bzoj 1907 树的路径覆盖 [贪心] [树的最小路径覆盖]

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这道题贪心DP都可以做，但是贪心的正确性容易证明，因为子树合并一定不会劣与合并，那么我就贪心了。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#ifdef WIN32
#define AUTO "%I64d"
#else
#define AUTO "%lld"
#endif
using namespace std;
#define smax(x,tmp) x=max((x),(tmp))
#define smin(x,tmp) x=min((x),(tmp))
#define maxx(x1,x2,x3) max(max(x1,x2),x3)
#define minn(x1,x2,x3) min(min(x1,x2),x3)
const int INF=0x3f3f3f3f;
const int maxn = 10005;
struct Edge
{
    int to,next;
}edge[maxn<<1];
int head[maxn];
int maxedge;
inline void addedge(int u,int v)
{
    edge[++maxedge] = (Edge) { v,head[u] };
    head[u] = maxedge;
    edge[++maxedge] = (Edge) { u,head[v] };
    head[v] = maxedge;
}
int n;
inline void init()
{
    scanf("%d",&n);
    memset(head,-1,sizeof(head));
    maxedge=-1;
    for(int i=1;iint u,v;
        scanf("%d%d",&u,&v);
        addedge(u,v);
    }
}
int f[maxn];
bool flag[maxn];
void dfs(int u,int father)
{
    int cnt=0;
    f[u]=1;
    for(int i=head[u];~i;i=edge[i].next)
    {
        int v=edge[i].to;
        if(v==father) continue;
        dfs(v,u);
        f[u]+=f[v];
        if(!flag[v]) cnt++;
    }
    if(cnt>=2) f[u]-=2,flag[u]=true;
    else if(cnt==1) f[u]-=1;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("road.in","r",stdin);
    freopen("road.out","w",stdout);
#endif
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        init();
        memset(flag,0,sizeof(flag));
        dfs((n+1)>>1,-1);
        printf("%d\n",f[(n+1)>>1]);
    }
    return 0;
}