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CQOI2016 不同的最小割
Date: Thu Mar 14 20:36:23 CST 2019 In Category: NoCategory

Description:

给一个无向图,求所有点对最小割的互不相同的个数。
$1\leqslant n\leqslant 850$

Solution:

最小割树模板题。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<set>
#include<queue>
#include<cctype>
#include<cstring>
using namespace std;
int n,m;
#define MAXN 1000
#define MAXM 10000
struct edges
{
int u,v,w;
}es[MAXM];
set<int> s;
namespace DINIC
{
struct edge
{
int to,nxt,f;
}e[MAXM << 2];
int edgenum = 0;
int lin[MAXN] = {0};
void add(int a,int b,int f)
{
e[edgenum] = (edge){b,lin[a],f};lin[a] = edgenum++;
e[edgenum] = (edge){a,lin[b],f};lin[b] = edgenum++;
return;
}
void init()
{
edgenum = 0;
memset(lin,-1,sizeof(lin));
return;
}
int s,t;
int ch[MAXN];
bool BFS()
{
memset(ch,-1,sizeof(ch));ch[s] = 0;
queue<int> q;q.push(s);
while(!q.empty())
{
int k = q.front();q.pop();
for(int i = lin[k];i != -1;i = e[i].nxt)
{
if(ch[e[i].to] == -1 && e[i].f)
{
ch[e[i].to] = ch[k] + 1;
q.push(e[i].to);
}
}
}
return (ch[t] != -1);
}
int flow(int k,int f)
{
if(k == t)return f;
int r = 0;
for(int i = lin[k];i != -1 && f > r;i = e[i].nxt)
{
if(ch[e[i].to] == ch[k] + 1 && e[i].f)
{
int l = flow(e[i].to,min(e[i].f,f - r));
e[i].f -= l;r += l;e[i ^ 1].f += l;
}
}
if(r == 0)ch[k] = -1;
return r;
}
#define INF 0x3f3f3f3f
int dinic()
{
int ans = 0,r;
while(BFS())while(r = flow(s,INF))ans += r;
return ans;
}
int tag[MAXN];
void mark(int k)
{
tag[k] = 1;
for(int i = lin[k];i != 0;i = e[i].nxt)
{
if(tag[e[i].to] || e[i].f == 0)continue;
mark(e[i].to);
}
return;
}
}
int p[MAXN],p_[MAXN];
void solve(int l,int r)
{
if(l == r)return;
for(int i = 1;i <= n;++i)DINIC::tag[i] = 0;
DINIC::s = p[l];DINIC::t = p[r];
DINIC::init();
for(int i = 1;i <= m;++i)
{
DINIC::add(es[i].u,es[i].v,es[i].w);
DINIC::add(es[i].v,es[i].u,es[i].w);
}
int val = DINIC::dinic();
s.insert(val);
DINIC::mark(DINIC::s);
int pl = l,pr = r;
for(int i = l;i <= r;++i)
{
if(DINIC::tag[p[i]])p_[pl++] = p[i],DINIC::tag[p[i]] = 0;
else p_[pr--] = p[i];
}
for(int i = l;i <= r;++i)p[i] = p_[i];
solve(l,pl - 1);solve(pr + 1,r);
return;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i = 1;i <= m;++i)scanf("%d%d%d",&es[i].u,&es[i].v,&es[i].w);
for(int i = 1;i <= n;++i)p[i] = i;
solve(1,n);
cout << s.size() << endl;
return 0;
}
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