最小割分治(最小割树):BZOJ2229 && BZOJ4519

定理：n个点的无向图的最小割最多n-1个。

可能从某种形式上形成了一棵树，不是很清楚。

最小割分治：先任选两个点求一边最小割，然后将两边分别递归，就能找到所有的最小割。

这两个题是一样的，直接搬dinic模板即可。

BZOJ2229:

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #define mem(a,k) memset(a,k,sizeof(a))
 #define rep(i,l,r) for (int i=l; i<=r; i++)
 #define For(i,x) for (int i=h[x],k; i; i=e[i].nxt)
 using namespace std;

 const int N=,inf=;
 int m,n,u,v,w,x,S,T,TT,Q,tot,cnt,tmp[N],a[N],b[N],d[N],q[N*],h[N],ans[N][N];
 struct E{ int to,nxt,v; }e[];
 bool mark[N];

 void add(int u,int v,int w){
    e[++cnt]=(E){v,h[u],w}; h[u]=cnt;
    e[++cnt]=(E){u,h[v],w}; h[v]=cnt;
 }

 bool bfs(){
     mem(d,); q[]=S; d[S]=;
     for (int st=,ed=; st!=ed; ){
         int x=q[++st];
         For(i,x) if (e[i].v && !d[k=e[i].to])
             d[k]=d[x]+,q[++ed]=k;
     }
     return d[T];
 }

 int dfs(int x,int lim){
     if (x==T) return lim;
     int t,c=;
     For(i,x) if (d[k=e[i].to]==d[x]+){
         t=dfs(k,min(lim-c,e[i].v));
         e[i].v-=t; e[i^].v+=t; c+=t;
         if (c==lim) return lim;
     }
     if (!c) d[x]=-; return c;
 }

 int dinic(){ int ans=; while(bfs()) ans+=dfs(S,inf); return ans; }

 void get(int x){
     mark[x]=; For(i,x) if (!mark[k=e[i].to] && e[i].v) get(k);
 }

 void solve(int l,int r){
     if (l==r) return;
     S=a[l]; T=a[r]; int t=dinic();
     mem(mark,); get(S); int p=l,p0;
     rep(i,,n) if (mark[i]) rep(j,,n) if (!mark[j]) ans[i][j]=ans[j][i]=min(ans[i][j],t);
     for (int i=; i<=cnt; i+=) e[i].v=e[i^].v=(e[i].v+e[i^].v)>>;
     rep(i,l,r) if (mark[a[i]]) tmp[p++]=a[i];
     p0=p;
     rep(i,l,r) if (!mark[a[i]]) tmp[p++]=a[i];
     rep(i,l,r) a[i]=tmp[i];
     solve(l,p0-); solve(p0,r);
 }

 int main(){
     freopen("bzoj2229.in","r",stdin);
     freopen("bzoj2229.out","w",stdout);
     for (scanf("%d",&TT); TT--; ){
         cnt=; mem(h,); mem(ans,0x3f);
         scanf("%d%d",&n,&m);
         rep(i,,m) scanf("%d%d%d",&u,&v,&w),add(u,v,w);
         rep(i,,n) a[i]=i; solve(,n);
         for (scanf("%d",&Q); Q--; ){
             scanf("%d",&x); tot=;
             rep(i,,n-) rep(j,i+,n) if (ans[i][j]<=x) tot++;
             printf("%d\n",tot);
         }
         puts("");
     }
     return ;
 }

BZOJ4519:

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #define rep(i,l,r) for (int i=l; i<=r; i++)
 #define For(i,x) for (int i=h[x],k; i; i=e[i].nxt)
 using namespace std;

 const int N=,inf=;
 int m,n,u,v,w,S,T,tot,cnt=,tmp[N],a[N],b[N],d[N],q[N],h[N];
 struct E{ int to,nxt,v; }e[];
 bool mark[N];

 void add(int u,int v,int w){
    e[++cnt]=(E){v,h[u],w}; h[u]=cnt;
    e[++cnt]=(E){u,h[v],w}; h[v]=cnt;
 }

 bool bfs(){
     memset(d,,sizeof(d)); q[]=S; d[S]=;
     for (int st=,ed=; st!=ed; ){
         int x=q[++st];
         For(i,x) if (e[i].v && !d[k=e[i].to])
             d[k]=d[x]+,q[++ed]=k;
     }
     return d[T];
 }

 int dfs(int x,int lim){
     if (x==T) return lim;
     int t,c=;
     For(i,x) if (d[k=e[i].to]==d[x]+){
         t=dfs(k,min(lim-c,e[i].v));
         e[i].v-=t; e[i^].v+=t; c+=t;
         if (c==lim) return lim;
     }
     if (!c) d[x]=-; return c;
 }

 int dinic(){ int ans=; while(bfs()) ans+=dfs(S,inf); return ans; }

 void get(int x){
     mark[x]=;
     For(i,x) if (!mark[k=e[i].to] && e[i].v) get(k);
 }

 void solve(int l,int r){
     if (l==r) return;
     S=a[l]; T=a[r]; b[++tot]=dinic();
     memset(mark,,sizeof(mark));
     get(S); int p=l,p0;
     for (int i=; i<=cnt; i+=) e[i].v=e[i^].v=(e[i].v+e[i^].v)>>;
     rep(i,l,r) if (mark[a[i]]) tmp[p++]=a[i];
     p0=p;
     rep(i,l,r) if (!mark[a[i]]) tmp[p++]=a[i];
     rep(i,l,r) a[i]=tmp[i];
     solve(l,p0-); solve(p0,r);
 }

 int main(){
     freopen("bzoj4519.in","r",stdin);
     freopen("bzoj4519.out","w",stdout);
     scanf("%d%d",&n,&m);
     rep(i,,m) scanf("%d%d%d",&u,&v,&w),add(u,v,w);
     rep(i,,n) a[i]=i;
     solve(,n); sort(b+,b+tot+); tot=unique(b+,b+tot+)-b-;
     printf("%d\n",tot);
     return ;
 }