y1));
vis[x1][y1]=;
while(!q.empty())
{
int x=q.front().first;
int y=q.front().second;
q.pop();
for(int i=;i<=;i++)
{
int dx1=x+dx[i];
int dy1=y+dy[i];
if(dx1<=||dx1>n||dy1<=||dy1>m)
{
continue;
}
if(d[dx1][dy1]!=-&&!vis[dx1][dy1])
{
vis[dx1][dy1]=;
cnt[dx1][dy1]=cnt[x][y]+;
q.push(make_pair(dx1,dy1));
}
if(d[dx1][dy1]>)
{
si[++sum].from=d[x1][y1];
si[sum].to=d[dx1][dy1];
si[sum].dis=cnt[dx1][dy1];
}
}
}
} int main()
{
scanf("%d",&ttt);
while(ttt--)
{
sum=;
ans=;
tmp=;
scanf("%d%d\n",&m,&n);
mem();
for(int i=;i<=n;i++)
{
gets(c[i]);
for(int j=;j<m;j++)
{
if(c[i][j]=='#')
{
d[i][j+]=-;
}
else
{
if(c[i][j]==' ')
{
d[i][j+]=;
}
else
{
if(c[i][j]=='A'||c[i][j]=='S')
{
d[i][j+]=++tmp;
}
}
}
}
}
for(int i=;i<=n;i++)
{
for(int j=;j<=m;j++)
{
if(d[i][j]>)
{
bfs(i,j);
}
}
}
sort(si+,si+sum+,cmp);
for(int i=;i<=sum;i++)
{
if(union_(si[i].from,si[i].to))
{
ans+=si[i].dis;
}
}
printf("%d\n",ans);
}
}
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