2014
01-05

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.

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

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

1
3 3 2
3 4
3 1
5 4
1 1
2 2
3 3

2.236068

/*最小支配集*/
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#define INF 0x3fffffff
#define EPS 1e-8
#define EP 1e-10
#define NN 55
struct POINT{
double x, y;
}f[NN], g[NN];
double dis[NN][NN];

int R[NN * NN], L[NN * NN], U[NN * NN], D[NN * NN];
int C[NN * NN];//记录每个节点所在列
int cntc[NN]; //记录每列包含的节点数

/*删除第c列*/
void remove(int c){
int i;
for (i = D[c]; i != c; i = D[i]){
R[L[i]] = R[i];
L[R[i]] = L[i];
}
}
/*恢复第c列*/
void resume(int c){
int i;
for (i = D[c]; i != c; i = D[i]){
R[L[i]] = i;
L[R[i]] = i;
}
}

int h(){
bool hash[NN];
memset(hash, 0, sizeof(hash));

int i, j, c;
int ans = 0;
if (!hash[c]){
ans ++;
for (i = D[c]; i != c; i = D[i]){
for (j = R[i]; j != i; j = R[j]){
hash[C[j]] = true;
}
}
}
}
return ans;
}

/*DLX主要部分*/
int dfs(int k){
if (k + h() > K) return 0;// A*剪枝

int i, j, c;
int Min = INF;
if (cntc[i] < Min){
Min = cntc[i];
c = i;
}
}
for (i = D[c]; i != c; i = D[i]){
remove(i);
for (j = R[i]; j != i; j = R[j]){
remove(j);
cntc[C[j]]--;
}
if (dfs(k + 1)) return 1;
for (j = L[i]; j != i; j = L[j]){
resume(j);
cntc[j]++;
}
resume(i);
}
return 0;
}
/*建图*/
int Build(){
int i, j, now, pre, first;
for (j = head; j < N; j++){
R[j] = j + 1;
L[j + 1] = j;
}

/*列双向链表*/
for (j = 1; j <= N; j++){
pre = j;
cntc[j] = 0;
for (i = 1; i <= M; i++){
now = i * N + j;
C[now] = j;
cntc[j]++;
D[pre] = now;
U[now] = pre;
pre = now;
}
}
now = j;
D[pre] = now;
U[now] = pre;
if (cntc[j] == 0) return 0;
}
/*行双向链表*/
for (i = 1; i <= M; i++){
pre = first = -1;
for (j = 1; j <= N; j++){
now = i * N + j;
if (pre != -1){
R[pre] = now;
L[now] = pre;
}else{
first = now;
}
pre = now;
}
}
if (first != -1){
now = first;
R[pre] = now;
L[now] = pre;
}
}
return 1;
}

/*判断*/
int OK(double mid){
int i, j;
for (i = 1; i <= M; i++){
for (j = 1; j <= N; j++){
if (dis[i][j] - mid < EP){
}
}
if (Build())
return dfs(0);
else return 0;
}

/*二分距离*/
double Binary(){
double low = 0;
double hig = 1500;
double mid, ans = -1;
while(hig - low > EPS){
mid = (low + hig) / 2;
if (OK(mid)){
ans = mid;
hig = mid;
}else low = mid;
}
return ans;
}
/*计算雷达i到城市j的距离*/
double Distance(int i, int j){
return sqrt((g[i].x - f[j].x) * (g[i].x - f[j].x)
+ (g[i].y - f[j].y) * (g[i].y - f[j].y));
}
/*初始化距离矩阵dis[M][N]*/
void Init(){
int i, j;
for (i = 1; i <= M; i++){
for (j = 1; j <= N; j++){
dis[i][j] = Distance(i, j);
}
}
}
int main()
{
int T, i;
scanf("%d", &T);
while(T--){
scanf("%d%d%d", &N, &M, &K);
for (i = 1; i <= N; i++){
scanf("%lf%lf", &f[i].x, &f[i].y);
}
for (i = 1; i <= M; i++){
scanf("%lf%lf", &g[i].x, &g[i].y);
}
Init();
printf("%.6lf\n", Binary());
}
return 0;
}

1. 为什么for循环找到的i一定是素数叻，而且约数定理说的是n=p1^a1*p2^a2*p3^a3*…*pk^ak，而你每次取余都用的是原来的m，也就是n