题目链接
题解
由题，一个环至少由三个点组成，一个点作为根时，可以单独成链，可以与其一个儿子成链，或者与其两个儿子成环，与其一个剩余链长度大于等于2的儿子成环。
那么我们设最小代价
\(f[u][0]\)表示以\(u\)为根全部成环
\(f[u][1]\)表示除\(u\)外全部成环
\(f[u][2]\)表示除\(u\)和一个儿子的一条长度至少为\(1\)的链外全部成环
转移就很容易想
记\(sum = \sum\limits_{(u,v) \in edge} f[v][0]\)
\[f[u][1] = sum\]
\[f[u][2] = min\{\sum\limits_{(u,v) \in edge} min(f[v][1],f[v][2]) + (sum - f[v][0]) \}\]
\[f[u][0] = min\{ \sum\limits_{(u,v) \in edge} \sum\limits_{(u,k) \in edge} min(f[v][1],f[v][2]) + min(f[k][1],f[k][2]) + (sum - f[v][0] - f[k][0]) + 1\}\]
\[f[u][0] = min\{ \sum\limits_{(u,v) \in edge} f[v][2] + (sum - f[v][0]) + 1 \}\]
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 105,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int h[maxn],ne = 2;
struct EDGE{int to,nxt;}ed[maxn << 1];
inline void build(int u,int v){
ed[ne] = (EDGE){v,h[u]}; h[u] = ne++;
ed[ne] = (EDGE){u,h[v]}; h[v] = ne++;
}
int n,fa[maxn],s[maxn];
LL f[maxn][3];
void dfs(int u){
f[u][0] = f[u][2] = INF; f[u][1] = 0; LL sum = 0;
Redge(u) if ((to = ed[k].to) != fa[u]){
fa[to] = u; dfs(to);
f[u][1] += f[to][0];
sum += f[to][0];
}
int cnt = 0; LL tmp;
Redge(u) if ((to = ed[k].to) != fa[u]){
s[++cnt] = to;
f[u][2] = min(f[u][2],min(f[to][1],f[to][2]) + sum - f[to][0]);
f[u][0] = min(f[u][0],f[to][2] + sum - f[to][0] + 1);
}
REP(i,cnt) REP(j,cnt) if (i != j){
tmp = min(f[s[i]][1],f[s[i]][2]) + min(f[s[j]][1],f[s[j]][2]) + sum - f[s[i]][0] - f[s[j]][0];
f[u][0] = min(f[u][0],tmp + 1);
}
}
int main(){
while (~scanf("%d",&n) && n){
ne = 2; memset(h,0,sizeof(h));
for (int i = 1; i < n; i++) build(read(),read());
dfs(1);
if (f[1][0] >= INF) puts("-1");
else printf("%lld\n",f[1][0]);
}
return 0;
}