tju 3219& hdu2295(Dancing Links重复覆盖模板题)

3219.    Radar Time Limit: 1.0 Seconds    Memory Limit: 65536K
Total Runs: 553    Accepted Runs: 157

N cities of the Java Kingdom need to be covered by radars for being in a state of war. Since the kingdom has  M radar stations but only  K operators, we can at most operate  K radars. All radars have the same circular coverage with a radius of  R. Our goal is to minimize  R while covering the entire city with no more than  K radars.

Input

The input consists of several test cases. The first line of the input consists of an integer  T, indicating the number of test cases. The first line of each test case consists of 3 integers:  N, M, K, representing the number of cities, the number of radar stations and the number of operators. Each of the following  N lines consists of the coordinate of a city.

Each of the last M lines consists of the coordinate of a radar station.

All coordinates are separated by one space.

Technical Specification

1. 1 ≤ T ≤ 20
2. 1 ≤ N, M ≤ 50
3. 1 ≤ K ≤ M
4. 0 ≤ X, Y ≤ 1000

Output

For each test case, output the radius on a single line, rounded to six fractional digits.

Sample Input

1
3 3 2
3 4
3 1
5 4
1 1
2 2
3 3

Sample Output

2.236068

Source: The 4th Baidu Cup Central China Invitational Programming Contest

题目：http://acm.tju.edu.cn/acm/showp3219.html

hdu上一样的题：http://acm.hdu.edu.cn/showproblem.php?pid=2295

分析：这题很明显可以转化为取最少的雷达，使得所有的城市都被覆盖，可以重复覆盖，只要二分半径，敲个重复覆盖的DLX。。。

最近眼力不行了，读入写错，找了一个小时。。。

代码：

#include<cstdio>
#define mm 3333
#define mn 55
#define dis(a,b) ((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))
struct place{int x,y;}city[mn],radar[mn];
int U[mm],D[mm],L[mm],R[mm],C[mm];
int H[mn],S[mn],d[mn][mn];
bool v[mn];
int m,n,K,size;
void prepare(int r,int c)
{
    for(int i=0;i<=c;++i)
    {
        S[i]=0;
        U[i]=D[i]=i;
        L[i+1]=i;
        R[i]=i+1;
    }
    R[c]=0;
    size=c;
    while(r)H[r--]=-1;
}
int f()
{
    int i,j,c,ret=0;
    for(c=R[0];c;c=R[c])v[c]=1;
    for(c=R[0];c;c=R[c])
        if(v[c])for(++ret,v[c]=0,i=D[c];i!=c;i=D[i])
            for(j=R[i];j!=i;j=R[j])v[C[j]]=0;
    return ret;
}
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 Dance(int k)
{
    if(k+f()>K)return 0;
    if(!R[0])return(k<=K);
    int i,j,c,tmp=mm;
    for(i=R[0];i;i=R[i])
        if(S[i]<tmp)tmp=S[c=i];
    for(i=D[c];i!=c;i=D[i])
    {
        remove(i);
        for(j=R[i];j!=i;j=R[j])remove(j);
        if(Dance(k+1))return 1;
        for(j=L[i];j!=i;j=L[j])resume(j);
        resume(i);
    }
    return 0;
}
void Link(int r,int c)
{
    ++S[C[++size]=c];
    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;
    }
}
int main()
{
    int i,j,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&K);
        for(i=1;i<=n;++i)scanf("%d%d",&city[i].x,&city[i].y);
        for(i=1;i<=m;++i)scanf("%d%d",&radar[i].x,&radar[i].y);
        double l=0,r=0,cur;
        for(i=1;i<=m;++i)
            for(j=1;j<=n;++j)
            {
                d[i][j]=dis(radar[i],city[j]);
                if(d[i][j]>r)r=d[i][j];
            }
        while(r-l>1e-7)
        {
            cur=(l+r)/2.0;
            prepare(m,n);
            for(i=1;i<=m;++i)
                for(j=1;j<=n;++j)
                    if(d[i][j]<=cur*cur)Link(i,j);
            if(Dance(0))r=cur;
            else l=cur;
        }
        printf("%.6lf\n",l);
    }
    return 0;
}