Codeforces Round #284 (Div. 1) C

C. Array and Operations

        题意：有n个数，a1~an；m对数i1/j1~im/jm(ik+jk为奇数)。每次操作取出其中一对ik/jk，使a[ik]和a[jk]同时除以一个不为1的正整数v。问最多能够进行几次操作。

        思路：很容易得出，每次操作的v越小，接下来还能进行的操作就越多，进一步得出，v应该选择素数。因为每对数的和是奇数，所以可以建一个二部图，一个集合是奇数，一个集合是偶数。然后枚举所有范围内的素数，对每个素数建边跑最大流就可以了。

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <map>#include <vector>#include <set>#include <string>#include <math.h>using namespace std;#define INF 100000010       #define maxn 110          struct Edge{              int u; int v;              int cap; int flow;              Edge(int u,int v,int c):u(u),v(v),cap(c),flow(0){}                Edge(){};          };          Edge edges[maxn*maxn];                  int ne;          int head[maxn];          int next[maxn*maxn];          int N,M;     int L[maxn*maxn];     int D[maxn];         void init(){              memset(head,-1,sizeof(head));           memset(D,0,sizeof(D));           ne=0;          }              int lv[maxn];          bool bfs(int s,int t){              memset(lv,-1,sizeof(lv)); lv[s]=0;              queue<int> que; que.push(s);               while(!que.empty()){                  int cur=que.front(); que.pop();                  for(int i=head[cur];i!=-1;i=next[i]){                      Edge& e=edges[i];                      if(lv[e.v]!=-1)continue;                      if(e.flow<e.cap){                          lv[e.v]=lv[cur]+1;                          que.push(e.v);                      }                  }                  if(lv[t]!=-1)return 1;              }              return 0;          }              int cur[maxn];    int dfs(int x,int a){              if(x==101||a==0)return a;     int re=0;              for(int& i=cur[x];i!=-1;i=next[i]){                  Edge& e=edges[i];                  int t;                  if(lv[e.v]==(lv[x]+1)&& (t=dfs( e.v , min(a,e.cap-e.flow))) ){                      edges[i].flow+=t;                      edges[i^1].flow-=t;                      re+=t;                      a-=t;                      if(a==0)break;                  }              }              return re;          }        int maxflow(int s,int t){              int flow=0;              while(bfs(s,t)){            memcpy(cur,head,sizeof(head));                  flow+=dfs(s,INF);          }              return flow;          }        void addedge(int u,int v,int c){              edges[ne]=Edge(u,v,c);                next[ne]=head[u];              head[u]=ne;              ne++;            edges[ne]=Edge(v,u,0);              next[ne]=head[v];              head[v]=ne;              ne++;          }     int a[maxn];int x[maxn];int y[maxn];vector<int> prime;int gcd(int a,int b){if(b==0)return a;return gcd(b,a%b);}int main(){//int n,m;cin>>n>>m;for(int i=1;i<=n;i++){cin>>a[i];}for(int i=1;i<=m;i++){cin>>x[i]>>y[i];if(x[i]%2==0)swap(x[i],y[i]);//确保x是奇数y是偶数 int g=gcd(a[x[i]],a[y[i]]);for(int j=2;j*j<=g;j++){if(g%j==0){prime.push_back(j);while(g%j==0)g/=j;}}if(g!=1)prime.push_back(g);}sort(prime.begin(),prime.end());vector<int>::iterator end;end=unique(prime.begin(),prime.end());int ans=0;//枚举素数for(vector<int>::iterator it=prime.begin() ;it!=end;it++){int k=*it;init();//源到奇数，偶数到汇 for(int i=1;i<=n;i++){int tmp=a[i];int cap=0;while(tmp%k==0){cap++;tmp/=k;}if(cap)if(i&1){addedge(0,i,cap);}else{addedge(i,101,cap);}}//奇数到偶数 for(int i=1;i<=m;i++){addedge(x[i],y[i],INF);} ans+=maxflow(0,101);}cout<<ans<<endl;return 0;}