一、内容
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
二、思路
- 从s出发（看作有很多人）意思可以从多个方向走，去寻找到达A的最短路径。如找到一个A，那么可以从当前位置出发这时候步数又为0了，又去寻找其他的A。
- 也可以直接暴力求解每个字母到达其他字母的路径进行建图，就从每个字母出发，进行多次BFS求解即可。
三、代码
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <iostream>
using namespace std;
const int N = 55, M = N * N;
int n, m, t, sx, sy, cnt, mcnt, path[N][N], id[N][N], p[N]; //代表某个字母到达(i, j)的距离
char g[N][N];
int dx[4] = {1, 0, 0, -1};
int dy[4] = {0, 1, -1, 0};
struct Node {
int x, y, sx, sy, d;
Node(int x, int y, int sx, int sy, int d): x(x), y(y), sx(sx), sy(sy), d(d){}
bool operator < (const Node&o) const {
return d > o.d;
}
};
struct E {
int u, v, w;
bool operator < (const E&o) const {
return w < o.w;
}
} e[M];
int find(int x){return x == p[x] ? x : (p[x] = find(p[x]));}
void bfs() {
priority_queue<Node> q;
path[sx][sy] = 0;
id[sx][sy] = ++cnt; //给每个字母标号
q.push(Node(sx, sy, sx, sy, 0));
while (!q.empty()) {
Node t = q.top();
q.pop();
for (int i = 0; i < 4; i++) {
int fx = t.x + dx[i];
int fy = t.y + dy[i];
if (fx >= 0 && fy >= 0 && fx < n && fy < m && g[fx][fy] != '#') {
if (t.d + 1 < path[fx][fy]) {
if (t.sx == fx && t.sy == fy) continue;//避免自己与自己连边
path[fx][fy] = t.d + 1;
if (g[fx][fy] == 'A') {
q.push(Node(fx, fy, fx, fy, 0));
if (!id[fx][fy]) id[fx][fy] = ++cnt;
//添加边
e[++mcnt].u = id[t.sx][t.sy];
e[mcnt].v = id[fx][fy];
e[mcnt].w = t.d + 1;
} else {
q.push(Node(fx, fy, t.sx, t.sy, path[fx][fy]));
}
}
}
}
}
}
int main() {
scanf("%d", &t);
while (t--) {
memset(path, 0x3f, sizeof(path)); cnt = mcnt = 0;
memset(id, 0, sizeof(id));
scanf("%d%d", &m, &n);
gets(g[0]); //消除上面的换行
for (int i = 0; i < n; i++) gets(g[i]);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (g[i][j] == 'S') sx = i, sy = j;
}
}
bfs(); //确定字母之间的关系
int ans = 0;
sort(e + 1, e + 1 + mcnt);
for (int i = 1; i <= cnt; i++) p[i] = i; //初始化并查集
for (int i = 1; i <= mcnt; i++) {
int fu = find(e[i].u), fv = find(e[i].v);
if (fu != fv) {
ans += e[i].w;
p[fu] = fv;
}
}
printf("%d\n", ans);
}
return 0;
}
