ACdream1032 Component 树形DP

来源：互联网发布：人工智能安防编辑：程序博客网时间：2024/05/18 18:15

思路：dp[i][j]表示以i为根结点有j个连通节点的最小和, 当进行状态转移时需要利用01背包，节点u下面有多个子节点，每个子节点可以最多可以贡献cnt[v]个节点，cnt[v]表示以v为根结点的树的节点总数。

AC代码

#include <cstdio>#include <cmath>#include <cctype>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> const int maxn = 2e3 + 5;int dp[maxn][maxn], w[maxn], cnt[maxn], p[maxn], ans[maxn];int n;vector<int>G[maxn];//dp[i][j]表示以i为根结点有j个连通节点的最小和 int dfs(int u, int pre) {int sum = 0;for(int i = 0; i < G[u].size(); ++i) {int v = G[u][i];if(v == pre) continue;cnt[v] += dfs(v, u);sum += cnt[v];}//p[i]表示凑足i个节点的最小和,利用了滚动数组 p[0] = 0;for(int i = 1; i <= sum; ++i) p[i] = inf;  for(int i = 0; i < G[u].size(); ++i) {int v = G[u][i];if(v == pre) continue;for(int j = sum; j >= 1; --j) {for(int k = 1; k <= cnt[v]; ++k) {if(j >= k) p[j] = min(p[j], p[j-k] + dp[v][k]);}}}for(int i = 0; i <= sum; ++i) {dp[u][i+1] = p[i] + w[u];ans[i+1] = min(ans[i+1], dp[u][i+1]);}return sum+1; }int main() {while(scanf("%d", &n) == 1) {for(int i = 1; i <= n; ++i) {scanf("%d", &w[i]);G[i].clear();ans[i] = inf;}int u, v;for(int i = 0; i < n-1; ++i) {scanf("%d%d", &u, &v);G[u].push_back(v);G[v].push_back(u);}dfs(1, -1);for(int i = 1; i <= n; ++i) {if(i == n) printf("%d\n", ans[i]);else printf("%d ", ans[i]);}}return 0;}

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