题目
传送门
\(d_i\) 表示初始使用的能量，如果你处理了这个点，就可以把相邻的点所需的能量减去 \(c_i\)
思路
前 \(50\%\) 的数据为 \(\max\{c_i\}=1,n\le10^5\)
因为最大的 \(c_i\) 只有 \(1\)，我们可以考虑贪心
首先，我们肯定先处理 \(c_i=1\) 的点，我们要考虑处理的顺序，不管顺序怎么变，迟早会把该节省的节省掉
int sum = 0;
for (int i = 1; i <= n; ++i)
{if (c[i] == 0)continue;vis[i] = 1;for (int v : e[i])if (!vis[v]) --d[v];
}
for (int i = 1; i <= n; ++i)sum += max(d[i], 0);
后 \(50\%\) 的数据，我们就要考虑使用树形dp (就是个背包)
定义
\(f[u][0/1]\) 表示当前节点是否在父节点之前处理完
\(g[i][0/1]\) 表示节省 \(i\) 元的最小花费
对于 \(g\) 的详细操作
int sz = c[fa];
for (int i = 0; i <= sz; ++i) g[i][0] = g[i][1] = inf;
g[0][1] = g[c[fa]][0] = 0;
for (int v : e[u])
{if (v == fa) continue;for (int i = sz + 1; i <= sz + c[v]; ++i) g[i][0] = g[i][1] = inf;sz += c[v];for (int i = sz; i >= 0; --i){if (i < c[v]){g[i][0] += f[v][0];g[i][1] += f[v][0];}else{g[i][0] = min(f[v][0] + g[i][0], f[v][1] + g[i - c[v]][0]);g[i][1] = min(f[v][0] + g[i][1], f[v][1] + g[i - c[v]][1]);}}
}
得到\(g\)之后就好处理了，我们可写出如下转移
\(\displaystyle f[u][0] = \min_{i=1}^{sz}\{g[i][0]+\max\{0,d[u]-i\}\}\)
\(\displaystyle f[u][1] = \min_{i=1}^{sz}\{g[i][1]+\max\{0,d[u]-i\}\}\)
代码
#include <bits/stdc++.h> #define ll long long
#define PII pair<int, int>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7, N = 1e5 + 5;
int n, d[N], c[N];
vector<int> e[N];
namespace pts1
{ int f[N][2], g[N][2]; void dfs(int u, int fa) { for (int v: e[u]) { if (v == fa) continue; dfs(v, u); } int sz = c[fa]; for (int i = 0; i <= sz; ++i) g[i][0] = g[i][1] = inf; g[0][1] = g[c[fa]][0] = 0; for (int v: e[u]) { if (v == fa) continue; for (int i = sz + 1; i <= sz + c[v]; ++i) g[i][0] = g[i][1] = inf; sz += c[v]; for (int i = sz; i >= 0; --i) { if (i < c[v]) { g[i][0] += f[v][0]; g[i][1] += f[v][0]; } else { g[i][0] = min(f[v][0] + g[i][0], f[v][1] + g[i - c[v]][0]); g[i][1] = min(f[v][0] + g[i][1], f[v][1] + g[i - c[v]][1]); } } } f[u][0] = g[0][0] + d[u], f[u][1] = g[0][1] + d[u]; for (int i = 1; i <= sz; ++i) { f[u][0] = min(f[u][0], g[i][0] + max(0, d[u] - i)); f[u][1] = min(f[u][1], g[i][1] + max(0, d[u] - i)); } } void solve() { dfs(1, 0); printf("%d", f[1][1]); }
}; namespace pts2
{ int vis[N]; void solve() { int sum = 0; for (int i = 1; i <= n; ++i) { if (c[i] == 0) continue; vis[i] = 1; for (int v: e[i]) if (!vis[v]) --d[v]; } for (int i = 1; i <= n; ++i) sum += max(d[i], 0); printf("%d", sum); }
}; signed main()
{ int maxx = 0; scanf("%d", &n); for (int i = 1; i <= n; ++i) scanf("%d", &d[i]); for (int i = 1; i <= n; ++i) scanf("%d", &c[i]), maxx = max(maxx, c[i]); for (int i = 1, u, v; i < n; ++i) { scanf("%d%d", &u, &v); e[u].push_back(v); e[v].push_back(u); } if (maxx <= 1) pts2::solve(); else pts1::solve(); return 0;
}
