洛谷$P3877\ [TJOI2010]$打扫房间 网络流

正解:网络流

解题报告:

传送门$QwQ$

昂考虑把题目的约束条件详细化?就说每个格点能向四连通连边,问能否做到每个格点度数等于2?

$umm$就先黑白染色建两排点呗,然后就$S$向左侧连流量为2的边右侧向$T$连流量为2的边,然后四连通之间连流量为1的边,跑个最大流看跑满没有,然后就做完了?$QwQ$

解释下趴还是,,,毕竟我之前没解释重新看一遍我题解都没想通$QAQ$

首先找出题目的所有约束条件$QwQ$

1)每个房间进出各一次

2)每扇门最多经过一次

3)环长大于2

昂条件一等价于每个点度数等于2嘛,然后其实有了条件二就一定能满足条件三鸭,所以现在就变成,每个点度数等于2,且相邻点之间最多经过一次.

所以就$ST$分别向点连流量为2的边相邻之间连流量为1的边就欧克了$QwQ$

#include<bits/stdc++.h>

using namespace std;

#define il inline

#define gc getchar()

#define t(i) edge[i].to

#define w(i) edge[i].wei

#define n(i) edge[i].nxt

#define ri register int

#define rb register int

#define rc register char

#define rp(i,x,y) for(ri i=x;i<=y;++i)

#define my(i,x,y) for(ri i=x;i>=y;--i)

#define e(i,x) for(ri i=head[x];~i;i=n(i))

const int N=+,M=+,inf=1e9;

char str[M];

bool zt[M][M];

struct ed{int to,nxt,wei;}edge[N<<];

int n,m,dep[N],head[N],cur[N],S,T,ed_cnt=-,mvx[]={,,,-},mvy[]={,-,,},cnt;

il int read()

{

    rc ch=gc;ri x=;rb y=;

    while(ch!='-' && (ch>'' || ch<''))ch=gc;

    if(ch=='-')ch=gc,y=;

    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;

    return y?x:-x;

}

il int nam(ri x,ri y){return (x-)*m+y;}

il void print(ri d)

{

    if(!d)return void(printf("S"));if(d>n*m)return void(printf("T"));

    ri y=d%m,x=d/m+;if(!y)y=m,--x;printf("(%d , %d)",x,y);

}

il void ad(ri x,ri y,ri z)

{//printf("%d -> %d : %d\n",y,x,z);

    //print(y);printf(" -> ");print(x);printf(" : %d\n",z);

    edge[++ed_cnt]=(ed){x,head[y],z};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x],};head[x]=ed_cnt;}

il bool bfs()

{

    queue<int>Q;Q.push(S);memset(dep,,sizeof(dep));dep[S]=;

    while(!Q.empty())

    {

        ri nw=Q.front();Q.pop();

        e(i,nw)if(w(i) && !dep[t(i)]){dep[t(i)]=dep[nw]+,Q.push(t(i));if(t(i)==T)return ;}

    }

    return ;

}

il int dfs(ri nw,ri flow)

{

    if(nw==T || !flow)return flow;ri ret=;

    for(ri &i=cur[nw];~i;i=n(i))

        if(w(i) && dep[t(i)]==dep[nw]+)

        {ri tmp=dfs(t(i),min(flow,w(i)));ret+=tmp,w(i)-=tmp;w(i^)+=tmp,flow-=tmp;}

    return ret;

}

il int dinic(){ri ret=;while(bfs()){rp(i,S,T)cur[i]=head[i];while(int d=dfs(S,inf))ret+=d;}return ret;}

int main()

{

    //freopen("3877.in","r",stdin);freopen("3877.out","w",stdout);

    ri tmp=read();

    while(tmp--)

    {

        ed_cnt=-;memset(head,-,sizeof(head));n=read();m=read();cnt=;

        rp(i,,n){scanf("%s",str);rp(j,,m)cnt+=(zt[i][j]=str[j-]=='.');}S=;T=n*m+;

        rp(i,,n)rp(j,,m)

        {

            if((i+j)& && zt[i][j])

                rp(k,,)

                {

                    ri tx=i+mvx[k],ty=j+mvy[k];

                    if(!tx || !ty || tx>n || ty>m)continue;

                    if(zt[tx][ty])ad(nam(tx,ty),nam(i,j),);

                }

        }

        rp(i,,n)rp(j,,m)if(zt[i][j]){if((i+j)&)ad(nam(i,j),S,);else ad(T,nam(i,j),);}

        if(cnt&){printf("NO\n");continue;}

        if(dinic()==cnt)printf("YES\n");else printf("NO\n");

    }

    return ;

}