2021 ICPC Southeastern Europe Regional Contest C. Werewolves

Posted 2022-03-05 TURNINING

传送门
思路：考虑对每个颜色分别进行计算。如果该节点的颜色为当前计算的颜色，令 $va{l}_i=1$,否则$va{l}_i=-1$。显然要使子树颜色相同的数量严格大于一半等价于这个子树和大于等于1。

求一个子树中和大于等于1的方案数，我们可以树上背包。

定义：$d{p}_1[u][s]$： 以u为根的子树，和为$s(s\ge 1)$的方案数。$d{p}_2[u][s]$：和为$-s(s\ge 1)$的方案数。$d{p}_3[u]$和为$0$的方案数

状态转移：

• $i+j<=m$: $d{p}_1[u][i+j]=d{p}_1[u][i]\ast d{p}_1[v][j]$

• $i>j$ : $d{p}_1[u][i-j]=d{p}_1[u][i]\ast d{p}_2[v][j]$

• $i<j$: $d{p}_1[u][j-i]=d{p}_2[u][i]\ast d{p}_1[v][j]$

$res={\sum }_u{\sum }_{i=1}^{m}d{p}_1[u][i]$

$d{p}_2$的转移类似，$d{p}_3$的转移也十分好推。

每次计算的复杂度:$O(n\ast m)$，总的复杂度:$O(n\ast \sum m)$ = $O(n^2)$，$m$为每个颜色的数量。

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

const int MAXN = 3000 + 10;
const int mod = 998244353;

vector<int> g[MAXN];
int n, m;
int val[MAXN], c[MAXN], vis[MAXN];

ll dp1[MAXN][MAXN], dp2[MAXN][MAXN], dp3[MAXN];
ll tmp1[MAXN][MAXN], tmp2[MAXN][MAXN], tmp3[MAXN];
ll res;

int dfs(int u, int fa) 
    int p = 1;
    if(val[u] == 1) dp1[u][1] = 1;
    else dp2[u][1] = 1;
    for(auto v : g[u]) 
        if(v == fa) continue;
        int sz = dfs(v, u);
        for(int i = 0; i <= min(p, m); i++) 
            tmp1[u][i] = dp1[u][i];
            tmp2[u][i] = dp2[u][i];
            tmp3[u] = dp3[u];
        
        dp3[u] = (dp3[u] + tmp3[u] * dp3[v]) % mod;
        for(int j = 1; j <= sz && j <= m; j++) 
            dp1[u][j] = (dp1[u][j] + tmp3[u] * dp1[v][j]) % mod;
            dp2[u][j] = (dp2[u][j] + tmp3[u] * dp2[v][j]) % mod;
        
        for(int i = 1; i <= min(p, m); i++) 
            dp1[u][i] = (dp1[u][i] + tmp1[u][i] * dp3[v]) % mod;
            dp2[u][i] = (dp2[u][i] + tmp2[u][i] * dp3[v]) % mod;
            for(int j = 1; j <= sz && j <= m; j++) 
                if(i+j <= m) 
                    dp1[u][i+j] = (dp1[u][i+j] + tmp1[u][i] * dp1[v][j]) % mod;
                    dp2[u][i+j] = (dp2[u][i+j] + tmp2[u][i] * dp2[v][j]) % mod;
                
                if(i-j >= 1) 
                    dp1[u][i-j] = (dp1[u][i-j] + tmp1[u][i] * dp2[v][j]) % mod;
                    dp2[u][i-j] = (dp2[u][i-j] + tmp2[u][i] * dp1[v][j]) % mod;
                
                if(j-i >= 1) 
                    dp1[u][j-i] = (dp1[u][j-i] + tmp2[u][i] * dp1[v][j]) % mod;
                    dp2[u][j-i] = (dp2[u][j-i] + tmp1[u][i] * dp2[v][j]) % mod;
                
                if(i == j) 
                    dp3[u] = (dp3[u] + tmp1[u][i] * dp2[v][j] + tmp2[u][i] * dp1[v][j]) % mod;
                
            
        
        p += sz;
    
    for(int i = 1; i <= min(p, m); i++) 
        res = (res + dp1[u][i]) % mod;
    
    return p;


void solve() 
    cin >> n;
    for(int i = 1; i <= n; i++) 
        cin >> c[i];
    
    for(int i = 1; i < n; i++) 
        int u, v; cin >> v >> u;
        g[u].push_back(v); 
        g[v].push_back(u);
    
    for(int i = 1; i <= n; i++) 
        if(vis[c[i]]) continue;
        vis[c[i]] = 1; m = 0;     
        for(int j = 1; j <= n; j++) 
            val[j] = (c[j] == c[i] ? 1 : -1);
            if(c[j] == c[i]) m++;
        
        for(int j = 1; j <= n; j++) 
            for2021 ICPC Southeastern Europe Regional Contest C. Werewolves

 2021 ICPC Southeastern Europe Regional Contest C. Werewolves
 2021 ICPC Southeastern Europe Regional Contest C. Werewolves
 2020-2021 ICPC Southeastern European Regional E. Divisible by 3(前缀和,优化暴力)
 ICPC Southeastern Europe Contest 2019
 2016-2017 ACM-ICPC Southeastern European Regional Programming Contest (SEERC 2016)