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 2100000000
using 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);
}