poj3026Borg Maze

Borg Maze

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10852		Accepted: 3583

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

首先要注意的一点是n m后面可能跟了空格,然后行数和列数超过50,
这题就是先bfs找出任意两个A和A,或者A和S之间的最短路程,然后利用最小生成树进行求解
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
#define Hash(a,b) (a-1)*m+b
typedef long long ll;
const int maxn=150;
int f[12510];
char mp[151][151];
int vis[101][101];
int n,m;
int dir[4][2]={-1,0,1,0,0,1,0,-1};
struct Edge{
    int from,to,dis;
}e[250*250];

bool cmp(Edge a,Edge b){
    return a.dis<b.dis;
}

struct node{
    int x,y,count1;
}ans[maxn];

bool over(int x,int y){
    if(x<=0||y<=0||x>n|y>m)
        return true;
    return false;
}

int find(int x){
    int k, j, r;
    r = x;
    while(r!=f[r])     //查找跟节点
        r = f[r];      //找到跟节点，用r记录下
    k = x;
    while(k != r){           //非递归路径压缩操作
        j = f[k];         //用j暂存parent[k]的父节点
        f[k] = r;        //f[x]指向跟节点
        k = j;                    //k移到父节点
    }
    return r;         //返回根节点的值
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&m,&n);
        int cnt=0;
        char c=getchar();
        while(c==' '){
            c=getchar();
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                mp[i][j]=getchar();
                if(mp[i][j]=='A'||mp[i][j]=='S'){
                    ans[cnt].x=i;
                    ans[cnt++].y=j;
                }
            }
            getchar();
        }
        int cnt1=0;
        for(int i=0;i<cnt;i++){
            mem0(vis);
            queue<node>Q;
            while(!Q.empty()){
                Q.pop();
            }
            vis[ans[i].x][ans[i].y]=1;
            Q.push((node){ans[i].x,ans[i].y,0});
            while(!Q.empty()){
                node a=Q.front();
                Q.pop();
                for(int j=0;j<4;j++){
                    int x=a.x+dir[j][0];
                    int y=a.y+dir[j][1];
                    if(over(x,y)||mp[x][y]=='#'||vis[x][y]==1)
                        continue;
                    if(mp[x][y]=='A'||mp[x][y]=='S'){
                        e[cnt1].from=Hash(ans[i].x,ans[i].y);
                        e[cnt1].to=Hash(x,y);
                        e[cnt1++].dis=a.count1+1;
                    }
                    vis[x][y]=1;
                    Q.push((node){x,y,a.count1+1});
                }
            }
        }
        for(int i=1;i<=n*n;i++)
            f[i]=i;
        sort(e,e+cnt1,cmp);
        int ans=0;
        for(int i=0;i<cnt1;i++){
            int u=e[i].from,v=e[i].to,w=e[i].dis;
            int t1=find(u);
            int t2=find(v);
            if(t1!=t2){
                ans+=w;
                f[t1]=t2;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}