wannaly挑战赛1 A Treepath（树形dp/dfs

题目连接：

题意：

直接无脑树形dp即可 dp[u][0]表示u的子树内的节点到u一共有几个奇数路径 dp[u][1]表示状态u的子树内到u一共有几个偶数路径
或者直接dfs, 将间隔的点染成黑白色,那么相隔的点 连起来 一定就是偶数路径了

树形dp

//    Created by Yishui
//    Time on 2018/10/
//    E-mail: Yishui_wyb@outlook
 
/*---------------------------------*/
 
#include 
using namespace std;
 
#define cpp_io() {ios::sync_with_stdio(false); cin.tie(NULL);}
#define rep(i,a,n) for (int i=a;i
#define repp(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define CLR(a,b) memset(a,(b),sizeof(a))
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ls o<<1
#define rs o<<1|1
 
 
typedef long long ll;
typedef vector<int> VI;
const int MAXN = (int)2e5+10;
const int INF = 0x3f3f3f3f;
const int mod = (int)1e9+7;
 
int dp[MAXN][2];
VI E[MAXN];
int a, b;
int cnt[MAXN];
 
void dfs(int u,int fa) {
    cnt[u]=cnt[fa]+1;
    if(cnt[u]&1) a++; else b++;
    rep(i,0,SZ(E[u])) {
        int v=E[u][i];
        if(v==fa) continue;
        dfs(v,u);
    }
}
 
int main() {
    cpp_io();
    int n; cin>>n;
    rep(i,1,n) {
        int u,v; cin>>u>>v;
        E[u].pb(v);
        E[v].pb(u);
    }
    dfs(1,0);
    cout<<1LL*a*(a-1)/2+1LL*b*(b-1)/2<<"\n";
    return 0;
}

DFS

//    Created by Yishui
//    Time on 2018/10/
//    E-mail: Yishui_wyb@outlook
 
/*---------------------------------*/
 
#include 
using namespace std;
 
#define cpp_io() {ios::sync_with_stdio(false); cin.tie(NULL);}
#define rep(i,a,n) for (int i=a;i
#define repp(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define CLR(a,b) memset(a,(b),sizeof(a))
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ls o<<1
#define rs o<<1|1
 
 
typedef long long ll;
typedef vector<int> VI;
const int MAXN = (int)2e5+10;
const int INF = 0x3f3f3f3f;
const int mod = (int)1e9+7;
 
int dp[MAXN][2];
VI E[MAXN];
int a, b;
int cnt[MAXN];
 
void dfs(int u,int fa) {
    cnt[u]=cnt[fa]+1;
    if(cnt[u]&1) a++; else b++;
    rep(i,0,SZ(E[u])) {
        int v=E[u][i];
        if(v==fa) continue;
        dfs(v,u);
    }
}
 
int main() {
    cpp_io();
    int n; cin>>n;
    rep(i,1,n) {
        int u,v; cin>>u>>v;
        E[u].pb(v);
        E[v].pb(u);
    }
    dfs(1,0);
    cout<<1LL*a*(a-1)/2+1LL*b*(b-1)/2<<"\n";
    return 0;
}