ZOJ 3209 Treasure Map dancing links

Posted 2020-06-24 shuguangzw

分析：dancing links 裸题 ，只要把每个格子看成一列就好了，精确不重复覆盖问题

#include<cstdio>
#include<cstring>
#include<queue>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
typedef long long LL;
const int N=500005;
const LL mod=1e9+7;
int n,m,sz;
int u[N],l[N],r[N],d[N];
int col[N],row[N];
int h[505],s[905];
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,row[sz]=x;
    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)
{
    l[r[y]]=l[y];
    r[l[y]]=r[y];
    for(int i=d[y]; i!=y; i=d[i])
    {
        for(int j=r[i]; j!=i; j=r[j])
        {
            u[d[j]]=u[j];
            d[u[j]]=d[j];
            --s[col[j]];
        }
    }
}
void resume(int y)
{
    for(int i=u[y]; i!=y; i=u[i])
    {
        for(int j=l[i]; j!=i; j=l[j])
        {
            d[u[j]]=u[d[j]]=j;
            ++s[col[j]];
        }
    }
    r[l[y]]=l[r[y]]=y;
}
int ans;
void dance(int pos)
{
    if(pos>=ans)return;
    if(!r[0])
    {
        if(pos<ans)ans=pos;
        return;
    }
    int t=r[0];
    for(int i=r[0]; i!=0; i=r[i])
        if(s[i]<s[t])t=i;
    del(t);
    for(int i=d[t]; i!=t; i=d[i])
    {
        for(int j=r[i]; j!=i; j=r[j])
            del(col[j]);
        dance(pos+1);
        for(int j=l[i]; j!=i; j=l[j])
            resume(col[j]);
    }
    resume(t);
}
int main()
{
    int T,n1,m1,p;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n1,&m1,&p);
        n=p;
        m=n1*m1;
        init();
        int x1,x2,y1,y2;
        for(int i=1; i<=p; ++i)
        {
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            for(int j=x1+1; j<=x2; ++j)
                for(int k=y1+1; k<=y2; ++k)
                    link(i,(k-1)*n1+j);
        }
        ans=505;
        dance(0);
        if(ans==505)printf("-1\n");
        else printf("%d\n",ans);
    }
    return 0;
}

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