题链:
http://www.lydsy.com/JudgeOnline/problem.php?id=2127
题解:
和 BZOJ 3984 建图类似(最小割模型)。
但是这个建图方法效率有点低。
另外这里这个解法比较高效,写得很好很清晰:
http://blog.csdn.net/hzj1054689699/article/details/53038620
代码:
#include<queue> #include<cstdio> #include<cstring> #include<iostream> #define MAXN 65000 #define MAXM 300000 #define INF 0x3f3f3f3f using namespace std; struct Edge{ int to[MAXM],cap[MAXM],nxt[MAXM],head[MAXN],ent; void Init(){ ent=2; memset(head,0,sizeof(head)); } void Adde(int u,int v,int w){ to[ent]=v; cap[ent]=w; nxt[ent]=head[u]; head[u]=ent++; to[ent]=u; cap[ent]=0; nxt[ent]=head[v]; head[v]=ent++; } int Next(int i,bool type){ return type?head[i]:nxt[i]; } }E; int cur[MAXN],d[MAXN]; int N,M,S,T,ans; int idx(int i,int j,int k){ return (i-1)*M+j+k*N*M; } bool bfs(){ queue<int>q; int u,v; memset(d,0,sizeof(d)); d[S]=1; q.push(S); while(!q.empty()){ u=q.front(); q.pop(); for(int i=E.Next(u,1);i;i=E.Next(i,0)){ v=E.to[i]; if(d[v]||!E.cap[i]) continue; d[v]=d[u]+1; q.push(v); } } return d[T]; } int dfs(int u,int reflow){ if(u==T||!reflow) return reflow; int flowout=0,f,v; for(int i=E.Next(u,1);i;i=E.Next(i,0)){ v=E.to[i]; if(d[v]!=d[u]+1) continue; f=dfs(v,min(reflow,E.cap[i])); flowout+=f; E.cap[i^1]+=f; reflow-=f; E.cap[i]-=f; if(!reflow) break; } if(!flowout) d[u]=0; return flowout; } int Dinic(){ int flow=0; while(bfs()){ memcpy(cur,E.head,sizeof(E.head)); flow+=dfs(S,INF); } return flow; } int main() { E.Init(); scanf("%d%d",&N,&M); S=N*M*5+1; T=N*M*5+2; for(int i=1,x;i<=N;i++) for(int j=1;j<=M;j++) scanf("%d",&x),ans+=x,E.Adde(S,idx(i,j,0),x); for(int i=1,x;i<=N;i++) for(int j=1;j<=M;j++) scanf("%d",&x),ans+=x,E.Adde(idx(i,j,0),T,x); for(int i=1,_i,_j,x;i<N;i++) for(int j=1;j<=M;j++){ scanf("%d",&x); ans+=x; E.Adde(S,idx(i,j,1),x); E.Adde(idx(i,j,1),idx(i,j,0),INF); E.Adde(idx(i,j,1),idx(i+1,j,0),INF); } for(int i=1,_i,_j,x;i<N;i++) for(int j=1;j<=M;j++){ scanf("%d",&x); ans+=x; E.Adde(idx(i,j,2),T,x); E.Adde(idx(i,j,0),idx(i,j,2),INF); E.Adde(idx(i+1,j,0),idx(i,j,2),INF); } for(int i=1,_i,_j,x;i<=N;i++) for(int j=1;j<M;j++){ scanf("%d",&x); ans+=x; E.Adde(S,idx(i,j,3),x); E.Adde(idx(i,j,3),idx(i,j,0),INF); E.Adde(idx(i,j,3),idx(i,j+1,0),INF); } for(int i=1,_i,_j,x;i<=N;i++) for(int j=1;j<M;j++){ scanf("%d",&x); ans+=x; E.Adde(idx(i,j,4),T,x); E.Adde(idx(i,j,0),idx(i,j,4),INF); E.Adde(idx(i,j+1,0),idx(i,j,4),INF); } int der=Dinic(); ans-=der; printf("%d",ans); return 0; }