HDU5046 Airport dancing links 重复覆盖+二分

Posted 2020-06-25 shuguangzw

这一道题和HDU2295是一样

是一个dancing links重复覆盖解决最小支配集的问题

在给定长度下求一个最小支配集，只要小于k就行

然后就是二分答案，每次求最小支配集

只不过HDU2295是浮点，这里是整数

我写的一个比较暴力

#include<cstdio>
#include<cstring>
#include<queue>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
typedef long long LL;
const int N=4e3;
int n,m,sz,k;
int u[N],l[N],r[N],d[N];
int h[65],s[65],col[N];
void init()
{
    for(int i=0; i<=m; ++i)
    {
        s[i]=0;
        u[i]=d[i]=i;
        l[i]=i-1;
        r[i]=i+1;
    }
    r[m]=0;
    l[0]=m;
    sz=m;
    for(int i=1; i<=n; ++i)
        h[i]=-1;
}
void link(int x,int y)
{
    ++sz;
    ++s[y],col[sz]=y;
    u[sz]=u[y],d[u[y]]=sz;
    d[sz]=y,u[y]=sz;
    if(h[x]==-1)h[x]=l[sz]=r[sz]=sz;
    {
        l[sz]=l[h[x]];
        r[l[h[x]]]=sz;
        r[sz]=h[x];
        l[h[x]]=sz;
    }
}
void del(int y)
{
    for(int i=d[y]; i!=y; i=d[i])
        r[l[i]]=r[i],l[r[i]]=l[i];
}
void resume(int y)
{
    for(int i=d[y]; i!=y; i=d[i])
        r[l[i]]=l[r[i]]=i;
}
bool vis[65];
int f()
{
    int ret=0;
    for(int i=r[0]; i; i=r[i])
        vis[i]=0;
    for(int i=r[0]; i; i=r[i])
    {
        if(vis[i])continue;
        vis[i]=1;
        ++ret;
        for(int j=d[i]; j!=i; j=d[j])
            for(int k=r[j]; k!=j; k=r[k])
                vis[col[k]]=1;
    }
    return ret;
}
bool dance(int pos)
{
    if(pos+f()>k)return 0;
    if(!r[0])
    {
        if(pos<=k) return 1;
        return 0;
    }
    int t=r[0];
    for(int i=r[0]; i!=0; i=r[i])
        if(s[i]<s[t])t=i;
    for(int i=d[t]; i!=t; i=d[i])
    {
        del(i);
        for(int j=r[i]; j!=i; j=r[j])
            del(j);
        if(dance(pos+1))return 1;
        for(int j=l[i]; j!=i; j=l[j])
            resume(j);
        resume(i);
    }
    return 0;
}
struct point
{
    LL x,y;
}o[65];
LL ABS(LL x,LL y)
{
    if(x>=y)return x-y;
    return y-x;
}
LL dis(point a,point b)
{
    return ABS(a.x,b.x)+ABS(a.y,b.y);
}
LL D[N];
int main()
{
    int T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
      scanf("%d%d",&n,&k),m=n;
      for(int i=1;i<=n;++i)
       scanf("%I64d%I64d",&o[i].x,&o[i].y);
      int cnt=0;
      for(int i=1;i<=n;++i)
           for(int j=i;j<=n;++j)
              D[++cnt]=dis(o[i],o[j]);
      sort(D+1,D+1+cnt);
      cnt=unique(D+1,D+1+cnt)-D-1;
      int high=cnt,low=0,mid;
      while(low<high)
      {
          mid=(low+high)>>1;
          init();
          for(int i=1;i<=n;++i)
           for(int j=1;j<=n;++j)
              if(dis(o[i],o[j])<=D[mid])
                link(i,j);
          if(dance(0))high=mid;
          else low=mid+1;
      }
      printf("Case #%d: %I64d\n",++cas,D[low]);
    }
    return 0;
}

View Code

 

然后另一个是离散化的

#include<cstdio>
#include<cstring>
#include<queue>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
typedef long long LL;
const int N=4e3;
int n,m,sz,k;
int u[N],l[N],r[N],d[N];
int h[65],s[65],col[N];
void init()
{
    for(int i=0; i<=m; ++i)
    {
        s[i]=0;
        u[i]=d[i]=i;
        l[i]=i-1;
        r[i]=i+1;
    }
    r[m]=0;
    l[0]=m;
    sz=m;
    for(int i=1; i<=n; ++i)
        h[i]=-1;
}
void link(int x,int y)
{
    ++sz;
    ++s[y],col[sz]=y;
    u[sz]=u[y],d[u[y]]=sz;
    d[sz]=y,u[y]=sz;
    if(h[x]==-1)h[x]=l[sz]=r[sz]=sz;
    {
        l[sz]=l[h[x]];
        r[l[h[x]]]=sz;
        r[sz]=h[x];
        l[h[x]]=sz;
    }
}
void del(int y)
{
    for(int i=d[y]; i!=y; i=d[i])
        r[l[i]]=r[i],l[r[i]]=l[i];
}
void resume(int y)
{
    for(int i=d[y]; i!=y; i=d[i])
        r[l[i]]=l[r[i]]=i;
}
bool vis[65];
int f()
{
    int ret=0;
    for(int i=r[0]; i; i=r[i])
        vis[i]=0;
    for(int i=r[0]; i; i=r[i])
    {
        if(vis[i])continue;
        vis[i]=1;
        ++ret;
        for(int j=d[i]; j!=i; j=d[j])
            for(int k=r[j]; k!=j; k=r[k])
                vis[col[k]]=1;
    }
    return ret;
}
bool dance(int pos)
{
    if(pos+f()>k)return 0;
    if(!r[0])
    {
        if(pos<=k) return 1;
        return 0;
    }
    int t=r[0];
    for(int i=r[0]; i!=0; i=r[i])
        if(s[i]<s[t])t=i;
    for(int i=d[t]; i!=t; i=d[i])
    {
        del(i);
        for(int j=r[i]; j!=i; j=r[j])
            del(j);
        if(dance(pos+1))return 1;
        for(int j=l[i]; j!=i; j=l[j])
            resume(j);
        resume(i);
    }
    return 0;
}
struct point
{
    LL x,y;
}o[65];
LL ABS(LL x,LL y)
{
    if(x>=y)return x-y;
    return y-x;
}
LL dis(point a,point b)
{
    return ABS(a.x,b.x)+ABS(a.y,b.y);
}
LL D[N];
int main()
{
    int T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
      scanf("%d%d",&n,&k),m=n;
      for(int i=1;i<=n;++i)
       scanf("%I64d%I64d",&o[i].x,&o[i].y);
      int cnt=0;
      for(int i=1;i<=n;++i)
           for(int j=i;j<=n;++j)
              D[++cnt]=dis(o[i],o[j]);
      sort(D+1,D+1+cnt);
      cnt=unique(D+1,D+1+cnt)-D-1;
      int high=cnt,low=0,mid;
      while(low<high)
      {
          mid=(low+high)>>1;
          init();
          for(int i=1;i<=n;++i)
           for(int j=1;j<=n;++j)
              if(dis(o[i],o[j])<=D[mid])
                link(i,j);
          if(dance(0))high=mid;
          else low=mid+1;
      }
      printf("Case #%d: %I64d\n",++cas,D[low]);
    }
    return 0;
}

View Code