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【最小割】【Dinic】bzoj3275 Number

每个点拆点,分别向源/汇连a[i]的边,满足条件的相互连INF的边,答案为sum-maxflow*2。

因为若有几个点不能同时被选,我们要贪心地选择其中和尽量大的部分,这可以由最小割来保证。

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;#define INF 2147483647#define MAXN 6005#define MAXM 600301int v[MAXM],cap[MAXM],en,first[MAXN],next[MAXM];int d[MAXN],cur[MAXN],A[3005],sumv;queue<int>q;int n,S,T;void Init_Dinic(){memset(first,-1,sizeof(first)); en=0; S=0; T=(n<<1|1);}void AddEdge(const int &U,const int &V,const int &W){v[en]=V; cap[en]=W; next[en]=first[U]; first[U]=en++;v[en]=U; next[en]=first[V]; first[V]=en++;}bool bfs(){    memset(d,-1,sizeof(d)); q.push(S); d[S]=0;    while(!q.empty())      {        int U=q.front(); q.pop();        for(int i=first[U];i!=-1;i=next[i])          if(d[v[i]]==-1 && cap[i])            {              d[v[i]]=d[U]+1;              q.push(v[i]);            }      }    return d[T]!=-1;}int dfs(int U,int a){    if(U==T || !a) return a;    int Flow=0,f;    for(int &i=cur[U];i!=-1;i=next[i])      if(d[U]+1==d[v[i]] && (f=dfs(v[i],min(a,cap[i]))))        {          cap[i]-=f; cap[i^1]+=f;          Flow+=f; a-=f; if(!a) break;        }    if(!Flow) d[U]=-1;    return Flow;}int max_flow(){    int Flow=0,tmp=0;    while(bfs())      {        memcpy(cur,first,((n<<1)+5)*sizeof(int));        while(tmp=dfs(S,INF)) Flow+=tmp;      }    return Flow;}int gcd(int a,int b){return b==0?a:gcd(b,a%b);}int sqr(const int &x){return x*x;}bool check(const int &a,const int &b){	int t=a*a+b*b;	return (sqr((int)sqrt(t))==t&&(gcd(a,b)==1));}int main(){	scanf("%d",&n);	for(int i=1;i<=n;++i)	  {	  	scanf("%d",&A[i]);	  	sumv+=A[i];	  }	Init_Dinic();	for(int i=1;i<=n;++i)	  for(int j=1;j<i;++j)	    if(check(A[i],A[j]))	      AddEdge(i,j+n,INF),		  AddEdge(j,i+n,INF);	for(int i=1;i<=n;++i)	  AddEdge(S,i,A[i]),	  AddEdge(i+n,T,A[i]);	printf("%d\n",sumv-(max_flow()>>1));	return 0;}

  

【最小割】【Dinic】bzoj3275 Number