POJ3057 Evacuation 二分图匹配+最短路

POJ3057 Evacuation 二分图匹配+最短路

题目描述

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

3
5 5
XXDXX
X...X
D...X
X...D
XXXXX
5 12
XXXXXXXXXXXX
X..........D
X.XXXXXXXXXX
X..........X
XXXXXXXXXXXX
5 5
XDXXX
X.X.D
XX.XX
D.X.X
XXXDX

Sample Output

3
21
impossible

分析

一句话题意：有一个X\(\times\)Y的房间，'X'代表墙壁，‘D’是门，‘.’代表人。这个房间着火了，人要跑出去，但是每一个时间点只有一个人可以从门出去。

问最后一个人逃出去的最短时间，如果不能逃出去，输出impossible。

其实这一道题和题库里外星人那一道题解法大致雷同，只是建边的方式有点不同，我们对比着来说

首先，外星人那一道题就是一个简单的两点之间距离公式求出最短路

但是这一道题肯定是不可以直接用公式的，因为题中会有墙

所以我们要用一个bfs预处理出每一个人到每一扇门的最短路

还有一个更大的不同点就是门的性质

在外星人那一道题中，一个人经过了一扇门后，门就会消失，而这一道题经过一扇门后，门不会消失

所以一扇门可以使用多次，但不能被多个人同时使用

所以我们考虑把一个门拆成若干个小门，每一个门代表一个时刻

我们从0时刻开始枚举，把0时刻每一个人可以到达的门和这个人连一条边

如果0时刻不能匹配成功，我们就把1时刻每一个人可以到达的门和这个人连一条边

以此类推，直到所有的时刻都枚举完毕

那么最大的时刻是多少呢

很显然，一个人如果可以从门里走出去的话，那么他到这个门所花费的时间不会超过\(n\times m\)

其中\(m\)为房间的长度，\(n\)为房间的宽度

我们假设题目中共有\(d\)个门，因为每一个门都要拆分到\(n\times m\)个时间段里，所以我们只要枚举到\(n\times m \times d\)就可以了

如果最后不能匹配成功，人就不能逃生成功

感性地理解一下，就是一个人被一圈墙围在中间

最后，人数为0的情况下要特判，不特判就会输出'impossible'

最后，我们再来讨论一下复杂度的问题

bfs最坏的情况下就是\(12\times12 \times (12 \times12+12 \times24)=62208\)

没什么影响

建边更是可以忽略不计

其实最主要的开销还是在匈牙利上

最坏的情况\(144\)个点，248832条边，总的复杂度为\(35831808\)

但是这样的强度是达不到的，因为你不可能门有144个，人也有144个，所以实际上是可行的

这道题有人可能会想到二分，但是我个人感觉二分是会使时间复杂度升高的

因为不二分的话，只跑一次匈牙利就可以，用二分的话，可能要跑多次

或者是有更好的二分方法我没有想到

代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int maxn=50005,maxm=15;
char s[maxm][maxm];
int n,m,tot;
int movex[maxm]={1,-1,0,0};
int movey[maxm]={0,0,-1,1};
int dis[maxm][maxm][maxm][maxm];
vector<int> g[maxn];
int match[maxn],vis[maxn];
vector<int> dx,dy,px,py;
int dfs(int xx){
    for(int i=0;i<g[xx].size();i++){
        int u=g[xx][i];
        if(!vis[u]){
            vis[u]=1;
            if(match[u]==-1 || dfs(match[u])){
                match[u]=xx;
                return 1;
            }
        }
    }
    return 0;
}
void Init(){
    memset(dis,0x3f,sizeof(dis));
    for(int i=0;i<maxn;i++){
        g[i].clear();
    }
    dx.clear(),dy.clear(),px.clear(),py.clear();
    memset(match,-1,sizeof(match));
    memset(vis,0,sizeof(vis));
}
void bfs(int xx,int yy){
    queue<int> qx,qy;
    qx.push(xx),qy.push(yy);
    dis[xx][yy][xx][yy]=0;
    while(!qx.empty()){
        int nx=qx.front();
        int ny=qy.front();
        qx.pop(),qy.pop();
        for(int i=0;i<4;i++){
            int mx=movex[i]+nx;
            int my=movey[i]+ny;
            if(mx>n || mx<=0 || my>m || my<=0 || s[mx][my]!='.') continue;
            if(dis[xx][yy][mx][my]>dis[xx][yy][nx][ny]+1){
                dis[xx][yy][mx][my]=dis[xx][yy][nx][ny]+1;
                qx.push(mx),qy.push(my);
            }
        }
    }
}
void solve(){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(s[i][j]=='D'){
                dx.push_back(i),dy.push_back(j);
                bfs(i,j);
            }
            else if(s[i][j]=='.') px.push_back(i),py.push_back(j);
        }
    }
    int d=dx.size(),p=px.size();
    for(int i=0;i<d;i++){
        for(int j=0;j<p;j++){
            int nows=dis[dx[i]][dy[i]][px[j]][py[j]];
            if(nows!=0x3f3f3f3f){
                for(int k=nows;k<=tot;k++){
                    g[k*d+i].push_back(j);
                }
            }
        }
    }
    if(p==0){
        printf("0\n");
        return;
    }
    int ans=0;
    for(int i=0;i<tot*d;i++){
        memset(vis,0,sizeof(vis));
        if(dfs(i)){
            if(++ans==p){
                printf("%d\n",i/d);
                return;
            }
        }
    }
    printf("impossible\n");
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        Init();
        scanf("%d%d",&n,&m);
        tot=n*m;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf(" %c",&s[i][j]);
            }
        }
        solve();
    }
    return 0;
}