POJ-3026

题意：给一个n*m的地图，里面有一个S和若干个A，问从S到所有的A的距离之和最短（S可以无限在任意时刻分身，但是每秒只能移动一个分身）
思路：由于S可以随时随地分身，那么其实就是求令所有的A和S连通的最小路径和；所以我们先用BFS求每个A或S到其他所有A或S的最短距离，然后求一个最小生成树即可

#include<iostream>
#include<cstdlib> 
#include<sstream>
#include<cstdio>
#include<stack>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long LL;
#define me(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
const int N=2111122;
const int M=20005;
const int H = 750;
int p[M],v[H][H],stemp[H][H],v_1[H][H];
int f1[4] = {0,0,1,-1};
int f2[4] = {1,-1,0,0};
char s[H][H];
int sum,o,n,m,cond;
struct node {
	int x,y;
};
struct node_1 {
	int l,r,ans;
}a[M];
bool cmp(node_1 a,node_1 b) {
	return a.ans < b.ans;
}
int find(int root) {
	return root == p[root]?root:p[root] = find(p[root]);
}
void kruskal() {
	int i,ct=0;
	for(i = 1 ; i <= cond ; i++)p[i] = i;
	sort(a+1,a+1+o,cmp);
	for(i = 1 ; i <= o ; i++) {
		int x = find(a[i].l);
		int y = find(a[i].r);
		if(x != y) {
			p[x] = y;
			ct ++;
			sum += a[i].ans;
			if(ct == cond-1)break;
		}
	}
}
queue<node> q;
void bfs(int l,int r) {
	memset(v_1,0,sizeof(v_1));
	v_1[l][r] = 1;
	int fx = l;
	int fy = r;
	int i;
	stemp[l][r] = 0;
	while(!q.empty())
	q.pop();
	node t;
	t.x = l;
	t.y = r;
	q.push(t);
	while(!q.empty())  {
		node g = q.front();
		q.pop();
		for(i = 0 ; i < 4 ; i ++) {
			int dx = g.x + f1[i];
			int dy = g.y + f2[i];
			if(dx>=0&&dx<m&&dy>=0&&dy<n) {
				if(s[dx][dy] != '#'&&v_1[dx][dy]==0) {
					v_1[dx][dy] = 1;
					t.x = dx;
					t.y = dy;
					stemp[dx][dy] = stemp[g.x][g.y]+1;
					q.push(t);
					if(s[dx][dy]!=' ') {
							a[++o].l = vis[dx][dy];
							a[o].r   = vis[fx][fy];
							a[o].ans = stemp[dx][dy];
					}
				}
			}
		}
	}
}
int main() {
	int tt,i,j;
	char ch[13];
	scanf("%d",&tt);
	while(tt--) {
		scanf("%d%d",&n,&m);
		gets(ch);
		cond = 0;
		sum=0;
		o=0;
		for(i = 0 ; i < m ; i++) gets(s[i]);
		for(i = 0 ; i < m ; i++) {
			for( j = 0 ; j < n; j++) {
			if(s[i][j] == 'S'||s[i][j] == 'A') {
				vis[i][j] = ++cond;//给每个A或S一个编号，后面求最小生成树用 
			}
		}
	  }
	  if(cond <= 1) {
	  	cout << 0 << endl;
	  	return 0;
	  }
	  for(i = 0 ; i < m ; i++) {
		for( j = 0 ; j < n; j++) {
			if(s[i][j] == 'S'||s[i][j] == 'A') {
				bfs(i,j);
			}
		}
	  }
	  kruskal();
	  printf("%d\n",sum);
	}
	return 0;
}