E - Red and Blue Tree Editorial by en_translator


First, we count how many times the piece passes each edge with algorithm like DFS (Depth-First Search). If it passes Edge \(i\) for \(C_i\) times, then the problem is equivalent to “how many ways are there to separate \(C_1,\ldots,C_{N-1}\) into two groups so that the sum of the former \(R\) and the sum of the latter \(B\) satisfies \(R-B=K\)?”

Moreover, with \(S=C_1+\ldots+C_{N-1}\), it is boiled down to “how many ways are there to choose some elements from \(C_1,\ldots,C_{N-1}\) so that its sum becomes \(\frac{K+S}{2}\)?”

Therefore, this is computed by a DP where

\(DP[i][j]=\) The number of ways to choose some elements from \(C_1,\ldots,C_i\) so that its sum becomes \(j\).

Note that in some cases \(\frac{K+S}{2}\) may not be an integer or falls below \(0\).

The DFS for finding \(C_i\) costs \(O(NM)\) time, and the DP costs \(O(N(NM+|K|))\) time, so the overall time complexity is \(O(N^2M+N|K|)\).

Sample code (C++)

#include<bits/stdc++.h>
using namespace std;
const int mod=998244353;

vector<vector<pair<int,int>>>G;
vector<int>C;

int dfs(int v,int pre,int goal){
	if(v==goal)return 1;
	for(auto [vv,i]:G[v])if(vv!=pre){
		if(dfs(vv,v,goal)){
			C[i]++;
			return 1;
		}
	}
	return 0;
}

int main(){
	int N, M, K;
	cin >> N >> M >> K;
	vector<int>A(M);
	
	for(int i=0;i<M;i++){
		cin >> A[i];
		A[i]--;
	}
	G.resize(N);
	for(int i=0;i<N-1;i++){
		int x,y;
		cin >> x >> y;
		x--,y--;
		G[x].push_back({y,i});
		G[y].push_back({x,i});
	}
	
	C.resize(N-1);
	for(int i=0;i<M-1;i++)dfs(A[i],-1,A[i+1]);
	
	int S=0;
	for(int i=0;i<N-1;i++)S+=C[i];
	if((S+K)%2 || S+K<0){
		cout << 0;
		return 0;
	}

	vector<int>dp(100001);
	dp[0]=1;
	for(int i=0;i<N-1;i++)for(int x=100000;x>=C[i];x--)(dp[x]+=dp[x-C[i]])%=mod;
	cout << dp[(S+K)/2];
}

posted:
last update: