传送门
题目描述
分析
去年写的题,今天看猫猫博客的时候又想起来了这道题,回顾一些细节
f
[
i
]
[
j
]
f[i][j]
f[i][j]表示以
i
i
i为根节点保留
j
j
j个节点所要砍掉的道路数量,然后跑一下树形背包即可
代码
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 200;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int h[N],ne[N],e[N],idx;
int in[N];
int f[N][N];
int sz[N];
int n,p;
void add(int x,int y){
ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}
void dfs(int u){
sz[u] = 1;
for(int i = h[u];~i;i = ne[i]){
int j = e[i];
dfs(j);
sz[u] += sz[j];
for(int k = sz[u];k;k--)
for(int l = 1;l < k;l++)
f[u][k] = min(f[u][k],f[u][k - l] + f[j][l] - 1);
}
}
int main() {
memset(h,-1,sizeof h);
read(n),read(p);
for(int i = 1;i < n;i++){
int x,y;
read(x),read(y);
add(x,y);
in[x]++;
}
memset(f,0x3f,sizeof f);
for(int i = 1;i <= n;i++) f[i][1] = in[i];
dfs(1);
int ans = f[1][p];
for(int i = 2;i <= n;i++) ans = min(ans,f[i][p] + 1);
di(ans);
return 0;
}
/**
* ┏┓ ┏┓+ +
* ┏┛┻━━━┛┻┓ + +
* ┃ ┃
* ┃ ━ ┃ ++ + + +
* ████━████+
* ◥██◤ ◥██◤ +
* ┃ ┻ ┃
* ┃ ┃ + +
* ┗━┓ ┏━┛
* ┃ ┃ + + + +Code is far away from
* ┃ ┃ + bug with the animal protecting
* ┃ ┗━━━┓ 神兽保佑,代码无bug
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛ + + + +
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛+ + + +
*/