HDU 2295.Radar (DLX重复覆盖)

2分答案+DLX判断可行

不使用的估计函数的可重复覆盖的搜索树将十分庞大

#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 = , MAXR = , MAXC = ;
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 = ; i <= n; i++)
            U[i] = D[i] = i, L[i] = i - , R[i] = i + ;
        R[n] = , L[] = n; //头尾相接
        Size = n + ;
        memset (S, , sizeof S);
    }
    //逐行添加
    void addRow (int r, int columns[]) {
        int first = Size;
        for (int i = ; i <= n ; i++) {
            if (columns[i] == ) continue;
            int c = i;
            L[Size] = Size - , R[Size] = Size + ;
            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 - ] = first, L[first] = Size - ; //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 = ;
        for (int c = R[]; c != ; c = R[c]) v[c] = true;
        for (int c = R[]; c != ; 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 ;
        if (R[] == ) {
            ncnt = d;
            return d <= k;
        }
        int c = R[];
        for (int i = R[]; i != ; 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 + ) ) return ;
            //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 ;
    }
    bool solve (vector<int> &v) {
        v.clear();
        if (!dfs () ) return ;
        for (int i = ; i < ncnt; i++) v.push_back (ans[i]);
        return ;
    }
} f;
struct node {
    int x, y;
} g[], Ra[];
int columns[][];
double dis[][];
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 = ; i <= m; i++)
        for (int j = ; j <= n; j++)
            columns[i][j] = dis[i][j] <= mid;
    for (int i = ; i <= m; i++)
        f.addRow (i, columns[i]);
    return f.dfs ();
}
int main() {
    scanf ("%d", &t);
    while (t--) {
        scanf ("%d %d %d", &n, &m, &k);
        for (int i = ; i <= n; i++)
            scanf ("%d %d", &g[i].x, &g[i].y);
        for (int i = ; i <= m; i++)
            scanf ("%d %d", &Ra[i].x, &Ra[i].y);
        double l = 1e9, r = ;
        for (int i = ; i <= m; i++)
            for (int j = ; 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 = -;
        while (r - l > 1e-) {
            double mid = (r + l) / .;
            if (make (mid) ) {
                ans = mid;
                r = mid - exp;
            }
            else
                l = mid + exp;
        }
        printf ("%.6f\n", ans);
    }
    return ;
}