hdu3656Fire station(DLX重复覆盖 + 二分)

题目请戳这里

题目大意：一个城市n个点，现在要建m个消防站，消防站建在给定的n个点中。求建m个消防站后，m个消防站要覆盖所有的n个点的覆盖半径最小。

题目分析：重复覆盖问题，DLX解决。不过要求覆盖半径最小，需要二分。虽然给的范围并不大，DLX毕竟还是暴力搜索，而且精度有6位小数，因此直接二分距离的话会TLE！解决方案是将图中任意2点的距离记录下来，去重后二分已知的距离。因为消防站建在给定的n个点中，那么最小覆盖半径一定在任意2点距离中产生。

DLX搜索的时候，一般习惯删除的时候从左往右，不过效率却不一定高。比如这题，从左向右删除和从右向左删除，跑的时间至少差了1s以上！可见用dancing links搜索的时候姿势还是很重要的。有时候换个姿势也许效率更高~

详情请见代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N = 51;
const int M = 30001;
const double eps = 1e-7;

double dis[N][N];
double tle[M];
int s[M],h[M],u[M],d[M],l[M],r[M],col[M],row[M];
int point[N][2];
int m,n,num,len;
int fs;
double getdis(int i,int j)
{
    return sqrt((double)(point[i][0] - point[j][0])*(point[i][0] - point[j][0])
               +(double)(point[i][1] - point[j][1])*(point[i][1] - point[j][1]));
}
void read()
{
    scanf("%d%d",&n,&m);
    int i,j;
    len = 1;
    tle[len ++] = 0;
    for(i = 1;i <= n;i ++)
    {
        dis[i][i] = 0.0;
        scanf("%d%d",&point[i][0],&point[i][1]);
        for(j = 1;j < i;j ++)
            dis[i][j] = dis[j][i] = getdis(i,j),tle[len ++] = dis[i][j];
    }
}
void init()
{
    memset(h,0,sizeof(h));
    memset(s,0,sizeof(s));
    for(int i = 0;i <= n;i ++)
    {
        u[i] = d[i] = i;
        l[i] = (i + n) % (n + 1);
        r[i] = (i + 1) % (n + 1);
    }
    num = n + 1;
}
void add(int i,int j)
{
    if(h[i])
    {
        r[num] = h[i];
        l[num] = l[h[i]];
        r[l[num]] = l[r[num]] = num;
    }
    else
        h[i] = l[num] = r[num] = num;
    s[j] ++;
    u[num] = u[j];
    d[num] = j;
    d[u[num]] = num;
    u[j] = num;
    col[num] = j;
    row[num] = i;
    num ++;
}
void build(double md)
{
    int i,j;
    init();
    for(i = 1;i <= n;i ++)
        for(j = 1;j <= n;j ++)
            if(md - dis[i][j] > -eps)
                add(i,j);
}
void remove(int c)
{
    for(int i = d[c];i != c;i = d[i])
        l[r[i]] = l[i],r[l[i]] = r[i],s[col[i]] --;
}
void resume(int c)
{
    for(int i = u[c];i != c;i = u[i])
        l[r[i]] = r[l[i]] = i,s[col[i]] ++;
}
int A()
{
    int i,j,k,ret = 0;
    bool vis[N];
    memset(vis,false,sizeof(vis));
    for(i = l[0];i;i = l[i])
    {
        if(vis[i] == false)
        {
            vis[i] = true;
            ret ++;
            for(j = d[i];j != i;j = d[j])
                for(k = r[j];k != j;k = r[k])
                    vis[col[k]] = true;
        }
    }
    return ret;
}
void dfs(int k)
{
    if(k + A() >= fs)
        return;
    int i,j;
    if(!r[0])
    {
        fs = min(fs,k);
        return;
    }
    int mn = 1000000;
    int c;
    for(i = l[0];i;i = l[i])
    {
        if(mn > s[i])
        {
            mn = s[i];
            c = i;
        }
    }
    for(i = d[c];i != c;i = d[i])
    {
        remove(i);
        for(j = l[i];j != i;j = l[j])
        {
            remove(j);
        }
        dfs(k + 1);
        for(j = r[i];j != i;j = r[j])
        {
            resume(j);
        }
        resume(i);
    }
}
void solve()
{
    int la,ra,mid,ans;
    sort(tle + 1,tle + len);
    int i = len;
    int j;
    len = 2;
    for(j = 2;j < i;j ++)
        if(fabs(tle[j] - tle[j - 1]) > eps)
            tle[len ++] = tle[j];
    len --;
    la = 1;ra = len;
    while(la <= ra)
    {
        mid = (la + ra)>>1;
        build(tle[mid]);
        fs = M;
        dfs(0);
        if(fs <= m)
        {
            ans = mid;
            ra = mid - 1;
        }
        else
            la = mid + 1;
    }
    printf("%f
",tle[ans]);
}
int main()
{
    int _;
    scanf("%d",&_);
    while(_ --)
    {
        read();
        solve();
    }
    return 0;
}
//1953MS  636K