分析：这题脑洞新奇...居然是最短路...将0到k-1看做k个点，第t个点向(10*t+0,1,2...,9)%k连一条长度为0,,1,2,..,9的边，然后枚举s=1,2,...,9,算出所有从s到0的最短路，答案就是最短路+s的最小值。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<vector> 5 #include<queue> 6 using namespace std; 7 const int maxn = 1e5 + 5, inf = 1e9; 8 struct Edge{ 9 int to,dis; 10 Edge(int _to, int _dis):to(_to), dis(_dis){} 11 }; 12 vector<Edge> G[maxn]; 13 bool inq[maxn]; 14 int dis[maxn]; 15 int n; 16 queue<int> Q; 17 int spfa(int s){ 18 for(int i = 0; i < n; i++)dis[i] = inf; 19 memset(inq, 0, sizeof(inq)); 20 Q.push(s); 21 dis[s] = 0; 22 inq[s] = true; 23 while(!Q.empty()){ 24 int u = Q.front();Q.pop(); 25 inq[u] = false; 26 for(int i = 0; i < G[u].size(); i++){ 27 Edge &e = G[u][i]; 28 if(dis[e.to] > dis[u] + e.dis){ 29 dis[e.to] = dis[u] + e.dis; 30 if(!inq[e.to])inq[e.to] = true, Q.push(e.to); 31 } 32 } 33 } 34 return s + dis[0]; 35 } 36 void build(){ 37 for(int i = 1; i < n; i++){ 38 for(int j = 0; j <= 9; j++){ 39 G[i].push_back(Edge((10 * i + j) % n, j)); 40 } 41 } 42 } 43 int main(){ 44 cin>>n; 45 build(); 46 int ans = inf; 47 for(int s = 1; s <= 9; s++)ans = min(ans, spfa(s)); 48 cout<<ans<<endl; 49 return 0; 50 }
