The question :

Yes n Nodes ,m side , Undirected acyclic graphs , Find the minimum point cover , And under the same number of points, the maximum number of changes covered twice is guaranteed

analysis :

1. Unified goals , There are two objectives to be optimized , A minimum number of lights a, A maximum number of double covering edges b, One big one small , It should be one ,a And the number of single covering edges c,\( x=Ma+c \) To minimize the goal ,\( M>\Delta c \)

2. Decision analysis , There are only two kinds of lights and no lights , If you don't put the light on, you need the parent node to put the light on , So we need the state of the parent node , set up \( f(i,j) \) For the node i In the parent node, the state is j It's the smallest x value ,j by 0 It means no lights ,j by 1 On behalf of

    sum\{d(k,0)|k by i All child nodes \}+(i Root node ?0:1)&,i No lights \\
    sum\{d(k,1)|k by i All child nodes \}+M+(i Is not a root node and j==0?1:0)&,i\ Light up

The code is as follows

university:China,Xidian University
**If you need to reprint,please indicate the source**
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
int n,m;
int first[];
int Next[],U[];
bool vis[][];
int ans[][];
int cnt;
int M=<<;
void add(int u,int v) // Two way side
Next[cnt]=first[u]; first[u]=cnt; U[cnt++]=v;
Next[cnt]=first[v]; first[v]=cnt; U[cnt++]=u;
int dfs(int i,int j,int f)//i Is the current node ,j by 0 It means the parent node is not lit , by 1 It's for lights ,f by -1 Represents the root node
if(vis[i][j])return ans[i][j];
int k,&Ans=ans[i][j],ans0=;
Ans+=dfs(U[k],,i);// Light up
else Ans++;
ans0+=dfs(U[k],,i); // No lights
return Ans;
int main()
int T;
int i,u,v;
int Ans=;
printf("%d %d %d\n",Ans>>,m-(Ans&(M-)),Ans&(M-));//Ans/M,m-Ans%M,Ans%M

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