CF 219D Choosing Capital for Treeland 树形DP 好题

  1 #include<cstdio>
  2 #include<cstring>
  3 
  4 using namespace std;
  5 
  6 const int maxn=200000+5;
  7 
  8 struct Edge
  9 {
 10     int to,next;
 11     bool flag;
 12 };
 13 Edge edge[maxn<<1];
 14 int head[maxn];
 15 int tot;
 16 int ans[maxn];
 17 int dp[maxn];
 18 int print[maxn];
 19 
 20 void init()
 21 {
 22     memset(head,-1,sizeof head);
 23     tot=0;
 24     memset(dp,0,sizeof dp);
 25 }
 26 
 27 void addedge(int u,int v,bool flag)
 28 {
 29     edge[tot].to=v;
 30     edge[tot].flag=flag;
 31     edge[tot].next=head[u];
 32     head[u]=tot++;
 33 }
 34 
 35 void dfs0(int ,int );
 36 void dfs1(int ,int );
 37 
 38 int main()
 39 {
 40     int n;
 41     init();
 42     bool cnt=true;
 43     scanf("%d",&n);
 44     for(int i=1;i<n;i++)
 45     {
 46         int u,v;
 47         scanf("%d %d",&u,&v);
 48         addedge(u,v,cnt);
 49         addedge(v,u,!cnt);
 50     }
 51     dfs0(1,-1);
 52     ans[1]=dp[1];
 53     dfs1(1,-1);
 54 
 55     int min=ans[1];
 56     for(int i=2;i<=n;i++)
 57     {
 58         if(ans[i]<min)
 59             min=ans[i];
 60     }
 61     tot=1;
 62     for(int i=1;i<=n;i++)
 63     {
 64         if(ans[i]==min)
 65         {
 66             print[tot++]=i;
 67         }
 68     }
 69 
 70     printf("%d\n",min);
 71     for(int i=1;i<tot-1;i++)
 72     {
 73         printf("%d ",print[i]);
 74     }
 75     printf("%d\n",print[tot-1]);
 76 
 77     return 0;
 78 }
 79 
 80 void dfs0(int u,int pre)
 81 {
 82     for(int i=head[u];~i;i=edge[i].next)
 83     {
 84         int v=edge[i].to;
 85         bool flag=edge[i].flag;
 86         if(v==pre)
 87             continue;
 88         dfs0(v,u);
 89         if(flag)
 90             dp[u]+=dp[v];
 91         else
 92             dp[u]+=(dp[v]+1);
 93     }
 94     return ;
 95 }
 96 
 97 void dfs1(int u,int pre)
 98 {
 99     for(int i=head[u];~i;i=edge[i].next)
100     {
101         int v=edge[i].to;
102         int flag=edge[i].flag;
103         if(v==pre)
104             continue;
105         if(flag)
106             ans[v]=ans[u]+1;
107         else
108             ans[v]=ans[u]-1;
109         dfs1(v,u);
110     }
111     return ;
112 }