Positive Solutions: Network Flow
Report problem solving:
A network flow look very expensive ,,, So the question now becomes how to build a $ QwQ $ Fig.
First, if there is only one requirement, that directly required to build and run a maximum flow diagram like.
Has now become, there are two requirements that must be met at the same time, consider how to solve?
Consider split point, the people split into two points, respectively, even food and hotel, and then run a maximum flow, finished $ QwQ $
$over$
#include<bits/stdc++.h> using namespace std; #define il inline #define lf double #define gc getchar() #define t(i) edge[i].to #define n(i) edge[i].nxt #define w(i) edge[i].wei #define ri register int #define rb register bool #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=edge[i].nxt) const int N=100+5,M=N*20,inf=1e9; int head[M],ed_cnt=-1,n,p,q,S,T,cur[M],nod_cnt,dep[M]; struct ed{int to,nxt,wei;}edge[M<<1]; il int read() { ri x=0;rb y=1;rc ch=gc; while(ch!='-' && (ch>'9' || ch<'0'))ch=gc; if(ch=='-')ch=gc,y=0; while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc; return y?x:-x; } il void ad(ri x,ri y,ri z){edge[++ed_cnt]=(ed){x,head[y],z};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x],0};head[x]=ed_cnt;} il bool bfs() { queue<int>Q;Q.push(S);memset(dep,0,sizeof(dep));dep[S]=1; while(!Q.empty()){ri nw=Q.front();Q.pop();e(i,nw)if(w(i) && !dep[t(i)])dep[t(i)]=dep[nw]+1,Q.push(t(i));} return dep[T]; } il int dfs(ri nw,ri flow) { if(nw==T || !flow)return flow;ri ret=0; for(ri &i=cur[nw];~i;i=n(i))if(w(i) && dep[t(i)]==dep[nw]+1){ri tmp=dfs(t(i),min(flow,w(i)));flow-=tmp,w(i)-=tmp,ret+=tmp,w(i^1)+=tmp;} return ret; } il int dinic(){ri ret=0;while(bfs()){rp(i,S,T)cur[i]=head[i];while(int d=dfs(S,inf))ret+=d;}return ret;} int main() { memset(head,-1,sizeof(head));n=read();p=read();q=read(); S=0;T=p+q+(n<<1)+1;rp(i,1,p)ad(i,S,1);rp(i,1,n)ad(i+p+n,i+p,1);rp(i,1,q)ad(T,i+p+(n<<1),1); rp(i,1,n)rp(j,1,p){ri tmp=read();if(tmp)ad(i+p,j,1);}rp(i,1,n)rp(j,1,q){ri tmp=read();if(tmp)ad(j+p+(n<<1),i+p+n,1);} printf("%d\n",dinic()); return 0; }