网络流&&费用流模板

ISAP

1、有源有汇有上界无下界最大流 code1（邻接矩阵版）：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int map[201][201],n;int lev[201],pre[201],gap[201],cur[201]; int ISAP(int vs,int vt){memset(gap,0,sizeof(gap));memset(pre,-1,sizeof(pre));memset(lev,0,sizeof(lev));int i,v,u=pre[vs]=vs,aug,maxt=0,minl;gap[0]=vt;for (i=vs;i<=vt;++i)  cur[i]=1;while (lev[vs]<vt)  {    for (v=cur[u];v<=vt;v++)      if (lev[u]==lev[v]+1&&map[u][v]>0)        {cur[u]=v; break;}    if (v<=vt)      {        pre[v]=u;        u=v;        if (v==vt)          {          aug=2100000000;            for (i=v;i!=vs;i=pre[i])              if (map[pre[i]][i]<aug)                aug=map[pre[i]][i];            maxt+=aug;            for (i=v;i!=vs;i=pre[i])              {                map[pre[i]][i]-=aug;                map[i][pre[i]]+=aug;              }            u=vs;          }      }    else      {        minl=vt;        for (v=1;v<=vt;++v)          if (map[u][v]>0&&lev[v]<minl)            minl=lev[v];        gap[lev[u]]--;        if (gap[lev[u]]==0) break;        lev[u]=minl+1;        cur[u]=1;        gap[lev[u]]++;        u=pre[u];      }  }return maxt;}int main(){int i,j;scanf("%d",&n);for (i=1;i<=n;++i)  for (j=1;j<=n;++j)    scanf("%d",&map[i][j]);printf("%d\n",ISAP(1,n));}

code2（next数组版）：

#include<iostream>#include<cstdio>#include<cstring>#include<cstdio>using namespace std;struct hp{int u,v,c;}a[60000];int map[2000][2000];int point[2000],next[30000];int pre[30000],gap[2000],lev[2000],cur[2000];int ISAP(int vs,int vt){memset(lev,0,sizeof(lev));memset(gap,0,sizeof(gap));memset(pre,0,sizeof(pre));int i,v,u=vs,maxt=0,minl,aug,c;bool f=false;gap[0]=vt-vs+1;while (lev[vs]<vt)  {  f=false;    for (v=cur[u];v!=0;v=next[v])      if (lev[u]==lev[a[v].v]+1&&a[v].c>0)        {f=true; cur[u]=v; break;}    if (f)      {        pre[a[v].v]=v;        u=a[v].v;        if (u==vt)          {            aug=2100000000;            for (i=v;i!=0;i=pre[a[i].u])              if (a[i].c<aug)                aug=a[i].c;            maxt+=aug;            for (i=v;i!=0;i=pre[a[i].u])              {                a[i].c-=aug;                a[i^1].c+=aug;              }            u=vs;          }      }    else      {        minl=vt;        for (i=point[u];i!=0;i=next[i])          if (minl>lev[a[i].v]&&a[i].c>0)            minl=lev[a[i].v];        gap[lev[u]]--;        if (gap[lev[u]]==0) break;        lev[u]=minl+1;        cur[u]=point[u];        gap[lev[u]]++;if (u!=vs) u=a[pre[u]].u;       }  }return maxt;}int main(){int i,n,j,c,e=1;scanf("%d",&n);for (i=1;i<=n;++i)  for (j=1;j<=n;++j)    {      scanf("%d",&c);      if (c!=0)        {          e++;          a[e].u=i; a[e].v=j; a[e].c=c;          next[e]=point[i];          cur[i]=point[i]=e;          e++;          a[e].u=j; a[e].v=i; a[e].c=0;          next[e]=point[j];          cur[j]=point[j]=e;        }    }    printf("%d\n",ISAP(1,n));}

2、（1）多源多汇有上界无下界最大流 ，设置一个超级源点，流向所有入度为0的原源点，容量为无穷，再设置一个超级汇点，使所有出度为0的原汇点，容量为无穷。求一遍最大流即可。

      （2）有源有汇有点界的最大流，对每个点插成两个点，求一遍最大流即可。

3、无源无汇有上界有下界最大流：构图时，依然设置超级源点和超级汇点，但如果该节点的下界是净流出（下界的出度减入度>0）的话，该点到超级汇点的权等于净流出量；如果该节点的下界是净流入的话（该点的出度减入度<0)的话，超级源点到该点的权等于净流入量。求一遍最大流，如果超级源点的出度满流的话，即存在最大流，输出即可。否则则不存在最大流 code：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int map[150][150],n;int lev[150],pre[150],gap[150];int ISAP(int vs,int vt){memset(lev,0,sizeof(lev));memset(gap,0,sizeof(gap));memset(pre,-1,sizeof(pre));int i,v,u=pre[vs]=vs,minl,aug,maxt=0;gap[0]=vt-vs+1;while (lev[vs]<vt)  {    for (v=vs;v<=vt;v++)      if (map[u][v]>0&&lev[u]==lev[v]+1)        break;    if (v<=vt)      {        pre[v]=u;        u=v;        if (v==vt)          {            aug=2100000000;            for (i=v;i!=vs;i=pre[i])              if (map[pre[i]][i]<aug)                aug=map[pre[i]][i];            maxt+=aug;            for (i=v;i!=vs;i=pre[i])              {                map[pre[i]][i]-=aug;                map[i][pre[i]]+=aug;              }            u=vs;          }      }    else      {        minl=vt;        for (v=vs;v<=vt;++v)          if (map[u][v]>0&&lev[v]<minl)            minl=lev[v];        gap[lev[u]]--;        if (gap[lev[u]]==0) break;        lev[u]=minl+1;        gap[lev[u]]++;        u=pre[u];      }  }return maxt;}int main(){int i,j,m,inf,outf,sum,ans,p=0;int pic[150][150][2];scanf("%d",&n);for (i=1;i<=n;++i)  for (j=1;j<=2*n;++j)    scanf("%d",&pic[i][(j+1)/2][1-j%2]);for (i=1;i<=n;++i)  {  inf=0; outf=0;    for (j=1;j<=n;++j)      {    inf+=pic[j][i][0];    outf+=pic[i][j][0];    map[i][j]=pic[i][j][1]-pic[i][j][0];  }sum=outf-inf;if (sum<0)  {    map[0][i]=abs(sum);    p+=abs(sum);  }else  map[i][n+1]=abs(sum);      }    ans=ISAP(0,n+1);    if (ans==p)      printf("%d\n",ans);    else      printf("0\n"); }

4、有源有汇有上界有下界最大流，使汇点到源点有一条上界为无穷，下界为0的边（注意，此边在构图之后加），使它变为无源无汇图，构图和判断同无源无汇有上界有下界最大流，但输出时，先删去汇点到源点的无穷边，再跑一遍原源点到原汇点的最大流，输出即可 code：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int map[150][150],n;int lev[150],pre[150],gap[150];int ISAP(int vs,int vt){memset(lev,0,sizeof(lev));memset(gap,0,sizeof(gap));memset(pre,-1,sizeof(pre));int i,v,u=pre[vs]=vs,minl,aug,maxt=0;gap[0]=vt-vs+1;while (lev[vs]<vt)  {    for (v=vs;v<=vt;v++)      if (map[u][v]>0&&lev[u]==lev[v]+1)        break;    if (v<=vt)      {        pre[v]=u;        u=v;        if (v==vt)          {            aug=2100000000;            for (i=v;i!=vs;i=pre[i])              if (map[pre[i]][i]<aug)                aug=map[pre[i]][i];            maxt+=aug;            for (i=v;i!=vs;i=pre[i])              {                map[pre[i]][i]-=aug;                map[i][pre[i]]+=aug;              }            u=vs;          }      }    else      {        minl=vt;        for (v=vs;v<=vt;++v)          if (map[u][v]>0&&lev[v]<minl)            minl=lev[v];        gap[lev[u]]--;        if (gap[lev[u]]==0) break;        lev[u]=minl+1;        gap[lev[u]]++;        u=pre[u];      }  }return maxt;}int main(){int i,j,m,inf,outf,sum,ans,p=0;int pic[150][150][2];scanf("%d",&n);for (i=1;i<=n;++i)  for (j=1;j<=2*n;++j)    scanf("%d",&pic[i][(j+1)/2][1-j%2]);for (i=1;i<=n;++i)  {  inf=0; outf=0;    for (j=1;j<=n;++j)      {    inf+=pic[j][i][0];    outf+=pic[i][j][0];    map[i][j]=pic[i][j][1]-pic[i][j][0];  }sum=outf-inf;if (sum<0)  {    map[0][i]=abs(sum);    p+=abs(sum);  }else  map[i][n+1]=abs(sum);      }    map[n][1]=2100000000;    ans=ISAP(0,n+1);    if (ans==p)      {      map[n][1]=0;      ans=ISAP(1,n);        printf("%d\n",ans);      }    else      printf("0\n"); }

5、费用流模板，先跑一遍最短路，再在最短路上进行增广，ans+=dis(vs,vt)*maxflow，直到无法求出最短路。输出ans即可。code：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int point[101],next[5000],queue[500],dis[101],minn[101],pre[101],e,head,tail,ans=0,n,st,en;bool exist[101];struct hp{int u,v,c,w;}a[5000];void add(int u,int v,int c,int w){e++;a[e].u=u; a[e].v=v; a[e].c=c; a[e].w=w; next[e]=point[u]; point[u]=e;e++;a[e].u=v; a[e].v=u; a[e].c=0; a[e].w=0-w; next[e]=point[v]; point[v]=e;}bool work(int vs,int vt){memset(exist,false,sizeof(exist));memset(dis,127,sizeof(dis));memset(pre,0,sizeof(pre));    int i,j,u,stan=dis[1]; minn[vs]=dis[1];    exist[vs]=true; dis[vs]=0; head=0; tail=1; queue[tail]=vs;while (head!=tail)  {    head=head%500+1;u=queue[head];exist[u]=false;for (i=point[u];i!=0;i=next[i])  {    if (a[i].c>0&&dis[a[i].v]>dis[u]+a[i].w)      {        dis[a[i].v]=dis[u]+a[i].w;        pre[a[i].v]=i;        minn[a[i].v]=min(minn[u],a[i].c);        if (!exist[a[i].v])          {            tail=tail%500+1;            queue[tail]=a[i].v;            exist[a[i].v]=true;          }      }  }    }if (dis[vt]==stan) return false;ans+=dis[vt]*minn[vt];for (i=pre[vt];i!=0;i=pre[a[i].u])  {    a[i].c-=minn[vt];    a[i^1].c+=minn[vt];  }return true;}int main(){int i,j,c,w;scanf("%d%d%d",&n,&st,&en);e=1;for (i=1;i<=n;++i)  for (j=1;j<=n;++j)    {      scanf("%d%d",&c,&w);      if (c) add(i,j,c,w);    }  while (work(st,en));printf("%d\n",ans);}

最大权闭合子图code：

#include<iostream>#include<cstdio>#include<cstring>#define inf 2100000000using namespace std;struct hp{int u,v,c;}a[500000];int e,n,m;int p[5001];int point[56000],next[500000],pre[56000],lev[56000],gap[56000];int ISAP(int vs,int vt){int v,i,u,maxf=0,aug,minl; bool f;gap[0]=vt-vs+1; u=vs;while (lev[vs]<vt)  {    f=false;for (v=point[u];v!=0;v=next[v])  if (lev[u]==lev[a[v].v]+1&&a[v].c>0)    {f=true; break;}if (f)  {    pre[a[v].v]=v;    u=a[v].v;    if (u==vt)      {        aug=inf;        for (i=v;i!=0;i=pre[a[i].u])          if (aug>a[i].c)            aug=a[i].c;        maxf+=aug;        for (i=v;i!=0;i=pre[a[i].u])          {            a[i].c-=aug;            a[i^1].c+=aug;          }        u=vs;      }  }else  {    minl=vt;for (i=point[u];i!=0;i=next[i])  if (minl>lev[a[i].v]&&a[i].c>0)    minl=lev[a[i].v];    gap[lev[u]]--;if (gap[lev[u]]==0) break;lev[u]=minl+1;gap[lev[u]]++;if (u!=vs) u=a[pre[u]].u;    }  }return maxf;}int main(){int i,x,y,c,ans,sum=0;freopen("profit9.in","r",stdin);freopen("profit.out","w",stdout);scanf("%d%d",&n,&m);e=1;for (i=1;i<=n;++i)  {    scanf("%d",&p[i]);    e++;     next[e]=point[i]; point[i]=e; a[e].u=i; a[e].v=n+m+1; a[e].c=p[i];    e++;    next[e]=point[n+m+1]; next[n+m+1]=e; a[e].u=n+m+1; a[e].v=i; a[e].c=0;      }for (i=1;i<=m;++i)  {    scanf("%d%d%d",&x,&y,&c); sum+=c;e++;next[e]=point[0]; point[0]=e; a[e].u=0; a[e].v=n+i; a[e].c=c;e++;next[e]=point[n+i]; point[n+i]=e; a[e].u=n+i; a[e].v=0; a[e].c=0;e++;next[e]=point[n+i]; point[n+i]=e; a[e].u=n+i; a[e].v=x; a[e].c=inf;e++;next[e]=point[x]; point[x]=e; a[e].u=x; a[e].v=n+i; a[e].c=0;    e++;    next[e]=point[n+i]; point[n+i]=e; a[e].u=n+i; a[e].v=y; a[e].c=inf;    e++;    next[e]=point[y]; point[y]=e; a[e].u=y; a[e].v=n+i; a[e].c=0;      }   ans=ISAP(0,n+m+1);   printf("%d\n",sum-ans);   fclose(stdin); fclose(stdout);}