poj 2251 Dungeon Master（多起点bfs）

You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides. 

Is an escape possible? If yes, how long will it take? 

Input

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size). 
L is the number of levels making up the dungeon. 
R and C are the number of rows and columns making up the plan of each level. 
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form 

Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape. 
If it is not possible to escape, print the line 

Trapped!

Sample Input

3 4 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!
思路：1 先寻找‘#’连通块的个数
      2 如果大于3块， -1 ，因为最多放两堆火。
      3 如果两块，每人放火烧一块，遍历所有“#”，两块各自取最小值，两最小值取较大作为结果。
      4 如果一块，就是双起点广搜了，排列组合选择两块同时开始，具体做法就是同时压入队列，所有  时间取最小值。
#include <cstdio>#include <queue>#include <map>#include <stack>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int mk[15][15];char mp[15][15];int m,n,T;typedef struct node{    int x,y;    int t;    node(int x_=0,int y_=0,int t_=0)    {        x=x_,y=y_,t=t_;    }}node;queue <node> q;int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};void bfs(int x,int y,int mark){    node tmp(x,y,0);    q.push(tmp);    mk[x][y] = mark;    T  = -1;    while(!q.empty())    {        tmp = q.front();        T = tmp.t;        q.pop();        for(int i = 0; i < 4; i++)        {            int x_ = tmp.x+dir[i][0];            int y_ = tmp.y+dir[i][1];            if(x_ < 1 || y_ < 1|| x_ > m || y_ > n) continue;            if(mp[x_][y_] == '.') continue;            if(mk[x_][y_] == mark) continue;            mk[x_][y_] = mark;            node nd(x_,y_,tmp.t+1);            q.push(nd);        }    }}int main(){    int C,index = 1;    cin >> C;    while(C--)    {        int mark = 0;//标记连通块个数        cin >> m >> n;        for(int i = 1; i <= m; i++)        for(int j = 1; j <= n; j++)            cin >> mp[i][j];        memset(mk,0,sizeof(mk));        for(int i = 1; i <= m; i++)        for(int j = 1; j <= n; j++)            if(mp[i][j] == '#' && !mk[i][j])            {                while(!q.empty())q.pop();                bfs(i,j,++mark);            }        if(mark >= 3)        {             printf("Case %d: -1\n",index++);             continue;        }        if(mark == 2)        {            int k1 = 1,k2 = 2;//作用是保证下次还能找到同一连通块的标志。            int a[2] ={110,110};            for(int i = 1; i <= m; i++)            for(int j = 1; j <= n; j++)            if(mp[i][j] == '#' && mk[i][j] == k1)            {                k1+=2;                while(!q.empty())q.pop();                bfs(i,j,k1);                a[0] = min(a[0],T);            }            else if(mp[i][j] == '#'&& mk[i][j] == k2)            {                k2+=2;                while(!q.empty())q.pop();                bfs(i,j,k2);                a[1] = min(a[1],T);            }            printf("Case %d: %d\n",index++,max(a[0],a[1]));        }        else        {            int k = 1,l = 0,time = 110;            node a[110];            for(int i = 1; i <= m; i++)            for(int j = 1; j <= n; j++)            if(mp[i][j] == '#' && mk[i][j] == k)            {                a[l].x =i,a[l++].y = j;            }            for(int i = 0; i < l; i++)            for(int j = i; j < l; j++)            {                while(!q.empty())q.pop();                if(j!=i) q.push(a[j]);                bfs(a[i].x,a[i].y,++k);                time = min(time,T);            }            printf("Case %d: %d\n",index++,time);        }    }    return 0;}