Atcoder 212E Safety Journey

Problem Statement

The Republic of AtCoder has N cities, called City 1, City 2, ……, City N. Initially, there was a bidirectional road between every pair of different cities, but M of these roads have become unusable due to deterioration over time. More specifically, for each 1≤i≤M, the road connecting City Ui and City Vi has become unusable.

Takahashi will go for a KK-day trip that starts and ends in City 1. Formally speaking, a KK-day trip that starts and ends in City 1 is a sequence of K+1 cities (A0,A1,…,AK) such that A0=AK=1 holds and for each 0≤i≤K−1, Ai and Ai+1 are different and there is still a usable road connecting City Ai and City Ai+1.

Print the number of different K-day trips that start and end in City 1, modulo 998244353. Here, two K-day trips (A0,A1,…,AK) and (B0,B1,…,BK) are said to be different when there exists an i such that Ai≠Bi.

Constraints

• 2≤N≤5000
• 0≤M≤min(N(N−1)2,5000)
• 2≤K≤5000
• 1≤Ui<Vi≤N
• All pairs (Ui,Vi) are pairwise distinct.
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

N M K
U1 V1
::
UM VM

Output

Print the answer.

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Sample Input 1 Copy

3 1 4
2 3

Sample Output 1 Copy

4

There are four different trips as follows.

• (1,2,1,2,1)
• (1,2,1,3,1)
• (1,3,1,2,1)
• (1,3,1,3,1)

No other trip is valid, so we should print 44.

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Sample Input 2 Copy

3 3 3
1 2
1 3
2 3

Sample Output 2 Copy

0

No road remains usable, so there is no valid trip.

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Sample Input 3 Copy

5 3 100
1 2
4 5
2 3

Sample Output 3 Copy

428417047

题目翻译

\(N\)座城市，每对城市间有双向道路，有\(M\)条道路无法使用

从城市\(1\)出发，旅行\(K\)天，最后一天回到\(1\)。旅游城市序列\(A_0,A_1,...A_k\)，其中\(A_0=A_k=1\)，共有多少不同旅行方案

\(2<=N<=5000\)

\(M<=min(\frac{N*(N-1)}{2},5000)\)

\(2<=K<=5000\)

题目解析

首先考虑最简单的做法，令\(f[j][i]\)表示第\(i\)天在城市\(j\)

枚举天数\(i\)，和停留的城市\(j\)进行转移

但题目给出的\(M\)条边是不可通行的道路，如果枚举所有可行道路，复杂度将达到\(O(K*N^2)\)，不符合数据范围

考虑到\(f[j][i]=\sum_{k=1}^{n}{f[k][i-1]}-\sum_{k不可达}{f[k][i-1]}-f[j][i-1]\)

\(\sum_{k=1}^{n}{f[k][i-1]}\)可以\(O(N)\)求出，\(\sum_{k不可达}{f[k][i-1]}\)可以\(O(M)\)求出

总复杂度为\(O(K*(N+M))\)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;
typedef long long ll;
struct Node{
    int next,to;
}edge[10005];
int n,m,k;
int head[5005],num,f[5005][5005],Mod=998244353;
void add(int u,int v){
    num++;
    edge[num].next=head[u];
    head[u]=num;
    edge[num].to=v;
}
int main(){
    int u,v;
    cin>>n>>m>>k;
    for (int i=1;i<=m;i++){
        scanf("%d%d",&u,&v);
        add(u,v);add(v,u);
    }
    memset(f,0,sizeof(f));
    f[1][0]=1;
    for (int i=1;i<=k;i++){
        int sum=0;
        for (int j=1;j<=n;j++)
        sum=(sum+f[j][i-1])%Mod;
        for (int j=1;j<=n;j++){
            f[j][i]=(sum-f[j][i-1]+Mod)%Mod;
            for (int k=head[j];k;k=edge[k].next){
                int v=edge[k].to;
                f[j][i]=(f[j][i]-f[v][i-1]+Mod)%Mod;
            }
        }
    }
    cout<<f[1][k]<<endl;
}