bzoj4205: 卡牌配对

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链接

　　http://www.lydsy.com/JudgeOnline/problem.php?id=4205

题解

　　边数107你告诉我网络流？打死我也不可能自己想到的吧。
　　博主很懒，不想打字了~
　　

代码

//网络流 #include <cstdio>#include <algorithm>#define maxn 500000#define maxe 30000000using namespace std;int head[maxn], to[maxe], c[maxe], nex[maxe], Exit, d[maxn], num[maxn], prime[60],    mark[maxn], last[maxn], W[maxn][3], n1, n2, tot=1, S, T;inline void adde(int a, int b, int cc){to[++tot]=b;c[tot]=cc;nex[tot]=head[a];head[a]=tot;}inline void adde2(int a, int b,int cc){adde(a,b,cc),adde(b,a,0);}int isap(int pos, int in){    int flow=0, t;    if(pos==T)return in;    for(int &p=last[pos];p;p=nex[p])        if(c[p] and d[to[p]]+1==d[pos])        {            flow+= t=isap(to[p],min(in-flow,c[p]));            c[p]-=t, c[p^1]+=t;            if(Exit or in==flow)return flow;        }    Exit=--num[d[pos]]==0;    ++num[++d[pos]];    last[pos]=head[pos];    return flow;}void shai(){    int i, j;    for(i=2;i<=200;i++)    {        if(!mark[i])prime[++prime[0]]=i;        for(j=1;j<=prime[0] and i*prime[j]<=200;j++)        {            mark[i*prime[j]]=1;            if(i%prime[j]==0)break;        }    }}inline int read(int x=0){    char c=getchar();    while(c<48 or c>57)c=getchar();    while(c>=48 and c<=57)x=(x<<1)+(x<<3)+c-48,c=getchar();    return x;}void input(){    int i;    n1=read(), n2=read();    for(i=1;i<=n1;i++)W[i][0]=read(), W[i][1]=read(), W[i][2]=read();    for(i=1;i<=n2;i++)W[n1+i][0]=read(), W[n1+i][1]=read(), W[n1+i][2]=read();}void build(){    int i, j, k, l, table[3][60][60], N;    int h[3][2]={{0,1},{0,2},{1,2}};    input();shai();    N=n1+n2;    for(i=1,k=N*4+10;i<=prime[0];i++)for(j=1;j<=prime[0];j++)            table[0][i][j]=++k, table[1][i][j]=++k, table[2][i][j]=++k;    S=k+10, T=S+1;    for(i=1;i<=n1;i++)    {        adde2(S,i,1);        adde2(i,i+N,1),adde2(i,i+N+N,1),adde2(i,i+N+N+N,1);        for(l=0;l<3;l++)for(j=1;j<=prime[0];j++)for(k=1;k<=prime[0];k++)            if(W[i][h[l][0]]%prime[j]==0 and W[i][h[l][1]]%prime[k]==0)                adde2(i+(l+1)*N,table[l][j][k],1);    }    for(i=n1+1;i<=n1+n2;i++)    {        adde2(i,T,1);        adde2(i+N,i,1),adde2(i+N+N,i,1),adde2(i+N+N+N,i,1);        for(l=0;l<3;l++)for(j=1;j<=prime[0];j++)for(k=1;k<=prime[0];k++)            if(W[i][h[l][0]]%prime[j]==0 and W[i][h[l][1]]%prime[k]==0)                adde2(table[l][j][k],i+(l+1)*N,1);    }}int main(){    int ans=0;    build();    while(!Exit)ans+=isap(S,0x3f3f3f3f);    printf("%d",ans);    return 0;}

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