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 = 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[n] = 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;

}

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