Network flow templates

Ford_fulkerson (O (f*e))

Don't feel like it, but just write it ...

Complexity: The maximum flow is F, at least 1 at a time, up to the augmented F, each time up to the E-bar (the entire picture).

#include <cstdio>#include<cstring>#include<vector>#include<algorithm>#defineRep (i,n) for (int i=0;i< (n); i++)#defineFor1 (i,a,n) for (int i= (a); i<= (n); i++)#defineCC (i,a) memset (i,a,sizeof (i))#defineRead (a) A=getnum ()#definePrint (a) printf ("%d", a)using namespacestd;Const intmaxn=205, inf=0x7fffffff;structEdge {intTo,cap,rev;};intM,n;BOOLVis[maxn];vector<edge>G[maxn];inlineintGetnum () {intR=0, k=1;CharC for(C=getchar ();c<'0'|| C>'9'; C=getchar ())if(c=='-') k=-1; for(; c>='0'&&c<='9'; C=getchar ()) r=r*Ten+c-'0';returnr*K;}voidAdd_edge (int  from,intTo,intcap) {g[ from].push_back (Edge) {to,cap,g[to].size ()}); G[to].push_back (Edge) { from,0, g[ from].size ()-1 });}intDfsintVintTintf) {    if(v==t)returnF; VIS[V]=true; Rep (I,g[v].size ()) {Edge&e=G[v][i]; if(!vis[e.to]&&e.cap>0)        {            intD=Dfs (E.to,t,min (f,e.cap)); if(d>0) {E.cap-=D; G[e.to][e.rev].cap+=D; returnD; }        }    }    return 0;}intMax_flow (intSintt) {CC (Vis,false); intflow=0, F;  while((F=dfs (s,t,inf)) >0) {Flow+=F; CC (Vis,false); }    returnflow;}voidinit () {read (m); Read (n); For1 (i,1, M) {        int  from, To,cap; Read ( from); Read (to);        Read (CAP); Add_edge ( from, To,cap); }}intMain () {init (); Print (Max_flow (1, N)); return 0;} 

Dinic (O (e*v^2))

Complexity: Each level chart runs, indicating the shortest augmentation path to become longer, re-BFS to construct a new hierarchical map of the shortest augmented road at least +1, the length of up to V-1, so there are up to (V-1) a hierarchy of graphs. Each of the augmented paths has at least one bottleneck in each hierarchy (the Small arc), after running the shortest circuit, the arc is gone (because the reverse arc does not conform to the level, will not go in this level chart), up to the e-arc, at least one elimination at a time, that up to run e times, each time up to run the V point, so in a hierarchy chart up to run E * V times, comprehensive, total complex The degree is O (e*v^2).

#include <cstdio>#include<cstring>#include<vector>#include<queue>#include<algorithm>#defineRep (i,n) for (int i=0;i<n;i++)#defineFor1 (i,a,n) for (int i= (a); i<= (n); i++)#defineCC (i,a) memset (i,a,sizeof (i))#defineRead (a) A=getnum ()#definePrint (a) printf ("%d\n", a)using namespacestd;Const intmaxn=205, inf=0x7fffffff;intM,n;intITER[MAXN],LEVEL[MAXN];structEdge {intTo,cap,rev;}; Vector<edge>G[maxn];inlineintGetnum () {intR=0, k=1;CharC for(C=getchar ();c<'0'|| C>'9'; C=getchar ())if(c==-'-') k=-1; for(; c>='0'&&c<='9'; C=getchar ()) r=r*Ten+c-'0';returnr*K;}voidAdd_edge (int  from,intTo,intcap) {g[ from].push_back (Edge) {to,cap,g[to].size ()}); G[to].push_back (Edge) { from,0, g[ from].size ()-1 });}voidBFsints) {CC (level,-1); Level[s]=1; Queue<int>Q;    Q.push (s);  while(!Q.empty ()) {        intt=Q.front (); Q.pop (); Rep (I,g[t].size ()) {Edge&e=G[t][i]; if(level[e.to]<0&&e.cap>0) {level[e.to]=level[t]+1;            Q.push (e.to); }        }    }}intDfsintVintTintf) {    if(v==t)returnF;  for(int&i=iter[v];i<g[v].size (); i++) {Edge&e=G[v][i]; if(e.cap>0&&level[v]<Level[e.to]) {            intD=Dfs (E.to,t,min (f,e.cap)); if(d>0) {E.cap-=D; G[e.to][e.rev].cap+=D; returnD; }        }    }    return 0;}intMax_flow (intSintt) {    intflow=0;    BFS (s);  while(level[t]>0) {CC (ITER,0); intF;  while((F=dfs (s,t,inf)) >0) flow+=F;    BFS (s); }    returnflow;}voidinit () {read (m); Read (n); For1 (i,1, M) {        int  from, To,cap; Read ( from); Read (to);        Read (CAP); Add_edge ( from, To,cap); }}intMain () {init (); Print (Max_flow (1, N)); return 0;} 

Network flow templates