Description
给图,求最大流
最大流模板题,这里用dinic
Code
#include <cstdio>
#include <cstring>
#include <algorithm>
#define Inf 0x7fffffff
#define N 210
using namespace std;
int g[N][N],d[N],q[N*10],h,t;
int n,m,Ans,tmp;
bool Bfs(){
memset(d,-1,sizeof(d));
d[1]=0;
h=0,t=1;
q[1]=1;
while(h<t){
int u=q[++h];
for(int v=1;v<=n;++v)
if(d[v]<0&&g[u][v]>0){
d[v]=d[u]+1;
q[++t]=v;
}
}
return d[n]>0;
}
int dfs(int u,int low){
if(u==n) return low;
int tmp;
for(int v=1;v<=n;++v){
if(g[u][v]>0&&d[v]==d[u]+1&&(tmp=dfs(v,min(low,g[u][v])))){
g[u][v]-=tmp;
g[v][u]+=tmp;
return tmp;
}
}
return 0;
}
int main(){
while(~scanf("%d%d",&m,&n)){
memset(g,0,sizeof(g));
for(int i=1,u,v,f;i<=m;++i){
scanf("%d%d%d",&u,&v,&f);
g[u][v]+=f;
}
Ans=0;
while(Bfs()) {
while(tmp=dfs(1,Inf)) Ans+=tmp;
}
printf("%d
",Ans);
}
return 0;
}
下面是sap算法,
#include <cstdio>
#include <algorithm>
#include <cstring>
#define Inf 0x7fffffff
#define N 110
using namespace std;
struct info{
int nex,fr,to,f;
}e[N*N];
int n,m,Ans,dis[N],head[N],cnt[N],S,T,tot;
void Link(int u,int v){
e[++tot].nex=head[u];
e[tot].f=1;e[tot].fr=u;e[tot].to=v;head[u]=tot;
e[++tot].nex=head[v];
e[tot].f=0;e[tot].fr=v;e[tot].to=u;head[v]=tot;
}
int sap(int u,int delta){
int sum=0,mins=n;
if(u==T) return delta;
for(int i=head[u];i;i=e[i].nex){
int v=e[i].to;
if(e[i].f>0&&dis[u]==dis[v]+1){
int save=sap(v,min(delta-sum,e[i].f));
sum+=save;
e[i].f-=save;
e[i^1].f+=save;
if(dis[S]>=n||sum==delta) return sum;
}
if(e[i].f>0) mins=min(mins,dis[v]);
}
if(sum==0){
if(!(--cnt[dis[u]])) dis[S]=n;
else ++cnt[dis[u]=mins+1];
}
return sum;
}
int main(){
scanf("%d%d",&m,&n);
S=0,T=n+1;tot=-1;//异或取反向边,所以从0开始编号
int u,v,tmp;
while(~scanf("%d%d",&u,&v)){
if(u+v==-2) break;
Link(u,v);
}
for (int i=1;i<=m;++i) Link(S,i);
for (int i=m+1;i<=n;++i) Link(i,T);
n+=2;cnt[0]=n;
while(dis[S]<n) Ans+=sap(S,Inf);
printf("%d
",Ans);
return 0;
}