BZOJ 2661 连连看

http://www.lydsy.com/JudgeOnline/problem.php?id=2661

思路:预处理出每个数字，然后若有x^2=y^2+z^2且z与y互质，

s->x 1 ,0

x+n-> t 1 , 0

x->y+n -> 1 , inf-x-y

y->x+n-> 1 ,inf-x-y

 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 #define inf 1000000
 int l,r,pd[],L[],R[],cnt,sz,b[];
 int tot,go[],next[],first[],cost[],flow[];
 int op[],dis[],c[],vis[],edge[],from[];
 int S,T,ans1,ans2;
 int gcd(int a,int b){if (b==) return a;else return gcd(b,a%b);}
 int read(){
     char ch=getchar();int t=,f=;
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 void insert(int x,int y,int z,int l){
     tot++;
     go[tot]=y;
     next[tot]=first[x];
     first[x]=tot;
     flow[tot]=z;
     cost[tot]=l;
 }
 void add(int x,int y,int z,int l){
     insert(x,y,z,l);op[tot]=tot+;
     insert(y,x,,-l);op[tot]=tot-;
 }
 bool spfa(){
     for (int i=;i<=T;i++)
      dis[i]=inf,vis[i]=;
     int h=,t=;c[]=S;vis[S]=;dis[S]=;
     while (h<=t){
         int now=c[h++];
         for (int i=first[now];i;i=next[i]){
             int pur=go[i];
             if (dis[pur]>dis[now]+cost[i]&&flow[i]){
                 dis[pur]=dis[now]+cost[i];
                 edge[pur]=i;
                 from[pur]=now;
                 if (vis[pur]) continue;
                 vis[pur]=;
                 c[++t]=pur;
             }
         }
         vis[now]=;
     }
     return dis[T]!=inf;
 }
 void updata(){
     int mn=0x3f3f3f3f;
     for (int i=T;i!=S;i=from[i]){
         mn=std::min(mn,flow[edge[i]]);
     }
     for (int i=T;i!=S;i=from[i]){
         ans2+=mn*cost[edge[i]];
         flow[edge[i]]-=mn;
         flow[op[edge[i]]]+=mn;
     }
     ans1+=mn;
 }
 int main(){
     l=read();r=read();
     if (l>r) std::swap(l,r);
     for (int i=;i<=;i++) pd[i*i]=i;
     for (int j=l;j<=r;j++)
      for (int i=j+;i<=r;i++){
             int k=i*i-j*j;
             if (pd[k]&&gcd(pd[k],j)==){
                 b[i]=b[j]=;
                 cnt++;
                 L[cnt]=j;
                 R[cnt]=i;
             }
      }
     for (int i=l;i<=r;i++)
      if (b[i])
       b[i]=++sz;
     S=;T=sz+sz+;
     for (int i=l;i<=r;i++)
      if (b[i])
       add(S,b[i],,),add(b[i]+sz,T,,);
     for (int i=;i<=cnt;i++){
         add(b[L[i]],b[R[i]]+sz,,inf-L[i]-R[i]);
         add(b[R[i]],b[L[i]]+sz,,inf-L[i]-R[i]);
     }
     while (spfa()) updata();
     printf("%d ",ans1/);
     printf("%d\n",(ans1*inf-ans2)/);
 }