CF343E Pumping Stations
\(n\) 个点 \(m\) 条边的带权无向图。
你需要构造一个排列，收益为 \(\sum_{i=2}^n \mathrm{mincut}(a_{i-1},a_i)\) 。
\(\mathrm{mincut}(S,T)\) 表示图中 \(S\) 为源点，\(T\) 为汇点的最小割。
求最大的收益，并输出方案。
$2\le n\le 200 $ , $ 1\le m\le 1000$
对无向图建出来最小割树，现在问题转化为了定义树上两点路径权值是路径上的边权最小值，构造一个排列，使得相邻两点的路径权值和最大。
我们考虑对当前的一棵树，取出边权最小的一条边，那么我们一定要只经过这条边一次，于是去掉这条边就变成了两个子问题，分治下去就可以构造出来了。
构造答案这部分实现的好是可以做到 \(O(n\log n\alpha(n))\) 的，而且我写的也是这种。
Code
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
const int N = 200;
const int inf = 1e9;
using namespace std;
struct edges
{
int u,v,w;
}edge[N + 5];
int n,m,t[N + 5],t1[N + 5],t2[N + 5],e1[N + 5],e2[N + 5],e[N + 5],edge_cnt,fa[N + 5],a[N + 5],ans;
namespace F
{
const int N = 3e6;
const long long inf = 2e18;
struct edges
{
int to;
long long cost;
}edge[N * 2 + 5],e[N * 2 + 5];
int nxt[N * 2 + 5],head[N + 5],edge_cnt = 1,dis[N + 5],q[N + 5],cur[N + 5],S,T;
void add_edge(int u,int v,long long w)
{
edge[++edge_cnt] = (edges){v,w};
nxt[edge_cnt] = head[u];
head[u] = edge_cnt;
}
void add(int u,int v,long long w)
{
add_edge(u,v,w);
add_edge(v,u,w);
}
int bfs()
{
for (int i = 1;i <= n;i++)
cur[i] = head[i],dis[i] = 0;
int l = 1,r = 0;
dis[S] = 1;
q[++r] = S;
while (l <= r)
{
int u = q[l++];
for (int i = head[u];i;i = nxt[i])
{
int v = edge[i].to,w = edge[i].cost;
if (w && !dis[v])
{
dis[v] = dis[u] + 1;
q[++r] = v;
}
}
}
return dis[T];
}
long long dfs(int u,long long flow)
{
if (u == T)
return flow;
long long sm = 0;
for (int &i = cur[u];i;i = nxt[i])
{
int v = edge[i].to;
long long w = edge[i].cost;
if (dis[v] == dis[u] + 1 && w)
{
long long res = dfs(v,min(w,flow));
edge[i].cost -= res;
edge[i ^ 1].cost += res;
sm += res;
flow -= res;
if (!flow)
break;
}
}
return sm;
}
void init()
{
for (int i = 2;i <= edge_cnt;i++)
e[i] = edge[i];
}
void clear()
{
for (int i = 2;i <= edge_cnt;i++)
edge[i] = e[i];
}
long long dinic(int s,int t)
{
S = s;T = t;
clear();
long long ans = 0;
while (bfs())
ans += dfs(S,inf);
return ans;
}
}
int find(int x)
{
if (fa[x] == x)
return x;
return fa[x] = find(fa[x]);
}
void build(int l,int r)
{
if (l >= r)
return;
int val = F::dinic(t[l],t[r]),cnt1 = 0,cnt2 = 0;
edge[++edge_cnt] = (edges){t[l],t[r],val};
for (int i = l;i <= r;i++)
if (F::dis[t[i]])
t1[++cnt1] = t[i];
else
t2[++cnt2] = t[i];
for (int i = 1;i <= cnt1;i++)
t[i] = t1[i];
for (int i = 1;i <= cnt2;i++)
t[i + cnt1] = t2[i];
build(1,cnt1);
build(cnt1 + 1,cnt2 + cnt1);
}
void solve(int l,int r,int ql,int qr)
{
if (l > r)
return;
if (l == r)
{
a[l] = t[l];
return;
}
for (int i = l;i <= r;i++)
fa[t[i]] = t[i];
int mi = inf,id = 0,cnt1 = 0,cnt2 = 0,q1 = 0,q2 = 0;
for (int i = ql;i <= qr;i++)
if (edge[e[i]].w < mi)
mi = edge[e[i]].w,id = e[i];
ans += mi;
for (int i = ql;i <= qr;i++)
if (e[i] != id)
fa[find(edge[e[i]].u)] = find(edge[e[i]].v);
int x = find(edge[id].u);
for (int i = l;i <= r;i++)
{
if (find(t[i]) == x)
t1[++cnt1] = t[i];
else
t2[++cnt2] = t[i];
}
for (int i = ql;i <= qr;i++)
if (e[i] != id)
{
if (find(edge[e[i]].u) == x)
e1[++q1] = e[i];
else
e2[++q2] = e[i];
}
for (int i = 1;i <= cnt1;i++)
t[i + l - 1] = t1[i];
for (int i = 1;i <= cnt2;i++)
t[l + i + cnt1 - 1] = t2[i];
for (int i = 1;i <= q1;i++)
e[i + ql - 1] = e1[i];
for (int i = 1;i <= q2;i++)
e[i + q1 + ql - 1] = e2[i];
solve(l,l + cnt1 - 1,ql,ql + q1 - 1);
solve(l + cnt1,r,ql + q1,ql + q1 + q2 - 1);
}
int main()
{
scanf("%d%d",&n,&m);
int u,v,w;
for (int i = 1;i <= m;i++)
{
scanf("%d%d%d",&u,&v,&w);
F::add(u,v,w);
}
F::init();
for (int i = 1;i <= n;i++)
t[i] = i;
build(1,n);
for (int i = 1;i <= n;i++)
t[i] = i;
for (int i = 1;i < n;i++)
e[i] = i;
solve(1,n,1,n - 1);
cout<<ans<<endl;
for (int i = 1;i <= n;i++)
printf("%d ",a[i]);
return 0;
}
