【bzoj4519】不同的最小割

  • 分治最小割系列第二题

有了上一题的铺垫,这道题的做法是显然的

几分钟写完代码后就1A了,真是舒畅啊

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#include<bits/stdc++.h>
#define FILE "read"
#define MAXN 855
#define INF 1000000000
using namespace std;
struct node{int y,next,v,rel;}e[17010];
int n,m,len,S,T,cnt,level[MAXN],Link[MAXN],val[MAXN],vis[MAXN],temp[MAXN],a[MAXN];
queue<int>q;
char buf[1<<15],*fs,*ft;
inline char getc(){return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;}
inline int read(){
int x=0,f=1; char ch=getc();
while(!isdigit(ch)) {if(ch=='-') f=-1; ch=getc();}
while(isdigit(ch)) {x=x*10+ch-'0'; ch=getc();}
return x*f;
}
void insert(int x,int y,int v){
e[++len].next=Link[x];Link[x]=len;e[len].y=y;e[len].v=v;e[len].rel=len+1;
e[++len].next=Link[y];Link[y]=len;e[len].y=x;e[len].v=v;e[len].rel=len-1;
}
bool bfs(){
memset(level,-1,sizeof(level));
q.push(S); level[S]=0;
while(!q.empty()){
int x=q.front(); q.pop();
for(int i=Link[x];i;i=e[i].next)if(e[i].v&&level[e[i].y]==-1)
q.push(e[i].y),level[e[i].y]=level[x]+1;
}
return level[T]>=0;
}
int MAXFLOW(int x,int flow){
if(x==T) return flow;
int d=0,maxflow=0;
for(int i=Link[x];i&&maxflow<flow;i=e[i].next)
if(level[e[i].y]==level[x]+1&&e[i].v)
if(d=MAXFLOW(e[i].y,min(e[i].v,flow-maxflow)))
e[i].v-=d,e[e[i].rel].v+=d,maxflow+=d;
if(!maxflow) level[x]=-1;
return maxflow;
}
void dfs(int x){
vis[x]=1;for(int i=Link[x];i;i=e[i].next)
if(e[i].v&&!vis[e[i].y]) dfs(e[i].y);
}
void solve(int l,int r){
if(l==r) return;
S=a[l]; T=a[r]; int ans=0,d=0;
for(int i=1;i<=len;++i) e[i].v=e[e[i].rel].v=(e[i].v+e[e[i].rel].v)>>1;
while(bfs()) while(d=MAXFLOW(S,INF)) ans+=d;
val[++cnt]=ans; memset(vis,0,sizeof(vis));
dfs(S); int ta=l,tb=r;
for(int i=l;i<=r;++i){
if(vis[a[i]]) temp[ta++]=a[i];
else temp[tb--]=a[i];
}
for(int i=l;i<=r;++i) a[i]=temp[i];
solve(l,ta-1); solve(tb+1,r);
}
int main(){
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
n=read(); m=read();
for(int i=1;i<=m;++i){
int x=read(),y=read(),v=read();
insert(x,y,v);
}
for(int i=1;i<=n;++i) a[i]=i;
solve(1,n); sort(val+1,val+cnt+1); int ans=0;
for(int i=1;i<=cnt;++i) if(val[i]!=val[i-1]) ans++;
printf("%d\n",ans);
return 0;
}
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