1.最大流(dinic)
1 struct MF { 2 static const int N=1e5+10,M=1e5+10; 3 int hd[N],ne,cur[N],n; 4 int d[N]; 5 struct E {int v,cp,nxt;} e[M]; 6 void init(int _n) {ne=0,n=_n; for(int i=1; i<=n; ++i)hd[i]=-1;} 7 void link(int u,int v,int cp) { 8 e[ne]= {v,cp,hd[u]},hd[u]=ne++; 9 e[ne]= {u,0,hd[v]},hd[v]=ne++; 10 } 11 int bfs(int s,int t) { 12 queue<int> q; 13 for(int i=1; i<=n; ++i)d[i]=-1; 14 d[s]=0,q.push(s); 15 while(q.size()) { 16 int u=q.front(); 17 q.pop(); 18 for(int i=hd[u]; ~i; i=e[i].nxt)if(e[i].cp) { 19 int v=e[i].v; 20 if(!~d[v])d[v]=d[u]+1,q.push(v); 21 } 22 } 23 return ~d[t]; 24 } 25 int dfs(int u,int t,int f) { 26 if(u==t||!f)return f; 27 int ret=0; 28 for(int& i=cur[u]; ~i; i=e[i].nxt) { 29 int v=e[i].v; 30 if(d[v]==d[u]+1) { 31 int df=dfs(v,t,min(f,e[i].cp)); 32 e[i].cp-=df,e[i^1].cp+=df,f-=df,ret+=df; 33 if(!f)return ret; 34 } 35 } 36 return ret; 37 } 38 int dinic(int s,int t) { 39 int ret=0; 40 while(bfs(s,t)) { 41 for(int i=1; i<=n; ++i)cur[i]=hd[i]; 42 ret+=dfs(s,t,inf); 43 } 44 return ret; 45 } 46 } mf;
2.最小费用最大流(轻量级)
1 struct MCMF { 2 static const int N=1e5+10,M=1e5+10; 3 int hd[N],ne,n,d[N],h[N],vis[N],cur[N],C,F,S,T; 4 struct E {int v,c,cp,nxt;} e[M]; 5 void init(int _n,int _S,int _T) { 6 n=_n,S=_S,T=_T,ne=0; 7 for(int i=1; i<=n; ++i)hd[i]=-1,vis[i]=0; 8 } 9 void link(int u,int v,int c,int cp) { 10 e[ne]= {v,c,cp,hd[u]},hd[u]=ne++; 11 e[ne]= {u,-c,0,hd[v]},hd[v]=ne++; 12 } 13 struct D { 14 int u,g; 15 bool operator<(const D& b)const {return g>b.g;} 16 }; 17 priority_queue<D> q; 18 void upd(int u,int c) {if(d[u]>c)d[u]=c,q.push({u,d[u]});} 19 int Dij() { 20 while(q.size())q.pop(); 21 for(int i=1; i<=n; ++i)d[i]=inf; 22 upd(T,0); 23 while(q.size()) { 24 int u=q.top().u,g=q.top().g; 25 q.pop(); 26 if(g>d[u])continue; 27 for(int i=hd[u]; ~i; i=e[i].nxt)if(e[i^1].cp) { 28 int v=e[i].v,c=e[i^1].c+h[u]-h[v]; 29 upd(v,g+c); 30 } 31 } 32 return d[S]<inf; 33 } 34 int dfs(int u,int f) { 35 if(u==T) {F+=f,C+=h[S]*f; return f;} 36 vis[u]=1; 37 int ret=0; 38 for(int& i=cur[u]; ~i; i=e[i].nxt)if(e[i].cp) { 39 int v=e[i].v,c=e[i].c; 40 if(!vis[v]&&h[u]==h[v]+c) { 41 int df=dfs(v,min(f,e[i].cp)); 42 e[i].cp-=df,e[i^1].cp+=df,f-=df,ret+=df; 43 if(!f)break; 44 } 45 } 46 vis[u]=0; 47 return ret; 48 } 49 void work() { 50 C=F=0; 51 while(Dij()) { 52 for(int i=1; i<=n; ++i)h[i]+=d[i],cur[i]=hd[i]; 53 while(dfs(S,inf)); 54 } 55 } 56 } mcmf;
3.最小费用最大流(dinic加强版)
1 struct MCMF { 2 static const int N=1e5+10,M=1e5+10; 3 int hd[N],ne,n,d[N],h[N],cur[N],g[N],C,F,S,T; 4 struct E {int v,c,cp,nxt;} e[M]; 5 void init(int _n,int _S,int _T) { 6 n=_n,S=_S,T=_T,ne=0; 7 for(int i=1; i<=n; ++i)hd[i]=-1; 8 } 9 void link(int u,int v,int c,int cp) { 10 e[ne]= {v,c,cp,hd[u]},hd[u]=ne++; 11 e[ne]= {u,-c,0,hd[v]},hd[v]=ne++; 12 } 13 struct D { 14 int u,g; 15 bool operator<(const D& b)const {return g>b.g;} 16 }; 17 priority_queue<D> q; 18 queue<int> qq; 19 void upd(int u,int c) {if(d[u]>c)d[u]=c,q.push({u,d[u]});} 20 int Dij() { 21 while(q.size())q.pop(); 22 for(int i=1; i<=n; ++i)d[i]=inf; 23 upd(T,0); 24 while(q.size()) { 25 int u=q.top().u,g=q.top().g; 26 q.pop(); 27 if(g>d[u])continue; 28 for(int i=hd[u]; ~i; i=e[i].nxt)if(e[i^1].cp) { 29 int v=e[i].v,c=e[i^1].c+h[u]-h[v]; 30 upd(v,g+c); 31 } 32 } 33 return d[S]<inf; 34 } 35 int bfs() { 36 for(int i=1; i<=n; ++i)g[i]=-1; 37 g[S]=0,qq.push(S); 38 while(qq.size()) { 39 int u=qq.front(); 40 qq.pop(); 41 for(int i=hd[u]; ~i; i=e[i].nxt)if(e[i].cp) { 42 int v=e[i].v,c=e[i].c; 43 if(!~g[v]&&h[u]==h[v]+c)g[v]=g[u]+1,qq.push(v); 44 } 45 } 46 return ~g[T]; 47 } 48 int dfs(int u,int f) { 49 if(u==T) {F+=f,C+=h[S]*f; return f;} 50 int ret=0; 51 for(int& i=cur[u]; ~i; i=e[i].nxt)if(e[i].cp) { 52 int v=e[i].v,c=e[i].c; 53 if(h[u]==h[v]+c&&g[v]==g[u]+1) { 54 int df=dfs(v,min(f,e[i].cp)); 55 e[i].cp-=df,e[i^1].cp+=df,f-=df,ret+=df; 56 if(!f)break; 57 } 58 } 59 return ret; 60 } 61 void work() { 62 C=F=0; 63 while(Dij()) { 64 for(int i=1; i<=n; ++i)h[i]+=d[i]; 65 while(bfs()) { 66 for(int i=1; i<=n; ++i)cur[i]=hd[i]; 67 dfs(S,inf); 68 } 69 } 70 } 71 } mcmf;
4.最小费用最大流(究极优化，手写堆和队列)
1 template<class D> 2 class Heap { 3 static const int N=1e5+10; 4 int n; 5 D a[N]; 6 int l(int u) {return u<<1;} 7 int r(int u) {return u<<1|1;} 8 int fa(int u) {return u>>1;} 9 public: 10 Heap() {n=0;} 11 int size() {return n;} 12 D top() {return a[1];} 13 void push(D x) { 14 a[++n]=x; 15 for(int u=n; u!=1&&a[fa(u)]<a[u]; u=fa(u))swap(a[u],a[fa(u)]); 16 } 17 void pop() { 18 a[1]=a[n--]; 19 for(int u=1;;) { 20 int t=u; 21 if(l(u)<=n&&a[t]<a[l(u)])t=l(u); 22 if(r(u)<=n&&a[t]<a[r(u)])t=r(u); 23 if(t==u)break; 24 swap(a[u],a[t]),u=t; 25 } 26 } 27 }; 28 template<class D> 29 class Queue { 30 static const int N=1e5+10; 31 int hd,tl; 32 D a[N]; 33 public: 34 Queue() {hd=tl=0;} 35 int size() {return tl>=hd?tl-hd:tl-hd+N;} 36 D front() {return a[hd];} 37 void push(D x) {a[tl++]=x; if(tl==N)tl-=N;} 38 void pop() {hd++; if(hd==N)hd-=N;} 39 }; 40 struct MCMF { 41 static const int N=1e5+10,M=1e5+10; 42 int hd[N],ne,n,d[N],h[N],cur[N],g[N],C,F,S,T; 43 struct E {int v,c,cp,nxt;} e[M]; 44 void init(int _n,int _S,int _T) { 45 n=_n,S=_S,T=_T,ne=0; 46 for(int i=1; i<=n; ++i)hd[i]=-1; 47 } 48 void link(int u,int v,int c,int cp) { 49 e[ne]= {v,c,cp,hd[u]},hd[u]=ne++; 50 e[ne]= {u,-c,0,hd[v]},hd[v]=ne++; 51 } 52 struct D { 53 int u,g; 54 bool operator<(const D& b)const {return g>b.g;} 55 }; 56 Heap<D> q; 57 Queue<int> qq; 58 void upd(int u,int c) {if(d[u]>c)d[u]=c,q.push({u,d[u]});} 59 int Dij() { 60 while(q.size())q.pop(); 61 for(int i=1; i<=n; ++i)d[i]=inf; 62 upd(T,0); 63 while(q.size()) { 64 int u=q.top().u,g=q.top().g; 65 q.pop(); 66 if(g>d[u])continue; 67 for(int i=hd[u]; ~i; i=e[i].nxt)if(e[i^1].cp) { 68 int v=e[i].v,c=e[i^1].c+h[u]-h[v]; 69 upd(v,g+c); 70 } 71 } 72 return d[S]<inf; 73 } 74 int bfs() { 75 for(int i=1; i<=n; ++i)g[i]=-1; 76 g[S]=0,qq.push(S); 77 while(qq.size()) { 78 int u=qq.front(); 79 qq.pop(); 80 for(int i=hd[u]; ~i; i=e[i].nxt)if(e[i].cp) { 81 int v=e[i].v,c=e[i].c; 82 if(!~g[v]&&h[u]==h[v]+c)g[v]=g[u]+1,qq.push(v); 83 } 84 } 85 return ~g[T]; 86 } 87 int dfs(int u,int f) { 88 if(u==T) {F+=f,C+=h[S]*f; return f;} 89 int ret=0; 90 for(int& i=cur[u]; ~i; i=e[i].nxt)if(e[i].cp) { 91 int v=e[i].v,c=e[i].c; 92 if(h[u]==h[v]+c&&g[v]==g[u]+1) { 93 int df=dfs(v,min(f,e[i].cp)); 94 e[i].cp-=df,e[i^1].cp+=df,f-=df,ret+=df; 95 if(!f)break; 96 } 97 } 98 return ret; 99 } 100 void work() { 101 C=F=0; 102 while(Dij()) { 103 for(int i=1; i<=n; ++i)h[i]+=d[i]; 104 while(bfs()) { 105 for(int i=1; i<=n; ++i)cur[i]=hd[i]; 106 dfs(S,inf); 107 } 108 } 109 } 110 } mcmf;