Codeforces 581F Zublicanes and Mumocrates 树形dp

Zublicanes and Mumocrates

dp[ i ][ j ][ k ] 表示 以 i 为根的子树， 占领 i 的 是 j 并且第一个人占了 i 子树的 k 个叶子节点的最小值。

然后随便d 1 d 就好了。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 5000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, leaf;
int dp[N][2][N];
int tmp[2][N];
int sum[N];
bool isleaf[N];
vector<int> G[N];

void dfs(int u, int fa) {
    if(isleaf[u]) {
        dp[u][0][1] = dp[u][1][0] = 0;
        sum[u] = 1;
        return;
    }
    dp[u][1][0] = dp[u][0][0] = 0;
    for(auto& v : G[u]) {
        if(v == fa) continue;
        dfs(v, u);
        memset(tmp, inf, sizeof(tmp));
        for(int i = 0; i < 2; i++) {
            for(int j = 0; j < 2; j++) {
                for(int c1 = 0; c1 <= sum[u]; c1++) {
                    for(int c2 = 0; c2 <= sum[v]; c2++) {
                        chkmin(tmp[i][c1 + c2], dp[u][i][c1] + dp[v][j][c2] + (i != j));
                    }
                }
            }
        }
        sum[u] += sum[v];
        memcpy(dp[u], tmp, sizeof(dp[u]));
    }
}

int main() {
    memset(dp, inf, sizeof(dp));
    scanf("%d", &n);
    for(int i = 1; i < n; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    if(n == 2) return puts("1"), 0;
    int root = -1;
    for(int i = 1; i <= n; i++) {
        if(SZ(G[i]) != 1) root = i;
        else leaf++, isleaf[i] = true;
    }
    dfs(root, 0);
    printf("%d
", min(dp[root][0][leaf >> 1], dp[root][1][leaf >> 1]));
    return 0;
}

/*
*/