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这题跟BZOJ 3275限制条件是一样的= =所以可以用相同的方法去做……只要把边的容量从a[i]改成b[i]就行了~
(果然不加当前弧优化要略快一点)
1 /************************************************************** 2 Problem: 3158 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:196 ms 7 Memory:13028 kb 8 ****************************************************************/ 9 10 //BZOJ 3158 11 #include<cmath> 12 #include<vector> 13 #include<cstdio> 14 #include<cstring> 15 #include<cstdlib> 16 #include<iostream> 17 #include<algorithm> 18 #define rep(i,n) for(int i=0;i<n;++i) 19 #define F(i,j,n) for(int i=j;i<=n;++i) 20 #define D(i,j,n) for(int i=j;i>=n;--i) 21 #define pb push_back 22 using namespace std; 23 inline int getint(){ 24 int v=0,sign=1; char ch=getchar(); 25 while(ch<‘0‘||ch>‘9‘){ if (ch==‘-‘) sign=-1; ch=getchar();} 26 while(ch>=‘0‘&&ch<=‘9‘){ v=v*10+ch-‘0‘; ch=getchar();} 27 return v*sign; 28 } 29 const int N=1100,M=1000000,INF=~0u>>2; 30 typedef long long LL; 31 /******************tamplate*********************/ 32 int n,m,tot,ans,a[N],b[N]; 33 struct edge{int to,v;}; 34 int gcd(int a,int b){return b ? gcd(b,a%b) : a;} 35 bool judge(LL a,LL b){ 36 LL s=a*a+b*b; 37 LL q=sqrt(s); 38 if (q*q!=s) return 0; 39 if (gcd(a,b)!=1) return 0; 40 return 1; 41 } 42 struct Net{ 43 edge E[M]; 44 int head[N],next[M],cnt; 45 void ins(int x,int y,int v){ 46 E[++cnt]=(edge){y,v}; 47 next[cnt]=head[x]; head[x]=cnt; 48 } 49 void add(int x,int y,int v){ 50 ins(x,y,v); ins(y,x,0); 51 } 52 int s,t,d[N],Q[N]; 53 void init(){ 54 n=getint(); cnt=1; 55 tot=ans=0; 56 s=0; t=n+1; 57 F(i,1,n) a[i]=getint(); 58 F(i,1,n){ 59 b[i]=getint();tot+=b[i]; 60 if (a[i]&1) add(s,i,b[i]); 61 else add(i,t,b[i]); 62 } 63 F(i,1,n) if (a[i]&1) 64 F(j,1,n) if((a[j]&1)==0) 65 if (judge(a[i],a[j])) add(i,j,INF); 66 } 67 bool mklevel(){ 68 memset(d,-1,sizeof d); 69 d[s]=0; 70 int l=0,r=-1; 71 Q[++r]=s; 72 while(l<=r){ 73 int x=Q[l++]; 74 for(int i=head[x];i;i=next[i]) 75 if (d[E[i].to]==-1 && E[i].v){ 76 d[E[i].to]=d[x]+1; 77 Q[++r]=E[i].to; 78 } 79 } 80 return d[t]!=-1; 81 } 82 int dfs(int x,int a){ 83 if (x==t) return a; 84 int flow=0; 85 for(int i=head[x];i && flow<a;i=next[i]) 86 if (E[i].v && d[E[i].to]==d[x]+1){ 87 int f=dfs(E[i].to,min(a-flow,E[i].v)); 88 E[i].v-=f; 89 E[i^1].v+=f; 90 flow+=f; 91 } 92 if (!flow) d[x]=-1; 93 return flow; 94 } 95 void Dinic(){ 96 while(mklevel()) 97 ans+=dfs(s,INF); 98 } 99 }G1; 100 101 int main(){ 102 #ifndef ONLINE_JUDGE 103 freopen("3158.in","r",stdin); 104 freopen("3158.out","w",stdout); 105 #endif 106 G1.init(); G1.Dinic(); 107 printf("%d\n",tot-ans); 108 return 0; 109 }
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原文地址:http://www.cnblogs.com/Tunix/p/4339925.html