题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5046
题意：
给定n个城市的坐标，要在城市中建k个飞机场，使城市距离最近的飞机场的最长距离最小，求这个最小距离。
分析：
最小化最大值，显然二分最大距离。然后我们将距离在范围内的两个城市建边，看能否选出k个城市，使得覆盖了所有城市。
将点之间的关系转化成01矩阵的覆盖问题，重复覆盖，建好边套个DLX即可。
看了鸟神博客，这里可以直接将所有距离保存在一个数组中，排序并去重，二分下标即可。这样快了很多很多！
hdu 2295 和这题一个套路，更裸一些。
跳舞链好强大，可惜只会用模板，这个讲的还挺清晰
代码：
/*************************************************************************
> File Name: O.cpp
> Author: jiangyuzhu
> Mail: 834138558@qq.com
> Created Time: Fri 08 Jul 2016 03:31:58 PM CST
************************************************************************/
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
#define sal(n) scanf("%I64d", &(n))
#define sa(n) scanf("%d", &(n))
typedef long long ll;
const int maxn = 3600 + 5, maxc = 60 + 5, maxr = 60 + 5, inf = 0x3f3f3f3f;
struct Node{ll x; ll y;}city[maxn];
int L[maxn], R[maxn], D[maxn], U[maxn], C[maxn];
int S[maxc], H[maxr], size;
int n, m, k;
ll d[maxr][maxc], t[maxn];
///不需要S域
void Link(int r, int c)
{
S[c]++; C[size] = c;
U[size] = U[c]; D[U[c]] = size;
D[size] = c; U[c] = size;
U[c] = size;
if(H[r] == -1) H[r] = L[size] = R[size] = size;
else {
L[size] = L[H[r]]; R[L[H[r]]] = size;
R[size] = H[r]; L[H[r]] = size;
}
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;
}
int h()
{///用精确覆盖去估算剪枝
int ret = 0;
bool vis[maxc];
memset (vis, false, sizeof(vis));
for(int i = R[0]; i; i = R[i]){
if(vis[i]) continue;
ret++;
vis[i] = true;
for (int j = D[i]; j != i; j = D[j])
for (int k = R[j]; k != j; k = R[k])
vis[C[k]] = true;
}
return ret;
}
int ans;
bool Dance(int a) //具体问题具体分析
{
if(a + h() > k) return 0;
if(!R[0]) return a <= k;
int c = R[0];
for (int i = R[0]; i; 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(a + 1)) return 1;
for (int j = L[i]; j != i; j = L[j])
resume(j);
resume(i);
}
return 0;
}
void initL(int x)///col is 1~x,row start from 1
{
for (int i = 0; i <= x; ++i){
S[i] = 0;
D[i] = U[i] = i;
L[i+1] = i; R[i] = i+1;
}///对列表头初始化
R[x] = 0;
size = x + 1;///真正的元素从m+1开始
memset (H, -1, sizeof(H));
///mark每个位置的名字
}
ll dist(int i, int j)
{
ll d = abs(city[i].x - city[j].x) + abs(city[i].y - city[j].y);
return d;
}
bool judge(ll mid)
{
initL(n);
ans = 0x3f3f3f3f;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
if(d[i][j] <= mid){
Link(i, j);
}
}
}
return Dance(0);
}
int main (void)
{
int T;sa(T);
for(int tt = 1; tt <= T; tt++){
sa(n);sa(k);
ll maxd = 0;
int cnt = 0;
for(int i = 1; i <= n; i++){
scanf("%I64d%I64d", &city[i].x, &city[i].y);
}
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
d[i][j] = dist(i, j);
t[cnt++] = d[i][j];
}
}
sort(t, t + cnt);
int ncnt = unique(t, t + cnt) - t;
ll l = -1, r = ncnt;
while(r - l > 1){
ll mid = (l + r) / 2;
if(judge(t[mid])) r = mid;
else l = mid;
}
printf("Case #%d: %I64d\n", tt, t[r]);
}
return 0;
}