HDU2295 Radar —— Dancing Links 可重复覆盖

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2295

Radar

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4106 Accepted Submission(s): 1576

Problem Description

N cities of the Java Kingdom need to be covered by radars for being in a state of war. Since the kingdom has M radar stations but only K operators, we can at most operate K radars. All radars have the same circular coverage with a radius of R. Our goal is to minimize R while covering the entire city with no more than K radars.

Input

The input consists of several test cases. The first line of the input consists of an integer T, indicating the number of test cases. The first line of each test case consists of 3 integers: N, M, K, representing the number of cities, the number of radar stations and the number of operators. Each of the following N lines consists of the coordinate of a city.
Each of the last M lines consists of the coordinate of a radar station.

All coordinates are separated by one space.
Technical Specification

1. 1 ≤ T ≤ 20
2. 1 ≤ N, M ≤ 50
3. 1 ≤ K ≤ M
4. 0 ≤ X, Y ≤ 1000

Output

For each test case, output the radius on a single line, rounded to six fractional digits.

Sample Input

Sample Output

2.236068

Source

The 4th Baidu Cup final

题解：

超时方法：

1.对于DLX的矩阵：行代表着雷达与城市的距离，列代表着城市。矩阵大小250*50。

2.Dancing跳起来，当R[0]==0时，取当前所选行中，距离的最大值dis（这样才能覆盖掉所有城市），然后再更新答案ans，ans = min（ans， dis）。

3.结果矩阵有点大，超时了。

4.错误思想分析：把雷达与城市的距离作为行，实际上是太明智的。因为题目说明了每个雷达的接收半径是相同的，而以上方法选出来的每个雷达的接收半径是相异的，然后又再取最大值，那为何不每次都取最大值（相同值）呢？如果取相同值，那么行就是雷达，列就是城市，矩阵的大小就减少了。但是又怎么确定雷达的接收半径呢？如下：

正确方法：

1.雷达作为行，城市作为列。

2.二分雷达的接收范围，每次二分都：根据接收半径更新矩阵中所含的元素，然后再进行一次Dance()，如果能覆盖掉所有城市，则缩小半径，否则扩大半径。

超时方法：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <queue>
  8 #include <stack>
  9 #include <map>
 10 #include <string>
 11 #include <set>
 12 #define ms(a,b) memset((a),(b),sizeof((a)))
 13 using namespace std;
 14 typedef long long LL;
 15 const int INF = 2e9;
 16 const int MAXN = 50+10;
 17 const int MAXM = 50+10;
 18 const int maxnode = 1e5+10;
 19 
 20 double city[MAXN][2], radar[MAXN][2];
 21 double r[MAXN*MAXM];
 22 int k;
 23 
 24 struct DLX
 25 {
 26     int n, m, size;
 27     int U[maxnode], D[maxnode], L[maxnode], R[maxnode], Row[maxnode], Col[maxnode];
 28     int H[MAXN*MAXM], S[MAXN*MAXM];
 29     double ansd, ans[MAXN*MAXM];
 30 
 31     void init(int _n, int _m)
 32     {
 33         n = _n;
 34         m = _m;
 35         for(int i = 0; i<=m; i++)
 36         {
 37             S[i] = 0;
 38             U[i] = D[i] = i;
 39             L[i] = i-1;
 40             R[i] = i+1;
 41         }
 42         R[m] = 0; L[0] = m;
 43         size = m;
 44         for(int i = 1; i<=n; i++) H[i] = -1;
 45     }
 46 
 47     void Link(int r, int c)
 48     {
 49         size++;
 50         Row[size] = r;
 51         Col[size] = c;
 52         S[Col[size]]++;
 53         D[size] = D[c];
 54         U[D[c]] = size;
 55         U[size] = c;
 56         D[c] = size;
 57         if(H[r]==-1) H[r] = L[size] = R[size] = size;
 58         else
 59         {
 60             R[size] = R[H[r]];
 61             L[R[H[r]]] = size;
 62             L[size] = H[r];
 63             R[H[r]] = size;
 64         }
 65     }
 66 
 67     void remove(int c)
 68     {
 69         for(int i = D[c]; i!=c; i = D[i])
 70             L[R[i]] = L[i], R[L[i]] = R[i];
 71     }
 72 
 73     bool v[MAXM];
 74     int f()
 75     {
 76         int ret = 0;
 77         for(int c = R[0]; c!=0; c = R[c])
 78             v[c] = true;
 79         for(int c = R[0]; c!=0; c = R[c])
 80         if(v[c])
 81         {
 82             ret++;
 83             v[c] = false;
 84             for(int i = D[c]; i!=c; i = D[i])
 85                 for(int j = R[i]; j!=i; j = R[j])
 86                     v[Col[j]] = false;
 87         }
 88         return ret;
 89     }
 90 
 91     void resume(int c)
 92     {
 93         for(int i = U[c]; i!=c; i = U[i])
 94             L[R[i]] = R[L[i]] = i;
 95     }
 96 
 97     void Dance(int d)
 98     {
 99         if(d+f()>k) return;
100         if(R[0]==0)
101         {
102             double tmp = -1.0;
103             for(int i = 0; i<d; i++)
104                 tmp = max(tmp, ans[i]);
105             ansd = min(tmp, ansd);
106             return;
107         }
108 
109         int c = R[0];
110         for(int i = R[0]; i!=0; i = R[i])
111             if(S[i]<S[c])
112                 c = i;
113         for(int i = D[c]; i!=c; i = D[i])
114         {
115             ans[d] = r[Row[i]];
116             remove(i);
117             for(int j = R[i]; j!=i; j = R[j]) remove(j);
118             Dance(d+1);
119             for(int j = L[i]; j!=i; j = L[j]) resume(j);
120             resume(i);
121         }
122     }
123 };
124 
125 double dis(double x1, double y1, double x2, double y2)
126 {
127     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
128 }
129 
130 DLX dlx;
131 int main()
132 {
133     int T;
134     int n, m;
135     scanf("%d", &T);
136     while(T--)
137     {
138         scanf("%d%d%d", &n, &m, &k);
139         dlx.init(n*m, n);
140         for(int i = 1; i<=n; i++)
141             scanf("%lf%lf",&city[i][0], &city[i][1]);
142         for(int i = 1; i<=m; i++)
143             scanf("%lf%lf",&radar[i][0], &radar[i][1]);
144 
145         for(int i = 1; i<=m; i++)
146         for(int j = 1; j<=n; j++)
147         {
148             double tmp = dis(radar[i][0], radar[i][1], city[j][0], city[j][1]);
149             r[(i-1)*n+j] = tmp;
150             for(int t = 1; t<=n; t++)
151                 if(dis(radar[i][0], radar[i][1], city[t][0], city[t][1])<=tmp)
152                     dlx.Link((i-1)*n+j, t);
153         }
154         dlx.ansd = 1.0*INF;
155         dlx.Dance(0);
156         printf("%.6f
", dlx.ansd);
157     }
158     return 0;
159 }

View Code

正确方法：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <queue>
  8 #include <stack>
  9 #include <map>
 10 #include <string>
 11 #include <set>
 12 #define ms(a,b) memset((a),(b),sizeof((a)))
 13 using namespace std;
 14 typedef long long LL;
 15 const int INF = 2e9;
 16 const double EPS = 1e-8;
 17 const int MAXN = 50+10;
 18 const int MAXM = 50+10;
 19 const int maxnode = 1e5+10;
 20 
 21 double city[MAXN][2], radar[MAXN][2];
 22 double r[MAXN*MAXM];
 23 int k;
 24 
 25 struct DLX
 26 {
 27     int n, m, size;
 28     int U[maxnode], D[maxnode], L[maxnode], R[maxnode], Row[maxnode], Col[maxnode];
 29     int H[MAXN*MAXM], S[MAXN*MAXM];
 30     double ansd, ans[MAXN*MAXM];
 31 
 32     void init(int _n, int _m)
 33     {
 34         n = _n;
 35         m = _m;
 36         for(int i = 0; i<=m; i++)
 37         {
 38             S[i] = 0;
 39             U[i] = D[i] = i;
 40             L[i] = i-1;
 41             R[i] = i+1;
 42         }
 43         R[m] = 0; L[0] = m;
 44         size = m;
 45         for(int i = 1; i<=n; i++) H[i] = -1;
 46     }
 47 
 48     void Link(int r, int c)
 49     {
 50         size++;
 51         Row[size] = r;
 52         Col[size] = c;
 53         S[Col[size]]++;
 54         D[size] = D[c];
 55         U[D[c]] = size;
 56         U[size] = c;
 57         D[c] = size;
 58         if(H[r]==-1) H[r] = L[size] = R[size] = size;
 59         else
 60         {
 61             R[size] = R[H[r]];
 62             L[R[H[r]]] = size;
 63             L[size] = H[r];
 64             R[H[r]] = size;
 65         }
 66     }
 67 
 68     void remove(int c)
 69     {
 70         for(int i = D[c]; i!=c; i = D[i])
 71             L[R[i]] = L[i], R[L[i]] = R[i];
 72     }
 73 
 74     void resume(int c)
 75     {
 76         for(int i = U[c]; i!=c; i = U[i])
 77             L[R[i]] = R[L[i]] = i;
 78     }
 79 
 80     bool v[MAXM];
 81     int f()
 82     {
 83         int ret = 0;
 84         for(int c = R[0]; c!=0; c = R[c])
 85             v[c] = true;
 86         for(int c = R[0]; c!=0; c = R[c])
 87         if(v[c])
 88         {
 89             ret++;
 90             v[c] = false;
 91             for(int i = D[c]; i!=c; i = D[i])
 92                 for(int j = R[i]; j!=i; j = R[j])
 93                     v[Col[j]] = false;
 94         }
 95         return ret;
 96     }
 97 
 98     bool Dance(int d)
 99     {
100         if(d+f()>k) return false;
101         if(R[0]==0) return true;
102 
103         int c = R[0];
104         for(int i = R[0]; i!=0; i = R[i])
105             if(S[i]<S[c]) c = i;
106         for(int i = D[c]; i!=c; i = D[i])
107         {
108             remove(i);
109             for(int j = R[i]; j!=i; j = R[j]) remove(j);
110             if(Dance(d+1))return true;
111             for(int j = L[i]; j!=i; j = L[j]) resume(j);
112             resume(i);
113         }
114         return false;
115     }
116 };
117 
118 double dis(double x1, double y1, double x2, double y2)
119 {
120     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
121 }
122 
123 DLX dlx;
124 int main()
125 {
126     int T;
127     int n, m;
128     scanf("%d", &T);
129     while(T--)
130     {
131         scanf("%d%d%d", &n, &m, &k);
132         for(int i = 1; i<=n; i++)
133             scanf("%lf%lf",&city[i][0], &city[i][1]);
134         for(int i = 1; i<=m; i++)
135             scanf("%lf%lf",&radar[i][0], &radar[i][1]);
136 
137         double l = 0.0, r = 2000.0;
138         while(l+EPS<=r)
139         {
140             double mid = (l+r)/2;
141             dlx.init(m, n);
142             for(int i = 1; i<=m; i++)
143             for(int j = 1; j<=n; j++)
144                 if(dis(radar[i][0], radar[i][1], city[j][0], city[j][1])<=mid)
145                     dlx.Link(i, j);
146             if(dlx.Dance(0))
147                 r = mid - EPS;
148             else
149                 l = mid + EPS;
150         }
151         printf("%.6lf
",l);
152     }
153     return 0;
154 }

View Code

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原文地址：https://www.cnblogs.com/DOLFAMINGO/p/7544669.html