KTU Programming Camp (Day 2) Problem I. Lazy mobile users - 树形dp

• 题意
  
  • 题目链接
  • 给一个有n个点的树，从1出发，每个点最多访问k次，求最多可以访问多少个点。
• 题解
  
  • 设 F(u) 为在满足每个点最多访问k次的条件下，从u这点出发并回到u点可以访问点数的最大值，G(u) 同样的条件下，从u这点出发但不返回u的最大值。
  • F(u)=∑F(v) v表示u的孩子中最大的k−1 个。
  • G(u)=G(i)+∑F(v) 通过枚举i计算。
• 代码

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <bitset>//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define ll long long#define SZ(x) ((int)(x).size()) #define ALL(v) (v).begin(), (v).end()#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)#define reveach(i, v) for (__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++ i) #define REP(i,a,n) for ( int i=a; i<int(n); i++ )#define FOR(i,a,n) for ( int i=n-1; i>= int(a);i-- )#define lson rt<<1, L, m#define rson rt<<1|1, m, Rtypedef pair<int, int> pii;typedef pair<ll, ll> pll;#define mp(x, y) make_pair(x, y)#define pb(x) push_back(x)#define fi first#define se second#define CLR(a, b) memset(a, b, sizeof(a))#define Max(a, b) a = max(a, b)#define Min(a, b) b = min(a, b)const int maxn = 1e5 + 7;int n, k;vector<int> g[maxn];int F[maxn], G[maxn];const bool cmp(const pii& a, const pii& b){    return a.fi > b.fi;}void dfs(int u, int pre){    F[u] = G[u] = 1;    vector<pii> t;    for(int v : g[u]){        dfs(v, u);        t.pb(mp(F[v], G[v]));    }    sort(t.begin(), t.end(), cmp);    //if(u == 1) REP(i, 0, t.size()) printf("%d %d\n", t[i].fi, t[i].se);    REP(i, 0, min((int)t.size(), k - 1)){        F[u] += t[i].fi;    }    REP(i, 0, t.size()){        int sum = 1 + t[i].se;        int cnt = 0;        REP(j, 0, t.size()){            if(cnt == k - 1) break;            if(j != i) sum += t[j].fi, cnt ++;        }        Max(G[u], sum);    }}int solve(){    dfs(1, 0);    return G[1];}int main(){#ifdef ac    freopen("in.txt","r",stdin);#endif    //freopen("out.txt","w",stdout);    scanf("%d%d", &n, &k);    REP(u, 1, n + 1){        int m;        scanf("%d", &m);        REP(j, 0, m){            int v;            scanf("%d", &v);            g[u].pb(v);        }    }    printf("%d\n", solve());    return 0;}