【BZOJ】【2661】【Beijing WC2012】连连看

网络流/费用流/二分图最大权匹配

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　　拆点费用流求最大权匹配……为什么我拿zyf和Hzwer的代码也交不过去……WA了那么多次……so sad

　　求路过的神牛指导啊>_<万分感谢

//BZOJ 2661
#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define rep(i,n) for(int i=0;i<n;++i)
#define F(i,j,n) for(int i=j;i<=n;++i)
#define D(i,j,n) for(int i=j;i>=n;--i)
#define pb push_back
using namespace std;
inline int getint(){
    int v=0,sign=1; char ch=getchar();
    while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();}
    return v*sign;
}
const int N=2010,M=2000000,INF=~0u>>2;
typedef long long LL;
/******************tamplate*********************/
int n,m,a,b,ans,flow;
inline int gcd(int a,int b){return b ? gcd(b,a%b) : a;}
bool judge(int x,int y){
    int q=x*x-y*y,s=sqrt(q);
    if (s*s==q && gcd(y,s)==1) return 1;
    return 0;
}
struct edge{int from,to,v,c;};
struct Net{
    edge E[M];
    int head[N],next[M],cnt;
    void ins(int x,int y,int z,int c){
        E[++cnt]=(edge){x,y,z,c};
        next[cnt]=head[x]; head[x]=cnt;
    }
    void add(int x,int y,int z,int c){
        ins(x,y,z,c); ins(y,x,0,-c);
    }
    int from[N],Q[M],d[N],S,T;
    bool inq[N];
    bool spfa(){
        int l=0,r=-1;
        F(i,0,T) d[i]=INF;
        d[S]=0; Q[++r]=S; inq[S]=1;
        while(l<=r){
            int x=Q[l++];
            inq[x]=0;
            for(int i=head[x];i;i=next[i])
                if(E[i].v>0 && d[x]+E[i].c<d[E[i].to]){
                    d[E[i].to]=d[x]+E[i].c;
                    from[E[i].to]=i;
                    if (!inq[E[i].to]){
                        Q[++r]=E[i].to;
                        inq[E[i].to]=1;
                    }
                }
        }
        return d[T]!=INF;
    }
    void mcf(){
        int x=INF,y,z;
        for(int i=from[T];i;i=from[E[i].from]){
//            if (E[i].from==S) y=E[i].to;
//            if (E[i].to==T) z=E[i].from-b;
            x=min(x,E[i].v);
        }
        for(int i=from[T];i;i=from[E[i].from]){
            E[i].v-=x;
            E[i^1].v+=x;
        }
        flow+=x;
        ans+=x*d[T];
    }
    void init(){
        scanf("%d%d",&a,&b);
        cnt=1; S=0; T=2001;
        F(i,a,b){
            add(S,i,1,0);
            add(i+b,T,1,0);
            F(j,a,i-1)
                if(judge(i,j)) add(i,j+b,1,-i-j);
        }
        while(spfa()) mcf();
        printf("%d %d",flow,-ans);
    }
}G1;

int main(){
#ifndef ONLINE_JUDGE
    freopen("2661.in","r",stdin);
    freopen("2661.out","w",stdout);
#endif
    G1.init();
    return 0;
}