HDU 2295.Radar (DLX重复覆盖)

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <vector>
using namespace std;

#define FOR(i,A,s)  for(int i = A[s]; i != s; i = A[i])
#define exp 1e-8

const int MAX = 111111, MAXR = 1111, MAXC = 1111;
int n, m, k, t;

struct DLX {
int n, Size;//Size为尾指针，真正大小
int row[MAX], col[MAX];//记录每个点的行列
int U[MAX], D[MAX], R[MAX], L[MAX]; //4个链表
int S[MAXC];//每列1的个数
int ncnt, ans[MAXR];
void init (int n) {
this->n = n;
//增加n+1个辅助链表，从0到n
for (int i = 0; i <= n; i++)
U[i] = D[i] = i, L[i] = i - 1, R[i] = i + 1;
R
= 0, L[0] = n; //头尾相接
Size = n + 1;
memset (S, 0, sizeof S);
}
//逐行添加
void addRow (int r, int columns[111]) {
int first = Size;
for (int i = 1; i <= n ; i++) {
if (columns[i] == 0) continue;
int c = i;
L[Size] = Size - 1, R[Size] = Size + 1;
U[Size] = U[c], D[Size] = c;//插入第c列
D[U[c]] = Size, U[c] = Size; //注意顺序!!!
row[Size] = r, col[Size] = c;
Size++, S[c]++;
}
if (Size > first)
R[Size - 1] = first, L[first] = Size - 1; //头尾相接
}
void Remove (int c) {
//精确覆盖
//        L[R[c]] = L[c], R[L[c]] = R[c];
//        FOR (i, D, c)
//                     FOR (j, R, i)
//                            U[D[j]] = U[j], D[U[j]] = D[j], --S[col[j]];
//重复覆盖
for (int i = D[c]; i != c; i = D[i])
L[R[i]] = L[i], R[L[i]] = R[i];
}
void Restore (int c) {
//        FOR (i, U, c)
//                     FOR (j, L, i)
//                            ++S[col[j]], U[D[j]] = j, D[U[j]] = j;
//        L[R[c]] = c, R[L[c]] = c;
//重复覆盖
for (int i = U[c]; i != c; i = U[i])
L[R[i]] = R[L[i]] = i;
}
bool v[MAX];
int ff()
{
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 dfs (int d) {
if (d + ff() > k) return 0;
if (R[0] == 0) {
ncnt = d;
return d <= k;
}
int c = R[0];
for (int i = R[0]; i != 0; i = R[i])
if (S[i] < S[c])
c = i;
//Remove (c);//精确覆盖
FOR (i, D, c) {
Remove (i);//重复覆盖
ans[d] = row[i];
//FOR (j, R, i) Remove (col[j]);
FOR (j, R, i) Remove (j);
if (dfs (d + 1) ) return 1;
//FOR (j, L, i) Restore (col[j]);
FOR (j, L, i) Restore (j);
Restore (i);//重复覆盖
}
//Restore (c);//精确覆盖
return 0;
}
bool solve (vector<int> &v) {
v.clear();
if (!dfs (0) ) return 0;
for (int i = 0; i < ncnt; i++) v.push_back (ans[i]);
return 1;
}
} f;
struct node {
int x, y;
} g[100], Ra[100];
int columns[111][111];
double dis[111][111];
inline double getdis (node a, node b) {
return sqrt (double ( (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y) ) );
}
bool make (double mid) {
f.init (n);
for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++)
columns[i][j] = dis[i][j] <= mid;
for (int i = 1; i <= m; i++)
f.addRow (i, columns[i]);
return f.dfs (0);
}
int main() {
scanf ("%d", &t);
while (t--) {
scanf ("%d %d %d", &n, &m, &k);
for (int i = 1; i <= n; i++)
scanf ("%d %d", &g[i].x, &g[i].y);
for (int i = 1; i <= m; i++)
scanf ("%d %d", &Ra[i].x, &Ra[i].y);
double l = 1e9, r = 0;
for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++) {
dis[i][j]  = getdis (Ra[i], g[j]);
l = min (dis[i][j], l), r = max (r, dis[i][j]);
}
double ans = -1;
while (r - l > 1e-7) {
double mid = (r + l) / 2.;
if (make (mid) ) {
ans = mid;
r = mid - exp;
}
else
l = mid + exp;
}
printf ("%.6f\n", ans);
}
return 0;
}