hdu 3498

DLX 重复覆盖模板题

# include<cstdio># include<cstring>#include<iostream>#include<vector>using namespace std;# define N 60# define V N*Nint L[V],R[V];//记录左右方向的双向链表int U[V],D[V];//记录上下方向的双向链表int C[V];//指向其列指针头的地址int H[N];//行指针头int S[N];//记录列链表中节点的总数int size,n,m,ak;int visit[N];vector<int >g[N];void Link(int r,int c){    S[c]++;C[size]=c;    U[size]=U[c];D[U[c]]=size;    D[size]=c;U[c]=size;    if(H[r]==-1) H[r]=L[size]=R[size]=size;    else    {        L[size]=L[H[r]];R[L[H[r]]]=size;        R[size]=H[r];L[H[r]]=size;    }    size++;}void remove(int Size){    int j;    for(j=D[Size];j!=Size;j=D[j])        L[R[j]]=L[j],R[L[j]]=R[j];}void resume(int Size){    int j;    for(j=D[Size];j!=Size;j=D[j])        L[R[j]]=R[L[j]]=j;}int h(){    int count=0,i,j,k;    memset(visit,0,sizeof(visit));    for(i=R[0];i;i=R[i])    {        if(visit[i]) continue;        count++;        for(j=D[i];j!=i;j=D[j])        {            for(k=R[j];k!=j;k=R[k])                visit[C[k]]=1;        }    }    return count;}void Dance(int k){    int i,j,min,c;    if(k+h()>=ak) return;    if(!R[0])    {        if(k<ak) ak=k;        return;    }    for(min=N,i=R[0];i;i=R[i])        if(min>S[i]) min=S[i],c=i;    for(i=D[c];i!=c;i=D[i])    {        remove(i);        for(j=R[i];j!=i;j=R[j])            remove(j);        Dance(k+1);        for(j=R[i];j!=i;j=R[j])            resume(j);        resume(i);    }}int main(){    int i,j;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=0;i<=n;i++)        {g[i].clear();g[i].push_back(i);            S[i]=0;            U[i]=D[i]=i;            L[i+1]=i;R[i]=i+1;        }R[n]=0;        size=n+1;memset(H,-1,sizeof(H));for(i=0;i<m;i++){int a,b;scanf("%d%d",&a,&b);g[a].push_back(b);g[b].push_back(a);}for(i=1;i<=n;i++){for(j=0;j<g[i].size();j++){Link(i,g[i][j]);}}ak=n;Dance(0);printf("%d\n",ak);    }    return 0;}