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ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#710028 | #6623. Perfect Matchings | hansue | WA | 0ms | 3864kb | C++20 | 2.6kb | 2024-11-04 18:07:44 | 2024-11-04 18:07:46 |
Judging History
answer
#pragma GCC optimize(2)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 4000;
const LL mod = 998244353;
int rt;
int n;
vector<int>g[N + 3];
int sz[N + 3];
LL f[N + 3][N + 3][2];
void dfs(int u, int fr){
sz[u] = 1;
f[u][0][0] = 1;
if((int)g[u].size() == 1&&u!=rt)
return ;
int v = g[u][0];
if(v == fr){
swap(g[u][0], g[u][1]);
v = g[u][0];
}
dfs(v, u);
sz[u] += sz[v];
for(int i = sz[u] / 2; i >= 1; i--){
f[u][i][0] = (f[v][i][0] + f[v][i][1]) % mod;
f[u][i][1] = f[v][i - 1][0];
}
for(int vi = 1; vi < g[u].size(); vi++){
v = g[u][vi];
if(v == fr)
continue;
dfs(v, u);
sz[u] += sz[v];
for(int i = sz[u] / 2; i >= 1; i--)
for(int j = min(i, sz[v] / 2); j >= 0; j--){
if(j > 0){
(f[u][i][0] += f[u][i - j][0] * (f[v][j][0] + f[v][j][1]) % mod) %= mod;
(f[u][i][1] += f[u][i - j][1] * (f[v][j][0] + f[v][j][1]) % mod) %= mod;
}
if(i - j >= 1)
(f[u][i][1] += f[u][i - j - 1][0] * f[v][j][0] % mod) %= mod;
}
}
}
LL d[N + 3];
int main(){
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
scanf("%d", &n);
for(int i = 1, u, v; i <= 2 * n - 1; i++){
scanf("%d%d", &u, &v);
g[u].push_back(v);
g[v].push_back(u);
}
srand(114514);
rt=rand()%n+1;
dfs(rt, 0);
// printf("\n");
// printf("[ ][ ]");
// for(int i = 0; i <= 2 * n - 1; i++)
// printf("%4d", i);
// printf("\n");
// for(int u = 1; u <= 2 * n; u++){
// printf("[%d][0]", u);
// for(int i = 0; i <= 2 * n - 1; i++)
// printf("%4lld", f[u][i][0]);
// printf("\n");
// printf("[%d][1]", u);
// for(int i = 0; i <= 2 * n - 1; i++)
// printf("%4lld", f[u][i][1]);
// printf("\n");
// }
// printf("\n");
d[0] = d[1] = 1;
for(int i = 2; i <= n; i++)
(d[i] = d[i - 1] * (2 * i - 1) % mod) %= mod;
// for(int i = 1; i <= n; i++)
// printf("%lld\n", d[i]);
// printf("\n");
LL ans = 0;
for(int i = 0; i <= n; i++)
(ans += ((i & 1)?(-1):(1)) * ((f[1][i][0] + f[1][i][1]) * d[n - i] % mod) + mod) %= mod;
printf("%lld", ans);
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 0ms
memory: 3864kb
input:
2 1 2 1 3 3 4
output:
1
result:
ok 1 number(s): "1"
Test #2:
score: -100
Wrong Answer
time: 0ms
memory: 3812kb
input:
3 1 2 2 3 3 4 4 5 5 6
output:
15
result:
wrong answer 1st numbers differ - expected: '5', found: '15'