POJ 3026 Borg Maze (最小生成树 + BFS)

Borg Maze

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6804		Accepted: 2293

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.

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 Mästerskap i Programmering/Norgesmesterskapet 2001

 

 

问题关键就是处理 任意两字母间的最短距离，由于存在了“墙#” ，这个距离不可能单纯地利用坐标加减去计算，必须额外考虑，推荐用BFS（广搜、宽搜），这是本题的唯一难点，因为prim根本直接套用就可以了

求 任意两字母间的最短距离 时不能直接用BFS求，

1、必须先把矩阵中每一个允许通行的格看做一个结点（就是在矩阵内所有非#的格都作为图M的一个顶点），对每一个结点i，分别用BFS求出它到其他所有结点的权值（包括其本身，为0），构造结点图M；

2、然后再加一个判断条件，从图M中抽取以字母为顶点的图，进而构造字母图N

这个判定条件就是当结点图M中的某点j为字母时，把i到j的权值再复制（不是抽离）出来，记录到字母图N的邻接矩阵中

3、剩下的就是对字母图N求最小生成树了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

const int INF=0x3f3f3f3f;

int col,row,point[120][120];
int n,cnt,ans,map[120][120],dis[120],vis[120];
int dis1[120][120],vis1[60][60];
char str[60][60];

int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};

struct node{
    int x,y;
};

void BFS(int si,int sj){
    queue<node> q;
    while(!q.empty())
        q.pop();
    memset(dis1,0,sizeof(dis1));
    memset(vis1,0,sizeof(vis1));
    vis1[si][sj]=1;
    node cur,next;
    cur.x=si;   cur.y=sj;
    q.push(cur);
    int x,y;
    while(!q.empty()){
        cur=q.front();
        q.pop();
        if(point[cur.x][cur.y])
            map[point[si][sj]][point[cur.x][cur.y]]=dis1[cur.x][cur.y];
        for(int i=0;i<4;i++){
            next.x=x=cur.x+dir[i][0];
            next.y=y=cur.y+dir[i][1];
            if(x>=1 && x<=row && y>=1 && y<=col && str[x][y]!='#' && !vis1[x][y]){
                vis1[x][y]=1;
                dis1[x][y]=dis1[cur.x][cur.y]+1;
                q.push(next);
            }
        }
    }
}

void Prim(){
    int i;
    for(i=1;i<=cnt;i++){
        dis[i]=map[1][i];
        vis[i]=0;
    }
    dis[1]=0;
    vis[1]=1;
    int j,k,tmp;
    for(i=1;i<=cnt;i++){
        tmp=INF;
        for(j=1;j<=cnt;j++)
            if(!vis[j] && tmp>dis[j]){
                tmp=dis[j];
                k=j;
            }
        if(tmp==INF)
            break;
        vis[k]=1;
        ans+=dis[k];
        for(j=1;j<=cnt;j++)
            if(!vis[j] && dis[j]>map[k][j])
                dis[j]=map[k][j];
    }
}

int main(){

    freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        memset(point,0,sizeof(point));
        cnt=0;
        cin>>col>>row;
        //printf("row=%d  col=%d
",row,col);
        gets(str[0]);
        for(int i=1;i<=row;i++){
            gets(str[i]+1);
            //puts(str[i]);
            //printf("%c
",str[i][1]);
            for(int j=1;j<=col;j++)
                if(str[i][j]=='A' || str[i][j]=='S')
                    point[i][j]=++cnt;
        }
        //printf("cnt=%d
",cnt);
        for(int i=1;i<=row;i++)
            for(int j=1;j<=col;j++)
                if(point[i][j])
                    BFS(i,j);
        /*
        printf("-------------------------
");
        for(int i=1;i<=cnt;i++){
            for(int j=1;j<=cnt;j++)
                printf("%d ",map[i][j]);
            printf("
");
        }
        printf("-------------------------
");
        */
        ans=0;
        Prim();
        printf("%d
",ans);
    }
    return 0;
}