poj 3057

       这道题是一道用二分图最大匹配算法解决的问题，关键在于如何构图，是一道难题。

       题目里说一个门每秒只能通过一个人，这一限制条件给解题造成很大的难度。解决这道题时，要把门和时间结合起来考虑，也就是说要区分1s时刻的门和2s时刻的门，即一个人在时间为1的情况下通过门和时间为2的情况下通过同一扇门是不同的匹配。

        简单来说，构图就是把人作为X集合中的点，把门和时间一起考虑，作为Y集合中的点，如果任何情况下人不能全被匹配，说明有一些人无法逃脱，否则肯定有一个最短时间能让所有人逃脱。而为了找到最短的时间，所以从时间1s开始，构出一个二分图，求最大匹配，如果能匹配所有的点，那么这就是最短的时间，否则时间自增1s，重新构图，再求最大匹配，再判断。如果超过时间上界，我的代码里取时间上界为n*m（n为行数，m为列数），那么则无法让所有人都逃脱。由于最大匹配算法可以从一个初始匹配开始运行，所以当我们在求时间为 t s二分图的最大匹配时，可以把 t-1 s时求得的最大匹配作为一个初始匹配。

       顺便提一下，这道题算法复杂度比较高，从题目给出的范围也可以感觉到，所以要尽可能节约时间，而不能图方便做多余的操作。从我写这道题的过程来看，开始写出来TLE，后来把MAX范围从25改到15，就AC了，时间是297ms，然后再少开一些空间，用以减少memeset函数赋值的时间，AC时间降到94ms，然后减少一个memeset函数的使用，时间又降到79ms，这足以说明一些小细节对于减少这道题通过时间甚至是否AC的重要性，这也是第一次见到MAX范围大了也TLE的情况，以后要多加注意。当然，由于没有多次提交实验，肯定有误差，请不要计较细节。

       这题在《挑战程序设计竞赛》（ 人民邮电出版社） P230也有详细解题过程，而我也是参考这个写这道题的，书中对于多维数组作为参数传递的方式上值得学习，虽然不能提高效率，但很简洁，下面的代码也有体现。

代码（C++）：

#include <cstdlib>#include <iostream>#include <queue>#include <vector>#define MAX 15#define INF (1<<30)using namespace std;//#define LOCALtypedef pair<int,int> pii;pii p[MAX*MAX],d[MAX*MAX];vector<int> G[MAX*MAX];const int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};int dist[MAX][MAX][MAX][MAX],cx[MAX*MAX],cy[2*MAX*MAX*MAX];int n,m,cp,cd,ans;bool vis[2*MAX*MAX*MAX];char map[MAX][MAX];void bfs(int x,int y,int dist[MAX][MAX]){     int a,b,i;     queue<pii> qi;     pii tmp;         dist[x][y]=0;     qi.push(make_pair(x,y));     while(!qi.empty())     {         tmp=qi.front();         qi.pop();              for(i=0;i<4;i++)         {             a=dir[i][0]+tmp.first;             b=dir[i][1]+tmp.second;              if(a>=0&&a<n&&b>=0&&b<m&&map[a][b]=='.'&&dist[a][b]==-1)              {                 dist[a][b]=dist[tmp.first][tmp.second]+1;                 qi.push(make_pair(a,b));                                }                   }             }      }void add_edge(int a,int b){     G[a].push_back(b);}bool dfs(int u){     int i,v;     for(i=0;i<G[u].size();i++)     {         v=G[u][i];         if(!vis[v])         {             vis[v]=true;             if(cy[v]==-1||dfs(cy[v]))             {                 cx[u]=v;                 cy[v]=u;                 return true;                                  }                }                           }     return false;}void Hungary(int k){     int i,j;          for(i=0;i<cp;i++)     {         if(cx[i]==-1)         {             //memset(vis,false,sizeof(vis));             for(j=0;j<cd*k;j++) vis[j]=false;             if(dfs(i)) ans++;                  }                  } }int main(int argc, char *argv[]){#ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);#endif    int t,i,j,k;    scanf("%d",&t);    while(t--)    {        scanf("%d %d",&n,&m);           for(i=0;i<n;i++) scanf("%s",map[i]);         cp=cd=0;        memset(dist,-1,sizeof(dist));        for(i=0;i<n;i++)        {            for(j=0;j<m;j++)            {                if(map[i][j]=='.') p[cp++]=make_pair(i,j);                else if(map[i][j]=='D')                {                       d[cd++]=make_pair(i,j);                     bfs(i,j,dist[i][j]);                                                }                        }                    }         if(cp==0)        {            printf("0\n");            continue;             }        ans=0;        memset(cx,-1,sizeof(cx));        memset(cy,-1,sizeof(cy));                for(k=1;k<n*m;k++)        {            for(i=0;i<cp;i++)            {                for(j=0;j<cd;j++)                {                     if(dist[d[j].first][d[j].second][p[i].first][p[i].second]!=-1&&dist[d[j].first][d[j].second][p[i].first][p[i].second]<=k)                            add_edge(i,j+cd*(k-1));                         }                         }            Hungary(k);            if(ans==cp)            {                printf("%d\n",k);                  break;            }         }         if(k==n*m) printf("impossible\n");               for(i=0;i<cp;i++) G[i].clear();     }    system("PAUSE");    return EXIT_SUCCESS;}

题目（http://poj.org/problem?id=3057）：

Evacuation

Time Limit: 1000MS Memory Limit: 65536K   

Description

Fires can be disastrous, especially when a fire breaks out in a room that is completely filled with people. Rooms usually have a couple of exits and emergency exits, but with everyone rushing out at the same time, it may take a while for everyone to escape. 

You are given the floorplan of a room and must find out how much time it will take for everyone to get out. Rooms consist of obstacles and walls, which are represented on the map by an 'X', empty squares, represented by a '.' and exit doors, which are represented by a 'D'. The boundary of the room consists only of doors and walls, and there are no doors inside the room. The interior of the room contains at least one empty square. 

Initially, there is one person on every empty square in the room and these persons should move to a door to exit. They can move one square per second to the North, South, East or West. While evacuating, multiple persons can be on a single square. The doors are narrow, however, and only one person can leave through a door per second. 

What is the minimal time necessary to evacuate everybody? A person is evacuated at the moment he or she enters a door square.

Input

The first line of the input contains a single number: the number of test cases to follow. Each test case has the following format: 
One line with two integers Y and X, separated by a single space, satisfying 3 <= Y, X <= 12: the size of the room 
Y lines with X characters, each character being either 'X', '.', or 'D': a valid description of a room

Output

For every test case in the input, the output should contain a single line with the minimal evacuation time in seconds, if evacuation is possible, or "impossible", if it is not. 

Sample Input

35 5XXDXXX...XD...XX...DXXXXX5 12XXXXXXXXXXXXX..........DX.XXXXXXXXXXX..........XXXXXXXXXXXXX5 5XDXXXX.X.DXX.XXD.X.XXXXDX

Sample Output

321impossible