卧薪尝胆,厚积薄发。
JSOI2013 旅行时的困惑
Date: Wed Mar 20 16:18:40 CST 2019 In Category: NoCategory

Description:

给一棵树,初始时每一条边有方向,要求加一些有方向的链使得两两可达。
$1\leqslant n\leqslant 10^5$

Solution1:

如果链能反向覆盖所有边就合法,于是跑最小流即可。

Solution2:

树形贪心,每次把尽可能多的链在子树内合并,否则向上传递还没有匹配的链的个数和种类。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cctype>
#include<cstring>
using namespace std;
int n;
#define MAXN 100010
inline int rd()
{
register int res = 0,f = 1;register char c = getchar();
while(!isdigit(c)){if(c == '-')f = -1;c = getchar();}
while(isdigit(c))res = (res << 1) + (res << 3) + c - '0',c = getchar();
return res * f;
}
struct edge
{
int to,nxt,f;
}e[MAXN * 12];
int edgenum = 0;
int lin[MAXN] = {0};
void add(int a,int b,int f)
{//cout << a << " " << b << " " << f << endl;
e[edgenum] = (edge){b,lin[a],f};lin[a] = edgenum++;
e[edgenum] = (edge){a,lin[b],0};lin[b] = edgenum++;
return;
}
int ind[MAXN];
#define INF 0x3f3f3f3f
int s,t,ss,tt;
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 cur[MAXN];
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;
}
lin[k] = i;
}
if(r == 0)ch[k] = -1;
return r;
}
#define INF 0x3f3f3f3f
int dinic()
{
int ans = 0,r;
memcpy(cur,lin,sizeof(lin));
while(BFS())
{
ans += flow(s,INF);
memcpy(lin,cur,sizeof(lin));
}//cout << ans << endl;
return ans;
}
int main()
{
memset(lin,-1,sizeof(lin));
scanf("%d",&n);
int a,b;
for(int i = 1;i < n;++i)
{
a = rd();b = rd();
++a;++b;
add(b,a,INF);
++ind[a];--ind[b];
}
ss = n + 1;tt = n + 2;s = n + 3;t = n + 4;
for(int i = 1;i <= n;++i)add(ss,i,INF),add(i,tt,INF);
for(int i = 1;i <= n;++i)
{
if(ind[i] > 0)add(s,i,ind[i]);
if(ind[i] < 0)add(i,t,-ind[i]);
}
dinic();
add(tt,ss,INF);
cout << dinic() << endl;
return 0;
}
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