POJ 3057 Evacuation 二分图匹配

每个门每个时间只能出一个人，那就把每个门拆成多个，对应每个时间。

不断增加时间，然后增广，直到最大匹配。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c)
{
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c)
{
    return max(max(a,b),max(a,c));
}
void debug()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in","r",stdin);
    // freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch()
{
    int ch;
    while((ch=getchar())!=EOF)
    {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}
 
int DX[] = {, , , -};
int DY[] = {, , -, };
 
const int maxn = ;
char grid[maxn][maxn];
int n, m;
int dist[maxn][maxn][maxn][maxn];
int vis[maxn][maxn];
vector<int> dx,dy,px,py;
 
void bfs(int X,int Y)
{
    queue<int> qx,qy;
    qx.push(X); qy.push(Y);
    memset(vis, , sizeof(vis));
    vis[X][Y] = true;
    while(!qx.empty())
    {
        int x = qx.front(); qx.pop();
        int y = qy.front(); qy.pop();
        for(int d=; d<; d++)
        {
            int nx = x + DX[d];
            int ny = y + DY[d];
            if(nx>=&&nx<=n&&ny>=&&ny<=m&&grid[nx][ny]=='.'&&!vis[nx][ny])
            {
                vis[nx][ny] = true;
                dist[X][Y][nx][ny] = dist[X][Y][x][y] + ;
                qx.push(nx);
                qy.push(ny);
            }
        }
    }
}
 
void prework()
{
    memset(dist, -, sizeof(dist));
    dx.clear(); dy.clear();
    px.clear(); py.clear();
 
    for(int i=; i<=n; i++)
        for(int j=; j<=m; j++)
        {
            if(grid[i][j]=='D')
            {
                dx.push_back(i);
                dy.push_back(j);
                dist[i][j][i][j]=;
                bfs(i, j);
            }else if(grid[i][j]=='.')
            {
                px.push_back(i);
                py.push_back(j);
            }
        }
    //printf("prework: %d %d\n",dx.size(), px.size());
}
 
const int maxv = * + ;
int id[maxn][maxn][];
bool used[maxv];
int match[maxv];
int vcnt;
vector<int> g[maxv];
int ID(int x, int y, int t)
{
    int &a = id[x][y][t];
    if(a==) a=++vcnt;
    return a;
}
void add(int u,int v)
{
    g[u].push_back(v);
    g[v].push_back(u);
}
void init()
{
    for(int i=; i<maxv; i++)
        g[i].clear();
    memset(id, , sizeof(id));
}
bool dfs(int u)
{
    used[u] = true;
    for(int i = ; i < g[u].size(); i++)
    {
        int v = g[u][i];
        int w = match[v];
        if(w<||!used[w]&&dfs(w))
        {
            match[u] = v;
            match[v] = u;
            return true;
        }
    }
    return false;
}
int solve()
{
    init();
    memset(match, -, sizeof(match));
    int res = ;
    vcnt = ;
    if(px.size() == ) return ;
    for(int t=; t<=; t++)
    {
        for(int i=; i<dx.size(); i++)
        {
            for(int j=; j<px.size(); j++)
            {
                int dis = dist[dx[i]][dy[i]][px[j]][py[j]];
                if(dis!=- && dis <= t)
                {
                    int u = ID(dx[i], dy[i], t);
                    int v = ID(px[j], py[j], );
                    add(u, v);
                    add(v, u);
                }
            }
        }
        for(int i=; i<dx.size(); i++)
        {
            int u = ID(dx[i], dy[i], t);
            memset(used, , sizeof(used));
            if(dfs(u))
                res++;
        }
        if(res == px.size()) return t;
    }
    return -;
}
int main()
{
    debug();
    int t;
    scanf("%d", &t);
    for(int ca=; ca<=t; ca++)
    {
        scanf("%d%d", &n, &m);
        for(int i=; i<=n; i++)
            scanf("%s", grid[i]+);
 
        prework();
 
        int ans = solve();
        if(ans != -)
            printf("%d\n", ans);
        else printf("impossible\n");
    }
    return ;
}