解题思路:从环里面取一条边,两个点a、b,从a点出发,用树形dp找最小点,但此时a与b是必须连接的,所以a必须要有士兵,取dp[a][1],还有一种情况考虑b点有士兵,取min(dp[a][0],dp[a][1]),因为此时b点已经与a点连接的,所以a点士兵可有可无。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double lf;
typedef unsigned long long ull;
typedef pair<int,int>P;
const int inf = 0x7f7f7f7f;
const ll INF = 1e16;
const int N = 1e6+10;
const ll mod = 998244353;
const double PI = acos(-1.0);
const double eps = 1e-4;
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}
inline string readstring(){string str;char s=getchar();while(s==' '||s=='\n'||s=='\r'){s=getchar();}while(s!=' '&&s!='\n'&&s!='\r'){str+=s;s=getchar();}return str;}
int random(int n){return (int)(rand()*rand())%n;}
void writestring(string s){int n = s.size();for(int i = 0;i < n;i++){printf("%c",s[i]);}}
vector<int>G[N];
int dp[N][2];
int a = 0,b = 0,flag = 1;
//dp[u][0]表示u不放士兵
//dp[u][1]表示放士兵
void dfs(int u,int fa){
for(int i = 0;i < G[u].size();i++){
int v = G[u][i];
if(v == fa) continue;
dfs(v,u);
dp[u][0] += dp[v][1];
dp[u][1] += min(dp[v][0],dp[v][1]);
}
dp[u][1]++;
if(u == b && flag == 2){//加入从b点与a点相连接,所以此时这里必须有士兵
dp[u][0] = dp[u][1];
}
}
int main(){
srand((unsigned)time(NULL));
int n = read(),m = read();
for(int i = 1;i <= n+m;i++){
int u = read(),v = read();
if(flag&&u<n&&v<n){//找到一个环的某条边
a = u;b = v;
flag = 0;
continue;
}
G[u].push_back(v);
G[v].push_back(u);
}
int ans = inf;
dfs(a,-1);
ans = min(ans,dp[a][1]);//此时对于a这个点是必须要放士兵的,因为要和b相连接
memset(dp,0,sizeof dp);
flag = 2;
dfs(a,-1);
ans = min(ans,min(dp[a][0],dp[a][1]));
cout<<ans<<endl;
return 0;
}