VK cup Div1 C. Vulnerable Kerbals （扩展gcd+DAG+最长路）

浅浅的花香味﹌ 2022-06-17 06:27 97阅读 0赞

题目链接：

[点击打开链接][Link 1]

http://codeforces.com/contest/800/problem/C

C. Vulnerable Kerbals

题意：

给定n个0~m-1内的数和m，构造一个尽可能长的所有元素都在0~m-1内的数列，并且使所有前缀积模m不相同且不在n个数中出现过。

题解：

如果前 i 个数的前缀积为x，前i+1个数的前缀积可以为y当且仅当ax-bm=y有解，即gcd(x,m)|y。如果我们让所有满足

这个条件的x，y，x向y连边，那么就变成求最长路，gcd(x,m)相同的x在一个连通块内，每个连通块gcd(x\_i,m)向连通

块gcd(x\_j,m)（gcd(x\_i,m) | gcd(x\_j,m)）连边，得到一个DAG，即将可以作为前缀乘积的数按照与m的gcd分类，

相同gcd值作为顶点，不同gcd值**满足整除关系**就建一条边，最终形成一个DAG。在DAG上DP一遍，跑一下最长路就可以了。

也可以看看zmy写的题解：[点击打开链接][Link 2]

AC代码：

#pragma comment(linker, "/STACK:102400000,102400000")
    #include <bits/stdc++.h>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    #include <cstring>
    #include <vector>
    #include <map>
    #include <cmath>
    #include <queue>
    #include <set>
    #include <bitset>
    #include <iomanip>
    #include <list>
    #include <stack>
    #include <utility> 
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<int,int> pii;
    typedef vector<int> vi;
    const double eps = 1e-8;  
    const int INF = 1e9+7; 
    const ll inf =(1LL<<62) ;
    const int MOD = 1e9 + 7;  
    const ll mod = (1LL<<32);
    const int N =1e6+6; 
    const int M=100010;
    const int maxn=1001;
    #define mst(a) memset(a, 0, sizeof(a))
    #define M_P(x,y) make_pair(x,y)
    #define pi acos(-1.0)
    #define in freopen("in.txt","r",stdin) 
    #define rep(i,j,k) for (int i = j; i <= k; i++)  
    #define per(i,j,k) for (int i = j; i >= k; i--)  
    #define lson x << 1, l, mid  
    #define rson x << 1 | 1, mid + 1, r  
    const int lowbit(int x) { return x&-x; }
    int read(){ int v = 0, f = 1;char c =getchar();
    while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}
    while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();
    return v*f;}
    ll exgcd(ll a,ll b,ll &x,ll &y)// 返回a,b的最大公约数
    {
        if(!a && !b)return -1;
        if(!b)return x=1,y=0,a;
        ll d=exgcd(b,a%b,y,x);
        return y-=a/b*x,d;
    }
    
    ll gcd(ll a,ll b)
    {
    	return b==0?a:gcd(b,a%b);
    }
    int main()
    {
      
      int n, m;
      n=read(),m=read();
      vector<int> a(m+1);
      vector<vector<int>> b(m+1);
      for (int i = 0; i < n; i++)
      {
        int t;
        t=read();
        a[t]++;
      }
      for (int i = 1; i < m; i++)
      {
        if (a[i]==0)
    	{
          b[gcd(i, m)].push_back(i);
        }
      }
      /*
      cout<<"----------------------"<<endl;
      for(int i=1;i<m;i++){
      	cout<<i<<" 子节点为"<<" "; 
      	for(int j=0;j<b[i].size();j++){
      		cout<<b[i][j]<<" ";
    	  }
    	  cout<<endl;
      }
      cout<<"----------------------"<<endl;
      */
      vector<vector<int>> g(m+1);
      for (int i = 1; i < m; i++)
      {
        if (!b[i].empty())
    	{
          for (int j = i + i; j < m; j += i) 
    	  {
            if (!b[j].empty())
    		{
              g[i].push_back(j);
            }
          }
        }
      }
      /*
      for(int i=1;i<m;i++){
      	cout<<i<<" 子节点为"<<" "; 
      	for(int j=0;j<g[i].size();j++){
      		cout<<g[i][j]<<" ";
    	  }
    	  cout<<endl;
      }
      cout<<"----------------------"<<endl;
      */
      vector<int> dp(m+1);
      vector<int> pre(m+1);
      int p = m - 1;
      for (int i = m - 1; i >= 1; i--)
      {
         dp[i] = b[i].size();
         for (auto v : g[i])
    	 {
    	//   cout<<"i="<<i<<" "<<"v="<<v<<endl;
           if (dp[v] + b[i].size() > dp[i])
    	   {
             dp[i] = dp[v] + b[i].size();
             pre[i] = v;
           }
         }
        
         if (dp[i] > dp[p])
    	 {
           p = i;
         }
      }
      
      vector<int> ans;
      for (int i = p; i; i = pre[i])
      {
        for (auto x : b[i]) {
          ans.push_back(x);
         // cout<<"x="<<x<<" ";
        }
      }
     // cout<<endl;
      if (!ans.empty())
      {
        ll now = ans[0];
        for (int i = 1; i < ans.size(); i++)
    	{
          ll x = 0, y = 0;
          int g = exgcd(now, m, x, y);
          x *= (ans[i] / g);
          x = (x % m + m) % m;
          ans[i] = x; //记录答案 
          now = now * x % m;
        }
      }
      
      if (!a[0])
      {
        ans.push_back(0);//hack
      }
      
      printf("%d\n",ans.size());
      for (auto x : ans) {
        printf("%d ",x);
      }
      return 0;
    }

[Link 1]: http://codeforces.com/contest/800/problem/C
[Link 2]: http://tieba.baidu.com/p/5076330276