sol
最小割树请转一道很相似完全一模一样的题
所以跑出所有点对之间的最小割然后暴力统计答案即可。
code
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
int gi()
{
int x=0,w=1;char ch=getchar();
while ((ch<'0'||ch>'9')&&ch!='0') ch=getchar();
if (ch=='-') w=0,ch=getchar();
while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
return w?x:-x;
}
const int N = 200;
const int inf = 1e9;
struct edge{int to,nxt,w;}a[N<<6];
int T,n,m,qq,head[N],cnt,dep[N],cur[N],q[N],tmp[N],cut[N][N];
queue<int>Q;
void link(int u,int v,int w)
{
a[++cnt]=(edge){v,head[u],w};
head[u]=cnt;
a[++cnt]=(edge){u,head[v],w};
head[v]=cnt;
}
bool bfs(int s,int t)
{
memset(dep,0,sizeof(dep));
dep[s]=1;Q.push(s);
while (!Q.empty())
{
int u=Q.front();Q.pop();
for (int e=head[u];e;e=a[e].nxt)
if (a[e].w&&!dep[a[e].to])
dep[a[e].to]=dep[u]+1,Q.push(a[e].to);
}
return dep[t];
}
int dfs(int u,int f,int t)
{
if (u==t) return f;
for (int &e=cur[u];e;e=a[e].nxt)
if (a[e].w&&dep[a[e].to]==dep[u]+1)
{
int tmp=dfs(a[e].to,min(a[e].w,f),t);
if (tmp) {a[e].w-=tmp;a[e^1].w+=tmp;return tmp;}
}
return 0;
}
int Dinic(int s,int t)
{
int res=0;
while (bfs(s,t))
{
for (int i=1;i<=n;++i) cur[i]=head[i];
while (int tmp=dfs(s,inf,t)) res+=tmp;
}
return res;
}
void solve(int L,int R)
{
if (L==R) return;
for (int i=3;i<=cnt;i+=2)
a[i].w=a[i^1].w=(a[i].w+a[i^1].w)>>1;
int res=Dinic(q[L],q[R]);
for (int i=1;i<=n;++i)
if (dep[i])
for (int j=1;j<=n;++j)
if (!dep[j])
cut[i][j]=cut[j][i]=min(cut[i][j],res);
int l=L-1,r=R+1;
for (int i=L;i<=R;++i)
if (dep[q[i]]) tmp[++l]=q[i];
else tmp[--r]=q[i];
for (int i=L;i<=R;++i) q[i]=tmp[i];
solve(L,l);solve(r,R);
}
int main()
{
T=gi();
while (T--)
{
n=gi();m=gi();
memset(head,0,sizeof(head));cnt=1;
for (int i=1;i<=m;++i)
{
int u=gi(),v=gi(),w=gi();
link(u,v,w);
}
for (int i=1;i<=n;++i)
for (int j=1;j<=n;++j)
cut[i][j]=inf;
for (int i=1;i<=n;++i) q[i]=i;
solve(1,n);qq=gi();
while (qq--)
{
int val=gi(),ans=0;
for (int i=1;i<n;++i)
for (int j=i+1;j<=n;++j)
ans+=(cut[i][j]<=val);
printf("%d\n",ans);
}
puts("");
}
return 0;
}