BZOJ 4514 费用流

思路:

懒得写了

http://blog.csdn.net/werkeytom_ftd/article/details/51277482

//By SiriusRen
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define int long long
#define mem(x,y) memset(x,y,sizeof(x))
const int M=405,N=444444,inf=1000000000000000ll;
int n,a[M],b[M],c[M],f[M],jy,T,ans1,ans2;
int edge[N],cost[N],v[N],next[N],first[M];
int with[M],vis[M],minn[M],dis[M],tot;
void Add(int x,int y,int C,int E){edge[tot]=E,cost[tot]=C,v[tot]=y,next[tot]=first[x],first[x]=tot++;}
void add(int x,int y,int C,int E){Add(x,y,C,E),Add(y,x,-C,0);}
bool tell(){
    mem(vis,0),mem(minn,0x3f),mem(dis,0x3f);
    queue<int>q;q.push(0),dis[0]=0;
    while(!q.empty()){
        int t=q.front();q.pop(),vis[t]=0;
        for(int i=first[t];~i;i=next[i])
            if(dis[v[i]]>dis[t]+cost[i]&&edge[i]){
                dis[v[i]]=dis[t]+cost[i],minn[v[i]]=min(minn[t],edge[i]),with[v[i]]=i;
                if(!vis[v[i]])vis[v[i]]=1,q.push(v[i]);
            }
    }return dis[T]<=10000000000000000ll;
}
void zeng(){
    for(int i=T;i;i=v[with[i]^1])edge[with[i]]-=minn[T],edge[with[i]^1]+=minn[T];
    ans1+=dis[T]*minn[T];ans2+=minn[T];
}
signed main(){
    mem(first,-1);
    scanf("%lld",&n),T=n+1;
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]),jy=a[i];
        int sqr=sqrt(jy);
        for(int j=2;j<=sqr;j++)
            while(jy%j==0)jy/=j,f[i]++;
        if(jy!=1)f[i]++;
    }
    for(int i=1;i<=n;i++)scanf("%lld",&b[i]),f[i]&1?add(0,i,0,b[i]):add(i,T,0,b[i]);
    for(int i=1;i<=n;i++)scanf("%lld",&c[i]);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            if(a[i]&&a[j]&&abs(f[j]-f[i])==1&&a[i]%a[j]==0)
                add((f[i]&1)?i:j,(f[i]&1)?j:i,-c[i]*c[j],inf);
    while(tell()){zeng();if(ans1>0){if(dis[T])ans2-=(ans1+dis[T]-1)/dis[T];printf("%lld\n",ans2);return 0;}}
    printf("%lld\n",ans2);
}

 

posted @ 2017-03-28 23:49  SiriusRen  阅读(122)  评论(0编辑  收藏  举报