HDU 2295 Radar【二分+Dancing Links重复覆盖】

重复覆盖不同于精确覆盖，要求是在0/1矩阵中选择最少的行使每一列至少有一个1。进行重复覆盖时要使用估价函数来剪枝。

这题二分雷达半径，找到能将所有城市覆盖的最小半径即可。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath> using namespace std;const int maxn = 50*50 + 10;const int oo = 1 << 30;const int maxrow = 55;const int maxcol = 55;int mtx[maxrow][maxcol];double dis[55][55];double cx[55], cy[55];double rx[55], ry[55];int t, n, m, k;int totRow, totCol, head, idx;int L[maxn], R[maxn], U[maxn], D[maxn];int RH[maxn], CH[maxn], S[maxn];void initMtx(double x){    memset(mtx, 0, sizeof(mtx));    for (int i = 0; i < m; ++i) {        for (int j = 0; j < n; ++j) {            if (dis[i][j] <= x) {                mtx[i][j] = 1;            }        }    }}int newNode(int up, int down, int left, int right){    U[idx] = up;    D[idx] = down;    L[idx] = left;  R[idx] = right;    U[down] = D[up] = L[right] = R[left] = idx;    return idx++;}void build(){    idx = maxn - 1;    head = newNode(idx, idx, idx, idx);    idx = 0;    for (int j = 0; j < totCol; ++j) {        newNode(idx, idx, L[head], head);        CH[j] = j;  S[j] = 0;    }    for (int i = 0; i < totRow; ++i) {        int k = -1;        for (int j = 0; j < totCol; ++j) {            if (!mtx[i][j]) continue;            if (-1 == k) {                k = newNode(U[CH[j]], CH[j], idx, idx);                RH[k] = i;  CH[k] = j;  S[j]++;            } else {                k = newNode(U[CH[j]], CH[j], k, R[k]);                RH[k] = i;  CH[k] = j;  S[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[CH[i]]--;*/    }}void resume(int c){    for (int i = U[c]; i != c; i = U[i]) {        L[R[i]] = R[L[i]] = i; /*S[CH[i]]++;*/     }}/*估价函数*/int h(){    bool vis[maxcol];    memset(vis, false, sizeof(vis));    int ret = 0;    for (int i = R[head]; i != head; i = R[i]) {        if (!vis[i]) {            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[CH[k]] = true;                }            }        }    }    return ret;}bool dance(int cnt){    if (cnt + h() > k) {        return false;    }    if (R[head] == head) {        return true;    }    int c, Min = oo;    for (int i = R[head]; i != head; i = R[i]) {        if (S[i] < Min) {            Min = S[i]; 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(cnt + 1)) {            return 1;        }         for (int j = L[i]; j != i; j = L[j]) {            resume(j);        }        resume(i);    }    return false; }int main(){    scanf("%d", &t);    while (t--) {        scanf("%d%d%d", &n, &m, &k);        totRow = m; totCol = n;        for (int i = 0; i < n; ++i) {            scanf("%lf%lf", &cx[i], &cy[i]);        }        for (int i = 0; i < m; ++i) {            scanf("%lf%lf", &rx[i], &ry[i]);         }        double l = 1e9, r = -1.0;        for (int i = 0; i < m; ++i) {            for (int j = 0; j < n; ++j) {                dis[i][j] = sqrt((rx[i] - cx[j])*(rx[i] - cx[j]) + (ry[i] - cy[j])*(ry[i] - cy[j]));                 l = min(dis[i][j], l);                r = max(dis[i][j], r);            }        }                while (r - l > 1e-7) {            double mid = (r + l) / 2.0;            initMtx(mid);            build();            if (dance(0)) {                r = mid;            } else {                l = mid;            }        }        printf("%.6lf\n", l);    }    return 0;}