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  • 【枚举】【二分答案】【分块答案】【BFS】【最大流】【Dinic】bzoj1189 [HNOI2007]紧急疏散evacuate

    【法一】枚举Time(0~N*M):

      S->'.'(1);

      'D'->T(Time);

      '.'->'D'(dis(用BFS预处理,注意一旦到达'D',BFS就不能继续扩展了,注意dis的初值0x7f)<=Time ? 1 : 0);

      判断是否满流;

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
    #define INF 2147483647
    #define MAXN 411
    #define MAXM 350001
    int v[MAXM],cap[MAXM],en,first[MAXN],next[MAXM];
    int d[MAXN],cur[MAXN];
    queue<int>q;
    int n,S,T,N,M;
    void Init_Dinic(){memset(first,-1,sizeof(first)); en=0; S=0; n=N*M+2; T=N*M+1;}
    void AddEdge(const int &U,const int &V,const int &W)
    {v[en]=V; cap[en]=W; next[en]=first[U]; first[U]=en++;
    v[en]=U; next[en]=first[V]; first[V]=en++;}
    bool bfs()
    {
        memset(d,-1,sizeof(d)); q.push(S); d[S]=0;
        while(!q.empty())
          {
            int U=q.front(); q.pop();
            for(int i=first[U];i!=-1;i=next[i])
              if(d[v[i]]==-1 && cap[i])
                {
                  d[v[i]]=d[U]+1;
                  q.push(v[i]);
                }
          }
        return d[T]!=-1;
    }
    int dfs(int U,int a)
    {
        if(U==T || !a) return a;
        int Flow=0,f;
        for(int &i=cur[U];i!=-1;i=next[i])
          if(d[U]+1==d[v[i]] && (f=dfs(v[i],min(a,cap[i]))))
            {
              cap[i]-=f; cap[i^1]+=f;
              Flow+=f; a-=f; if(!a) break;
            }
        if(!Flow) d[U]=-1;
        return Flow;
    }
    int max_flow()
    {
        int Flow=0,tmp=0;
        while(bfs())
          {
            memcpy(cur,first,(n+5)*sizeof(int));
            while(tmp=dfs(S,INF)) Flow+=tmp;
          }
        return Flow;
    }
    int dis[22*22][22*22];
    bool vis[22][22];
    char map[22][22];
    int num[22][22];
    struct Node
    {
    	int x,y,d;
    	Node(const int &a,const int &b,const int &c)
    	{x=a;y=b;d=c;}
    	Node(){}
    };
    const int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
    int man[22*22],door[22*22],summ,sumd;
    int main()
    {
    	scanf("%d%d",&N,&M);
    	for(int i=1;i<=N;++i) scanf("%s",map[i]+1);
    	for(int i=1;i<=N;++i)
    	  for(int j=1;j<=M;++j)
    	    {
    	      num[i][j]=++en;
    	      if(map[i][j]=='.') man[++summ]=en;
    	      else if(map[i][j]=='D') door[++sumd]=en;
    	    }
    	memset(dis,0x7f,sizeof(dis));
        queue<Node>q;
    	for(int i=1;i<=N;++i)
    	  for(int j=1;j<=M;++j)
    	    if(map[i][j]=='.')
    	      {
    	      	memset(vis,0,sizeof(vis));
    	      	q.push(Node(i,j,0));
    	      	vis[i][j]=1;
    	      	while(!q.empty())
    	      	  {
    	      	  	Node U=q.front(); q.pop();
    	      	  	for(int k=0;k<4;++k)
    	      	  	  {
    	      	  	  	int tx=U.x+dx[k],ty=U.y+dy[k];
    	      	  	  	if(tx>=1 && tx<=N && ty>=1 && ty<=M && map[tx][ty]!='X' && (!vis[tx][ty]))
    	      	  	  	  {
    	      	  	  	  	if(map[tx][ty]=='D')
    						  {
    						  	dis[num[i][j]][num[tx][ty]]=U.d+1;
    						  	continue;//注意:到了门立刻离开,不能继续。 
    						  }
    	      	  	  	  	vis[tx][ty]=1;
    						q.push(Node(tx,ty,U.d+1));
    	      	  	  	  }
    	      	  	  }
    	      	  }
    	      }
    	for(int Time=0;Time<=N*M;++Time)
    	  {
    	  	Init_Dinic();
    	  	for(int i=1;i<=summ;++i) AddEdge(S,man[i],1);
    	  	for(int i=1;i<=sumd;++i) AddEdge(door[i],T,Time);
    	  	for(int i=1;i<=summ;++i)
    	  	  for(int j=1;j<=sumd;++j)
    	  	    if(dis[man[i]][door[j]]<=Time)
    	  	      AddEdge(man[i],door[j],1);
    	  	if(max_flow()==summ)
    	  	  {
    	  	  	printf("%d
    ",Time);
    	  	  	return 0;
    	  	  }
    	  }
    	puts("impossible");
    	return 0;
    }
    

      

    【法二】可以二分答案,但是边界总是挂……分块答案。

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #include<queue>
    using namespace std;
    #define INF 2147483647
    #define MAXN 411
    #define MAXM 350001
    int v[MAXM],cap[MAXM],en,first[MAXN],next[MAXM];
    int d[MAXN],cur[MAXN];
    queue<int>q;
    int n,S,T,N,M;
    void Init_Dinic(){memset(first,-1,sizeof(first)); en=0; S=0; n=N*M+2; T=N*M+1;}
    void AddEdge(const int &U,const int &V,const int &W)
    {v[en]=V; cap[en]=W; next[en]=first[U]; first[U]=en++;
    v[en]=U; next[en]=first[V]; first[V]=en++;}
    bool bfs()
    {
        memset(d,-1,sizeof(d)); q.push(S); d[S]=0;
        while(!q.empty())
          {
            int U=q.front(); q.pop();
            for(int i=first[U];i!=-1;i=next[i])
              if(d[v[i]]==-1 && cap[i])
                {
                  d[v[i]]=d[U]+1;
                  q.push(v[i]);
                }
          }
        return d[T]!=-1;
    }
    int dfs(int U,int a)
    {
        if(U==T || !a) return a;
        int Flow=0,f;
        for(int &i=cur[U];i!=-1;i=next[i])
          if(d[U]+1==d[v[i]] && (f=dfs(v[i],min(a,cap[i]))))
            {
              cap[i]-=f; cap[i^1]+=f;
              Flow+=f; a-=f; if(!a) break;
            }
        if(!Flow) d[U]=-1;
        return Flow;
    }
    int max_flow()
    {
        int Flow=0,tmp=0;
        while(bfs())
          {
            memcpy(cur,first,(n+5)*sizeof(int));
            while(tmp=dfs(S,INF)) Flow+=tmp;
          }
        return Flow;
    }
    int dis[22*22][22*22];
    bool vis[22][22];
    char map[22][22];
    int num[22][22];
    struct Node
    {
    	int x,y,d;
    	Node(const int &a,const int &b,const int &c)
    	{x=a;y=b;d=c;}
    	Node(){}
    };
    const int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
    int man[22*22],door[22*22],summ,sumd;
    void Rebuild(const int &Time)
    {
    	Init_Dinic();
    	for(int i=1;i<=summ;++i) AddEdge(S,man[i],1);
    	for(int i=1;i<=sumd;++i) AddEdge(door[i],T,Time);
    	for(int i=1;i<=summ;++i)
    	  for(int j=1;j<=sumd;++j)
    	  	if(dis[man[i]][door[j]]<=Time)
    	  	  AddEdge(man[i],door[j],1);
    }
    int main()
    {
    	scanf("%d%d",&N,&M);
    	for(int i=1;i<=N;++i) scanf("%s",map[i]+1);
    	for(int i=1;i<=N;++i)
    	  for(int j=1;j<=M;++j)
    	    {
    	      num[i][j]=++en;
    	      if(map[i][j]=='.') man[++summ]=en;
    	      else if(map[i][j]=='D') door[++sumd]=en;
    	    }
    	memset(dis,0x7f,sizeof(dis));
        queue<Node>q;
    	for(int i=1;i<=N;++i)
    	  for(int j=1;j<=M;++j)
    	    if(map[i][j]=='.')
    	      {
    	      	memset(vis,0,sizeof(vis));
    	      	q.push(Node(i,j,0));
    	      	vis[i][j]=1;
    	      	while(!q.empty())
    	      	  {
    	      	  	Node U=q.front(); q.pop();
    	      	  	for(int k=0;k<4;++k)
    	      	  	  {
    	      	  	  	int tx=U.x+dx[k],ty=U.y+dy[k];
    	      	  	  	if(tx>=1 && tx<=N && ty>=1 && ty<=M && map[tx][ty]!='X' && (!vis[tx][ty]))
    	      	  	  	  {
    	      	  	  	  	if(map[tx][ty]=='D')
    						  {
    						  	dis[num[i][j]][num[tx][ty]]=U.d+1;
    						  	continue;//注意:到了门立刻离开,不能继续。 
    						  }
    	      	  	  	  	vis[tx][ty]=1;
    						q.push(Node(tx,ty,U.d+1));
    	      	  	  	  }
    	      	  	  }
    	      	  }
    	      }
    	int sz=sqrt(N*M),last=0;
    	for(int Time=0;last<=N*M;Time+=sz)
    	  {
    	  	Rebuild(Time);
    	  	if(max_flow()>=summ)
    	  	  {
    	  	  	for(int i=last+1;i<=Time;++i)
    	  	  	  {
    	  	  	  	Rebuild(i);
    	  	  	  	if(max_flow()==summ)
    	  	  	  	{printf("%d
    ",i); return 0;}
    	  	  	  }
    	  	  	return 0;
    	  	  } last=Time;
    	  }
    	puts("impossible");
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/autsky-jadek/p/4175085.html
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