POJ3026 Borg Maze BFS+最小生成树

该题题义是某个人从S点出发，去寻找所有的A，他可以直接到达每个A，也可以通过分身来到达，具体视那种方法所走的路程短而定。换句话说就是可以从A点再走到A点来寻找下一个A，而不选择再从S出发。

首先将任意两点之间（A或者S）的距离求出来（通过BFS）然后再建立最小生成树即可。注意输入数据中x,y后面不只一个空格。

代码如下:

#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <queue>
#define MAXN 105
using namespace std;

int N, M, pos, pnum, set[MAXN];
int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
char G[55][55], hash[55][55];

struct Node
{
    int x, y, dist;
    bool operator < (Node t) const
    {
        return dist < t.dist;
    }
}e[MAXN*MAXN+5];

struct Point
{
    int x, y;
}p[MAXN];

struct Info
{
    int x, y, step;
}info;

bool judge(int x, int y) 
{
    if (x >= 1 && x <= N && y >= 1 && y <= M) {
        return true;
    }
    else {
        return false;
    }
}

int bfs(int x)
{
    memset(hash, 0, sizeof (hash));
    struct Info obj;
    queue<Info>q;
    info.step = 0;
    info.x = p[x].x, info.y = p[x].y;
    q.push(info);
    while (!q.empty()) {
        obj = q.front();
        q.pop();
        for (int i = 0; i < 4; ++i) {
            int xx = obj.x+dir[i][0], yy = obj.y+dir[i][1];
            if (judge(xx, yy) && !hash[xx][yy] && G[xx][yy] != -2) {
                if (G[xx][yy] != -1 && G[xx][yy] > x) {  // 在这里建边
                    ++pos;
                    e[pos].x = x, e[pos].y = G[xx][yy];
                    e[pos].dist = obj.step+1;
                }
                hash[xx][yy] = 1;
                info.x = xx, info.y = yy, info.step = obj.step+1;
                q.push(info);
            }
        }
    }
}

int find(int x)
{
    return set[x] = x == set[x] ? x : find(set[x]);
}

void merge(int a, int b)
{
    set[a] = b;
}

int main()
{
    int T, length, cnt, ans;
    char cc;
    scanf("%d", &T);
    for (int t = 1; t <= T; ++t) {
        pos = pnum = cnt = ans = 0;
        scanf("%d %d", &M, &N); 
        gets(G[0]);
        for (int i = 1; i <= N; ++i) {
            gets(G[i]+1);
            for (int j = 1; j <= M; ++j) {
                if (G[i][j] == 'S' || G[i][j] == 'A') {
                    ++pnum;
                    G[i][j] = pnum;
                    // 直接将G图中的值赋为点的编号
                    p[pnum].x = i, p[pnum].y = j;
                }
                else if (G[i][j] == ' ') {
                    G[i][j] = -1;
                    // 由于‘#’和‘ ’的ascii码值与点的编号一样，一直WA
                }
                else if (G[i][j] == '#') {
                    G[i][j] = -2;
                }
            }
        } 
        for (int i = 1; i <= pnum; ++i) {
            set[i] = i;
            bfs(i);
        }
        sort(e+1, e+1+pos);
        for (int i = 1; i <= pos; ++i) {
            int a = find(e[i].x), b = find(e[i].y);
            if (a != b) {
                merge(a, b);
                ans += e[i].dist;
                ++cnt;
                if (cnt == pnum-1) {
                    break;
                }
            }
        }
        printf("%d\n", ans);
    }        
    return 0;
}