定理：n個點的無向圖的最小割最多n-1個。

可能從某種形式上形成了一棵樹，不是很清楚。

最小割分治：先任選兩個點求一邊最小割，然後將兩邊分別遞歸，就能找到所有的最小割。

這兩個題是一樣的，直接搬dinic模板即可。

BZOJ2229:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<vector>
 5 #define mem(a,k) memset(a,k,sizeof(a))
 6 #define rep(i,l,r) for (int i=l; i<=r; i++)
 7
 
 #define For(i,x) for (int i=h[x],k; i; i=e[i].nxt)
 8 using namespace std;
 9 
10 const int N=160,inf=1000000000;
11 int m,n,u,v,w,x,S,T,TT,Q,tot,cnt,tmp[N],a[N],b[N],d[N],q[N*10],h[N],ans[N][N];
12 struct E{ int to,nxt,v; }e[7010];
13 bool mark[N];
14 
15 void add(int u,int v,int w){
16    e[++cnt]=(E){v,h[u],w}; h[u]=cnt;
 
17    e[++cnt]=(E){u,h[v],w}; h[v]=cnt;
18 }
19 
20 bool bfs(){
21     mem(d,0); q[1]=S; d[S]=1;
22     for (int st=0,ed=1; st!=ed; ){
23         int x=q[++st];
24         For(i,x) if (e[i].v && !d[k=e[i].to])
25             d[k]=d[x]+1,q[++ed]=k;
26     }
27     return d[T];
28 }
29
 
 
30 int dfs(int x,int lim){
31     if (x==T) return lim;
32     int t,c=0;
33     For(i,x) if (d[k=e[i].to]==d[x]+1){
34         t=dfs(k,min(lim-c,e[i].v));
35         e[i].v-=t; e[i^1].v+=t; c+=t;
36         if (c==lim) return lim;
37     }
38     if (!c) d[x]=-1; return c;
39 }
40 
41 int dinic(){ int ans=0; while(bfs()) ans+=dfs(S,inf); return ans; }
42 
43 void get(int x){
44     mark[x]=1; For(i,x) if (!mark[k=e[i].to] && e[i].v) get(k);
45 }
46 
47 void solve(int l,int r){
48     if (l==r) return;
49     S=a[l]; T=a[r]; int t=dinic();
50     mem(mark,0); get(S); int p=l,p0;
51     rep(i,1,n) if (mark[i]) rep(j,1,n) if (!mark[j]) ans[i][j]=ans[j][i]=min(ans[i][j],t);
52     for (int i=2; i<=cnt; i+=2) e[i].v=e[i^1].v=(e[i].v+e[i^1].v)>>1;
53     rep(i,l,r) if (mark[a[i]]) tmp[p++]=a[i];
54     p0=p;
55     rep(i,l,r) if (!mark[a[i]]) tmp[p++]=a[i];
56     rep(i,l,r) a[i]=tmp[i];
57     solve(l,p0-1); solve(p0,r);
58 }
59 
60 int main(){
61     freopen("bzoj2229.in","r",stdin);
62     freopen("bzoj2229.out","w",stdout);
63     for (scanf("%d",&TT); TT--; ){
64         cnt=1; mem(h,0); mem(ans,0x3f);
65         scanf("%d%d",&n,&m);
66         rep(i,1,m) scanf("%d%d%d",&u,&v,&w),add(u,v,w);
67         rep(i,1,n) a[i]=i; solve(1,n);
68         for (scanf("%d",&Q); Q--; ){
69             scanf("%d",&x); tot=0;
70             rep(i,1,n-1) rep(j,i+1,n) if (ans[i][j]<=x) tot++;
71             printf("%d\n",tot);
72         }
73         puts("");
74     }
75     return 0;
76 }

BZOJ4519:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<vector>
 5 #define rep(i,l,r) for (int i=l; i<=r; i++)
 6 #define For(i,x) for (int i=h[x],k; i; i=e[i].nxt)
 7 using namespace std;
 8 
 9 const int N=1010,inf=1000000000;
10 int m,n,u,v,w,S,T,tot,cnt=1,tmp[N],a[N],b[N],d[N],q[N],h[N];
11 struct E{ int to,nxt,v; }e[20010];
12 bool mark[N];
13 
14 void add(int u,int v,int w){
15    e[++cnt]=(E){v,h[u],w}; h[u]=cnt;
16    e[++cnt]=(E){u,h[v],w}; h[v]=cnt;
17 }
18 
19 bool bfs(){
20     memset(d,0,sizeof(d)); q[1]=S; d[S]=1;
21     for (int st=0,ed=1; st!=ed; ){
22         int x=q[++st];
23         For(i,x) if (e[i].v && !d[k=e[i].to])
24             d[k]=d[x]+1,q[++ed]=k;
25     }
26     return d[T];
27 }
28 
29 int dfs(int x,int lim){
30     if (x==T) return lim;
31     int t,c=0;
32     For(i,x) if (d[k=e[i].to]==d[x]+1){
33         t=dfs(k,min(lim-c,e[i].v));
34         e[i].v-=t; e[i^1].v+=t; c+=t;
35         if (c==lim) return lim;
36     }
37     if (!c) d[x]=-1; return c;
38 }
39 
40 int dinic(){ int ans=0; while(bfs()) ans+=dfs(S,inf); return ans; }
41 
42 void get(int x){
43     mark[x]=1;
44     For(i,x) if (!mark[k=e[i].to] && e[i].v) get(k);
45 }
46 
47 void solve(int l,int r){
48     if (l==r) return;
49     S=a[l]; T=a[r]; b[++tot]=dinic();
50     memset(mark,0,sizeof(mark));
51     get(S); int p=l,p0;
52     for (int i=2; i<=cnt; i+=2) e[i].v=e[i^1].v=(e[i].v+e[i^1].v)>>1;
53     rep(i,l,r) if (mark[a[i]]) tmp[p++]=a[i];
54     p0=p;
55     rep(i,l,r) if (!mark[a[i]]) tmp[p++]=a[i];
56     rep(i,l,r) a[i]=tmp[i];
57     solve(l,p0-1); solve(p0,r);
58 }
59 
60 int main(){
61     freopen("bzoj4519.in","r",stdin);
62     freopen("bzoj4519.out","w",stdout);
63     scanf("%d%d",&n,&m);
64     rep(i,1,m) scanf("%d%d%d",&u,&v,&w),add(u,v,w);
65     rep(i,1,n) a[i]=i;
66     solve(1,n); sort(b+1,b+tot+1); tot=unique(b+1,b+tot+1)-b-1;
67     printf("%d\n",tot);
68     return 0;
69 }

最小割分治(最小割樹):BZOJ2229 && BZOJ4519