Question chain :

Answer key :

First, take a look at this blog :
Very good , We can have a simple perceptual knowledge of minimal cut trees .

It's very troublesome to find the minimum cut tree , And the number of points in this question is not large ,
So there is no need to construct a minimum cut tree , Just figure out all the ans[i][j]:i->j The minimum cut .
That is, divide and conquer , Find out n-1 A minimal cut ,
And after every minimal cut , Try to update the ans that will do .

Code :

#define MAXN 200
#define MAXM 7000
#define INF 0x3f3f3f3f
#define filein(x) freopen(#x".in","r",stdin);
#define fileout(x) freopen(#x".out","w",stdout);
using namespace std;
struct Edge{
int to[MAXM],cap[MAXM],next[MAXM],head[MAXN],ent;
void Init(){
ent=2; memset(head,0,sizeof(head));
void Adde(int u,int v,int w){
to[ent]=v; cap[ent]=w; next[ent]=head[u]; head[u]=ent++;
to[ent]=u; cap[ent]=0; next[ent]=head[v]; head[v]=ent++;
void reset(){
for(int i=2;i<ent;i+=2){
bool mark[MAXN];
int a[MAXN],cur[MAXN],d[MAXN],ans[MAXN][MAXN];
int N,M,Q;
bool bfs(int S,int T){
static queue<int> q; int u,v;
d[S]=1; q.push(S);
u=q.front(); q.pop();
for(int i=E.head[u];i;[i]){[i];
if(d[v]||!E.cap[i]) continue;
d[v]=d[u]+1; q.push(v);
return d[T];
int dfs(int u,int reflow,const int &T){
if(u==T||!reflow) return reflow;
int flowout=0,f,v;
for(int &i=cur[u];i;[i]){[i];
if(d[v]!=d[u]+1) continue;
flowout+=f; E.cap[i^1]+=f;
reflow-=f; E.cap[i]-=f;
if(!reflow) break;
if(!flowout) d[u]=0;
return flowout;
int Dinic(int S,int T){
int flow=0;
return flow;
void dfs(int u){
for(int i=E.head[u];i;[i])
if(!mark[[i]]&&E.cap[i]) dfs([i]);
void Partition(int l,int r){
static int tmp[MAXN],S,T,cut;
if(l>=r) return;
E.reset(); S=a[l]; T=a[r];
memset(mark,0,sizeof(mark)); dfs(S);
for(int i=1;i<=N;i++)
for(int j=1;j<i;j++) if(mark[a[i]]^mark[a[j]])
int L=l,R=r;
for(int i=l;i<=r;i++)
if(mark[a[i]]) tmp[L++]=a[i];
else tmp[R--]=a[i];
for(int i=l;i<L;i++) a[i]=tmp[i];
for(int i=r;i>R;i--) a[i]=tmp[i];
int main()
filein(mincut); fileout(mincut);
int T; scanf("%d",&T);
for(int i=1;i<=N;i++) a[i]=i;
for(int i=1,u,v,w;i<=M;i++){
E.reset(); Partition(1,N);
for(int k=1,c,num;k<=Q;k++){
scanf("%d",&c); num=0;
for(int i=1;i<=N;i++)
for(int j=1;j<i;j++)
if(ans[i][j]<=c) num++;
} return 0;

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