題目鏈接

題解

樹形dp

\(f_{i, j}\)表示以\(i\)為根的子樹切出聯(lián)通塊大小為\(j\)的最小答案
顯然\(f[i][1]\)為與\(i\)連的邊數(shù)

設(shè)\(v\)是\(u\)的兒子
那么有\(f[u][i]=f[u][i-j]+f[v][j]-2\)，因?yàn)?span id="ozvdkddzhkzd" class="math inline">\(u->v\)這條邊算了兩次

• 注意\(i\)要從大到小枚舉

Code

#include<bits/stdc++.h> #define LL long long #define RG register using namespace std;inline int gi() {int f = 1, s = 0;char c = getchar();while (c != '-' && (c < '0' || c > '9')) c = getchar();if (c == '-') f = -1, c = getchar();while (c >= '0' && c <= '9') s = s*10+c-'0', c = getchar();return f == 1 ? s : -s; } int n, p; const int N = 200; int f[N][N]; struct node {int to, next; }g[N<<1]; int last[N], gl, siz[N]; inline void add(int x, int y) {g[++gl] = (node) {y, last[x]};last[x] = gl;return ; } #define Min(a, b) (a>b?b:a) inline void dfs(int u, int fa) {siz[u] = 1;f[u][1] = 0;for (int i = last[u]; i; i = g[i].next) {int v = g[i].to;f[u][1]++;if (v == fa) continue;dfs(v, u);siz[u] += siz[v];}for (int i = last[u]; i; i = g[i].next) {int v = g[i].to;if (v == fa) continue;for (register int j = siz[u]; j > 1; j--)for (register int k = 1; k < j; k++)f[u][j] = min(f[u][j], f[v][k]+f[u][j-k]-2);}return ; }int main() {n = gi(), p = gi();for (int i = 1; i < n; i++) {int u = gi(), v = gi();add(u, v); add(v, u);}memset(f, 0x3f, sizeof(f));dfs(1, 0);int ans = f[1][p];for (int i = 2; i <= n; i++)ans = Min(ans, f[i][p]);printf("%d\n", ans);return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/zzy2005/p/10161211.html

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