POJ-3026 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;

}