POJ3057 Evacuation Problem Solving report

The difficulty of this problem should be to think that this is a binary map matching

When I think of this, it's a map.

//by baobaopangzi88#include <iostream>#include<sstream>#include<ios>#include<iomanip>#include<functional>#include<algorithm>#include<vector>#include<string>#include<list>#include<queue>#include<deque>#include<stack>#include<Set>#include<map>#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<climits>#include<cctype>#defineINF 0x3f3f3f3f#defineMP (x, y) make_pair (x, y)#definePB (x) push_back (x)#defineREP (x,n) for (int x=0; x<n; X + +)#defineREP2 (X,L,R) for (int x=l; x<=r; X + +)#defineDEP (x,r,l) for (int x=r; x>=l; x--)#defineCLR (a,x) memset (a,x,sizeof (A))#defineIT iterator#definePI 3.14159265358979323846#define_ Ios_base::sync_with_stdio (0); Cin.tie (0);#defineX First#defineY Second#defineMax_v 10101#defineMAXN 13using namespaceStd;typedefLong LongLl;typedef pair<int,int>PII;intn,m,v;CharM[maxn][maxn];vector<int> g[30000];vector<int>dx,dy,px,py;intD[MAXN][MAXN][MAXN][MAXN];intdx[4]={-1,1,0,0},dy[4]={0,0,-1,1};intmatch[30000];BOOLused[30000];voidAdd_edge (int  from,intTo ) {g[ from].    PB (to); G[to]. PB ( from);}BOOLDfsintv) {Used[v]=1;  for(intI=0; I<g[v].size (); + +i) {        intto=g[v][i],w=Match[to]; if(w<0|| !USED[W] &&DFS (W)) {Match[v]=to ; Match[to]=v; return true; }    }    return false;}voidBFsintXintYintD[MAXN][MAXN]) {D[x][y]=0; Queue<int>qx,qy;    Qx.push (x);    Qy.push (y);  while(!Qx.empty ()) {x=Qx.front (); Qx.pop (); Y=Qy.front (); Qy.pop ();  for(intI=0;i<4;++i) {            intnx=x+dx[i],ny=y+Dy[i]; if(nx>=0&& nx<n && ny>=0&&NY<m && m[nx][ny]=='.'&& d[nx][ny]<0) {D[nx][ny]=d[x][y]+1;                Qx.push (NX);            Qy.push (NY); }        }    }}intsolve () {dx.clear ();d y.clear ();    Px.clear ();p y.clear (); memset (D,-1,sizeof(D));  for(intI=0; i<n;++i) for(intj=0; j<m;++j) {        if(m[i][j]=='D') {dx.            PB (i); Dy.            PB (j);        BFS (I,j,d[i][j]); }        Else if(m[i][j]=='.') {px.            PB (i); Py.        PB (j); }    }    //Build_graph    intTemp=n*m;//the maximum number of times    intD=dx.size (), p=px.size (); V=n*m*d+p;  for(intv=0; v<v;++v) g[v].clear ();  for(intI=0; i<d;++i) for(intj=0; j<p;++j) for(intt=0; t<temp;++6) {        if(d[dx[i]][dy[i]][px[j]][py[j]]>0&& d[dx[i]][dy[i]][px[j]][py[j]]<=t+1) {Add_edge (t*d+i,temp*d+j); }    }    if(p==0)return 0; intres=0; memset (Match,-1,sizeof(match));  for(intt=0; t<temp;++t) {         for(intv=t*d;v< (t+1) *d;++v) {            if(match[v]<0) {memset (used,0,sizeof(used)); if(Dfs (v)) {res++; }            }                }        if(res==p)returnt+1; }    return-1;}intMain () {intT; CIN>>u;  while(t--) {cin>>n>>m;  for(intI=0; i<n;++i) Cin>>M[i]; inttemp=solve (); if(temp<0) cout<<"Impossible"<<Endl; Elsecout<<temp<<Endl; }    return 0;}

POJ3057 Evacuation Problem Solving report