hdu2295（DLX重复覆盖）

来源：互联网发布：linux如何修改用户组编辑：程序博客网时间：2024/05/21 14:41
很容易想到，二分答案。
每次根据半径建好0，1矩阵。
然后就是DLX可重复覆盖的模板
#include<cstdio>#include<cstring>#include<cmath>double x[55],y[55],X[55],Y[55];const int MN=1005;const int MM=1005;const int MNN=1e5+5+MM; //最大点数const double eps = 1e-9;int n,m,k;struct DLX{    int n,m,size;    int U[MM],D[MM],R[MM],L[MM],Row[MM],Col[MM];    int H[MM],S[MM];    int ands,ans[MM];    void init(int _n,int _m)    {        n = _n;        m = _m;        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;        size = m;        for(int i = 1;i <= n;i++)            H[i] = -1;    }    void Link(int r,int c)    {        ++S[Col[++size]=c];        Row[size] = r;        D[size] = D[c];        U[D[c]] = size;        U[size] = c;        D[c] = size;        if(H[r] < 0)H[r] = L[size] = R[size] = size;        else        {            R[size] = R[H[r]];            L[R[H[r]]] = size;            L[size] = H[r];            R[H[r]] = size;        }    }    void remove(int c)    {        for(int i = D[c];i != c;i = D[i])            L[R[i]] = L[i], R[L[i]] = R[i];    }    void resume(int c)    {        for(int i = U[c];i != c;i = U[i])            L[R[i]]=R[L[i]]=i;    }    bool v[MM];    int f()    {        int ret = 0;        for(int c = R[0];c != 0;c = R[c])v[c] = true;        for(int c = R[0];c != 0;c = R[c])            if(v[c])            {                ret++;                v[c] = false;                for(int i = D[c];i != c;i = D[i])                    for(int j = R[i];j != i;j = R[j])                        v[Col[j]] = false;            }        return ret;    }    bool Dance(int d)    {        if(d + f() > k)return false;        if(R[0] == 0)return d <= k;        int c = R[0];        for(int i = R[0];i != 0;i = R[i])            if(S[i] < S[c])                c = i;        for(int i = D[c];i != c;i = D[i])        {            remove(i);            for(int j = R[i];j != i;j = R[j])remove(j);            if(Dance(d+1))return true;            for(int j = L[i];j != i;j = L[j])resume(j);            resume(i);        }        return false;    }}dlx;double cla(int a,int b){    return sqrt((X[a]-x[b])*(X[a]-x[b])+(Y[a]-y[b])*(Y[a]-y[b]));}bool check(double mid){    dlx.init(m,n);    for(int i=1;i<=m;i++)    {        for(int j=1;j<=n;j++)        {            if(cla(i,j)-mid<=eps)                dlx.Link(i,j);        }    }    dlx.ands = -1;    if(dlx.Dance(0)) return true;    return false;}int main(){    int cases;    scanf("%d",&cases);    while(cases--)    {        scanf("%d%d%d",&n,&m,&k);        for(int i=1;i<=n;i++)            scanf("%lf%lf",&x[i],&y[i]);        for(int i=1;i<=m;i++)            scanf("%lf%lf",&X[i],&Y[i]);        double l = 0,r = 10000000,mid;        for(int i=0;i<100;i++)        {            mid = (l+r)/2;            bool flag = check(mid);            //printf("%f %d\n",mid,flag);            if(flag) r = mid-eps;            else l = mid+eps;        }        printf("%.6lf\n",r);    }    return 0;}
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