看到这种题,应该一眼就能知道考的是最小割
没错这题就是如此简单,跑两遍最大流(最小割=最大流),一次边权为题目所给,一次边权为1
还有一种优化,优化后只需跑一次最大流,把每条边的权值改成w*MOD+1(MOD为常数,珂以取八位质数233)
答案为maxflow/MOD和maxflow%MOD
基础版本
#include <bits/stdc++.h>
#define N 40
#define M 2005
#define inf 0x3f3f3f3f
#define getchar nc
using namespace std;
inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{
register int x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
return x*f;
}
inline void write(register int x)
{
if(!x)putchar('0');if(x<0)x=-x,putchar('-');
static int sta[20];register int tot=0;
while(x)sta[tot++]=x%10,x/=10;
while(tot)putchar(sta[--tot]+48);
}
inline int Min(register int a,register int b)
{
return a<b?a:b;
}
struct edge{
int to,nxt,v;
}e[M];
int head[N],cnt=1;
inline void add(register int u,register int v,register int w)
{
e[++cnt]=(edge){v,head[u],w};
head[u]=cnt;
}
int n,m,s,t,maxflow=0;
int cur[N],dep[N],gap[N];
inline void bfs()
{
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
dep[t]=0;
++gap[dep[t]];
queue <int> q;
q.push(t);
while(!q.empty())
{
int u=q.front();
q.pop();
for(register int i=head[u];i;i=e[i].nxt)
{
int v=e[i].to;
if(dep[v]!=-1)
continue;
q.push(v);
dep[v]=dep[u]+1;
++gap[dep[v]];
}
}
}
inline int dfs(register int u,register int flow)
{
if(u==t)
{
maxflow+=flow;
return flow;
}
int used=0;
for(register int i=cur[u];i;i=e[i].nxt)
{
cur[u]=i;
int v=e[i].to;
if(e[i].v&&dep[u]==dep[v]+1)
{
int tmp=dfs(v,Min(flow-used,e[i].v));
if(tmp)
e[i].v-=tmp,e[i^1].v+=tmp,used+=tmp;
}
if(used==flow)
return used;
}
--gap[dep[u]++]==0?dep[s]=n+1:++gap[dep[u]];
return used;
}
inline void ISAP()
{
maxflow=0;
bfs();
while(dep[s]<=n)
{
memcpy(cur,head,sizeof(head));
dfs(s,inf);
}
}
int main()
{
n=read(),m=read();
s=1,t=n;
for(register int i=1;i<=m;++i)
{
int u=read(),v=read(),w=read();
add(u,v,w),add(v,u,0);
}
ISAP();
write(maxflow),putchar(' ');
for(register int i=2;i<=cnt;++i)
e[i].v=i%2?0:1;
ISAP();
write(maxflow);
return 0;
}
优化版本
#include <bits/stdc++.h>
#define N 40
#define M 2005
#define inf 123456789123456789LL
#define ll long long
#define mod 19260817LL
#define getchar nc
using namespace std;
inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline ll read()
{
register ll x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
return x*f;
}
inline void write(register ll x)
{
if(!x)putchar('0');if(x<0)x=-x,putchar('-');
static int sta[20];register int tot=0;
while(x)sta[tot++]=x%10,x/=10;
while(tot)putchar(sta[--tot]+48);
}
inline ll Min(register ll a,register ll b)
{
return a<b?a:b;
}
struct edge{
int to,nxt;
ll v;
}e[M];
int head[N],cnt=1;
inline void add(register int u,register int v,register ll w)
{
e[++cnt]=(edge){v,head[u],w};
head[u]=cnt;
}
int n,m,s,t;
ll maxflow=0;
int cur[N],dep[N],gap[N];
inline void bfs()
{
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
dep[t]=0;
++gap[dep[t]];
queue <int> q;
q.push(t);
while(!q.empty())
{
int u=q.front();
q.pop();
for(register int i=head[u];i;i=e[i].nxt)
{
int v=e[i].to;
if(dep[v]!=-1)
continue;
q.push(v);
dep[v]=dep[u]+1;
++gap[dep[v]];
}
}
}
inline ll dfs(register int u,register ll flow)
{
if(u==t)
{
maxflow+=flow;
return flow;
}
ll used=0;
for(register int i=cur[u];i;i=e[i].nxt)
{
cur[u]=i;
int v=e[i].to;
if(e[i].v&&dep[u]==dep[v]+1)
{
ll tmp=dfs(v,Min(flow-used,e[i].v));
if(tmp)
e[i].v-=tmp,e[i^1].v+=tmp,used+=tmp;
}
if(used==flow)
return used;
}
--gap[dep[u]++]==0?dep[s]=n+1:++gap[dep[u]];
return used;
}
inline void ISAP()
{
maxflow=0;
bfs();
while(dep[s]<n)
{
memcpy(cur,head,sizeof(head));
dfs(s,inf);
}
}
int main()
{
n=read(),m=read();
s=1,t=n;
for(register int i=1;i<=m;++i)
{
int u=read(),v=read(),w=read();
add(u,v,w*mod+1),add(v,u,0);
}
ISAP();
write(maxflow/mod),putchar(' '),write(maxflow%mod);
return 0;
}