题解

这显然是一道题拆成两道

然后我胡乱分析了一波，决定第一题就用点度贪心（反正散播的能量肯定能被使用），然后过了

第二题开始mengbier
设\(f_u\)表示第u个点在父亲发动之后才发动的最小价值
\(g_u\)表示第u个点在父亲发动之前发动最小价值

转移的时候
\(son_f\)表示在父亲发动之后才发动的儿子
\(son_g\)表示在儿子发动之后才发动的父亲
\(f_u = \sum_{v \in son_f} f_v + \sum_{t \in son_g} g_t + d_u - c_{fa} - \sum_{t \in son_g} c_t\)
\(g_u = \sum_{v \in son_f} f_v + \sum_{t \in son_g} g_t + d_u - \sum_{t \in son_g} c_t\)
只需要用一个背包\(h[i][j]\)求出来选到第i个儿子能给父亲的额外能量为j的最小代价
分组背包，f和g只能选一个且必须选，转移比较显然

代码

#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define MAXN 100005
#define pb push_back
#define mp make_pair
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 + c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) out(x / 10);putchar('0' + x % 10);
}
struct node {int to,next;
}E[MAXN * 2];
int head[MAXN],sumE,N,C[MAXN],D[MAXN];
void add(int u,int v) {E[++sumE].to = v;E[sumE].next = head[u];head[u] = sumE;
}
void Init() {read(N);for(int i = 1 ; i <= N ; ++i) read(D[i]);for(int i = 1 ; i <= N ; ++i) read(C[i]);int u,v;for(int i = 1 ; i < N ; ++i) {read(u);read(v);add(u,v);add(v,u);}
}
namespace task1 {int ind[MAXN];set<pii > S;void spread(int u) {D[u] = 0;for(int i = head[u] ; i ; i = E[i].next) {int v = E[i].to;if(D[v]) {--ind[v];if(C[u]) --D[v];if(!D[v]) spread(v);else if(C[v]) S.insert(mp(ind[v],v));}}}void Solve() {for(int u = 1 ; u <= N ; ++u) {for(int i = head[u] ; i ; i = E[i].next) {++ind[u];}}for(int u = 1 ; u <= N ; ++u) {if(C[u] == 1) S.insert(mp(ind[u],u));}int ans = 0;while(S.size()) {pii p = *S.begin();S.erase(S.begin());if(p.fi != ind[p.se] || !D[p.se]) continue;ans += D[p.se];D[p.se] = 0;spread(p.se);}for(int u = 1 ; u <= N ; ++u) ans += D[u];out(ans);enter;}
}
namespace task2 {int f[2005],g[2005],h[10005];void dfs(int u,int fa) {for(int i = head[u] ; i ; i = E[i].next) {int v = E[i].to;if(v != fa) dfs(v,u);}int siz = 0;h[0] = 0;for(int i = head[u] ; i ; i = E[i].next) {int v = E[i].to;if(v != fa) {for(int j = siz + 1; j <= siz + C[v] ; ++j) h[j] = 1000000000;siz += C[v];for(int k = siz ; k >= 0 ; --k) {if(k >= C[v]) h[k] = min(h[k - C[v]] + g[v],h[k] + f[v]);else h[k] = h[k] + f[v];}}}f[u] = g[u] = 1000000000;for(int i = 0 ; i <= siz ; ++i) {f[u] = min(f[u],h[i] + max(D[u] - i - C[fa],0));g[u] = min(g[u],h[i] + max(D[u] - i,0));}}void Solve() {dfs(1,0);out(min(f[1],g[1]));enter;}
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifInit();int maxc = 0;for(int i = 1 ; i <= N ; ++i) maxc = max(maxc,C[i]);if(maxc <= 1) task1::Solve();else task2::Solve();return 0;
}