Description

Input

Output

　　一行一個數，最多進行多少次配對

Sample Input

Sample Output

HINT

　　 n≤200，ai≤10^9，bi≤10^5，∣ci∣≤10^5

Solution

　　顯然是一個費用流，並且這可以是一個二分圖，由於 ai/aj 要是質數，那顯然可以根據質因子個數的奇偶

　　然後跑一跑最大費用最大流。判斷一下值要>=0即可統計入答案。mmpBearChild查了一個下午錯，發現是INF開小了qaq。

Code

  1 #include<iostream>    
  2 #include<string>    
  3 #include<algorithm>    
  4 #include<cstdio>    
  5 #include<cstring>    
  6 #include<cstdlib>
  7 #include<cmath>
  8
 
 #include<map>
  9 using namespace std;  
 10 typedef long long s64;
 11   
 12 const int ONE = 205*205;
 13 const int INF = 2147483640;
 14  
 15 int n;
 16 int a[ONE], b[ONE], c[ONE];
 17 s64 dist[ONE];
 18 int vis[ONE], q[ONE*10], pre[ONE];
 19 int first[ONE], go[ONE], next[ONE], pas[ONE], from[ONE], tot;
 20 int pNum[ONE];
 21 int odd[ONE],Onum, eve[ONE],Enum;
 22 int S, T, Ans;
 23 int prime[ONE], isp[ONE], p;
 24 s64 Val, w[ONE];
 25   
 26 int get()
 27 {    
 28         int res=1,Q=1;char c;    
 29         while( (c=getchar())<48 || c>57 ) 
 30         if(c==‘-‘)Q=-1; 
 31         res=c-48;     
 32         while( (c=getchar())>=48 && c<=57 )    
 33         res=res*10+c-48;    
 34         return res*Q;
 35 }
 36 
 37 void Getp(int MaxN)
 38 {
 39         for(int i=2; i <= MaxN; i++)
 40         {
 41             if(!isp[i]) prime[++p] = i;
 42             for(int j=1; j <= prime[p], i*prime[j] <= MaxN; j++)
 43             {
 44                 isp[i * prime[j]] = 1;
 45                 if(i % prime[j] == 0) break;
 46             }
 47         }
 48 }
 49   
 50 int Factor(int x)
 51 {
 52         int res = 0;
 53         while(x != 1)
 54         {
 55             for(int i=1; i<=p; i++)
 56             if(x % prime[i] == 0)
 57             {
 58                 x /= prime[i];
 59                 res++;
 60                 break;
 61             }
 62         }
 63         return res;
 64 }
 65   
 66 int PD(int x, int y)
 67 {
 68         if(x > y) swap(x, y);
 69         if(x==0 || y==0 || y%x!=0) return 0;
 70         x = y / x;
 71         for(int i=2; i<x; i++)
 72         if(x % i == 0) return 0;
 73         return 1;
 74 }
 75   
 76 int Add(int u, int v, int flow, s64 z)
 77 {
 78         next[++tot]=first[u];   first[u]=tot;   go[tot]=v;  pas[tot]=flow;  w[tot]=z;   from[tot]=u;
 79         next[++tot]=first[v];   first[v]=tot;   go[tot]=u;  pas[tot]=0;     w[tot]=-z;  from[tot]=v;
 80 }
 81    
 82 bool Bfs()
 83 {
 84         for(int i=S; i<=T; i++) dist[i] = -1e18;
 85         int tou = 0, wei = 1;
 86         q[1] = S;   vis[S] = 1; dist[S] = 0;
 87         while(tou < wei)
 88         {
 89             int u = q[++tou];
 90             for(int e=first[u]; e; e=next[e])
 91             {
 92                 int v=go[e];
 93                 if(dist[v] < dist[u] + w[e] && pas[e])
 94                 {
 95                     dist[v] = dist[u] + w[e];
 96                     pre[v] = e;
 97                     if(!vis[v])
 98                     {
 99                         q[++wei] = v;
100                         vis[v] = 1;
101                     }
102                 }
103             }
104             vis[u] = 0;
105         }
106         return dist[T] != -1e18;
107 }
108   
109 void Deal()
110 {
111         int x = INF;
112         for(int e=pre[T]; e; e=pre[from[e]]) x = min(x, pas[e]);
113         if(Val + dist[T] * x >= 0)
114         {
115             for(int e=pre[T]; e; e=pre[from[e]])
116             {
117                 pas[e] -= x;
118                 pas[((e-1)^1)+1] += x;
119             }
120             Val += dist[T] * x;
121             Ans += x;
122             return;
123         }
124         printf("%d", Ans - Val / dist[T]);
125         exit(0);
126 }
127   
128 int main()
129 {
130         Getp(ONE);
131         n = get();
132         for(int i=1; i<=n; i++) a[i] = get(), pNum[i] = Factor(a[i]);
133         for(int i=1; i<=n; i++) b[i] = get();
134         for(int i=1; i<=n; i++) c[i] = get();
135           
136         S = 0; T = n+1;
137         for(int i=1; i<=n; i++)
138         {
139             if(pNum[i] & 1) Add(S,i, b[i],0), odd[++Onum] = i;
140             else Add(i,T, b[i],0), eve[++Enum] = i;
141         }
142           
143         for(int i=1; i<=Onum; i++)
144         for(int j=1; j<=Enum; j++)
145         {
146             int x = odd[i], y = eve[j];
147             if( PD(a[x], a[y]) )
148                 Add(x,y, INF,(s64)c[x]*c[y]);
149         }
150         
151         while(Bfs()) Deal();
152         printf("%d", Ans - Val / dist[T]);
153 }

【BZOJ4514】【SDOI2016】數字配對 [費用流]