hdu3656Fire station(DLX重复覆盖 + 二分)
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\n",tle[ans]);
}
int main()
{
int _;
scanf("%d",&_);
while(_ --)
{
read();
solve();
}
return 0;
}
//1953MS 636K