bzoj
luogu

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;
}

原文地址：https://www.cnblogs.com/zhoushuyu/p/8613492.html