Morris' Blog 我在摸索我存在的意、生命的意、涵及值。 有真正神手般的害，套用模式是大部分的情。 那害的我存在的意何，托出神的存在？ 有候不能怪人不我，我已始省思， 什我有人的原因，而是全要靠人。 有候我我不同，那是因我做不到很多人能到的事情。 我不到、玩不起、得不明、理解得不多， 我缺少的文能力，就是一切一切能的判定。 此，我不得能做些人有助的事情。 我知道我自己是不服的性格，但在我不得不。 我有能力，因此我必。 我到底世界作什？充不定性， 而我是法出拔萃，想一般人似乎也回不了了。 越大的挑需要越力的支持，而我的支持是什？ 最近人，但都是失的局，人能接受， 不需要言，不需要通，一切所想都表不出。

49的鼓 6站台

Problem E: A Star not a Tree?

Unfortunately, Luke cannot use his existing cabling. The 100mbs system uses 100baseT (twisted pair) cables. Each 100baseT cable connects only two devices: either two network cards or a network card and a hub. (A hub is an electronic device that interconnects several cables.) Luke has a choice: He can buy 2N-2 network cards and connect his N computers together by inserting one or more cards into each computer and connecting them all together. Or he can buy N network cards and a hub and connect each of his N computers to the hub. The first approach would require that Luke configure his operating system to forward network traffic. However, with the installation of Winux 2007.2, Luke discovered that network forwarding no longer worked. He couldn't figure out how to re-enable forwarding, and he had never heard of Prim or Kruskal, so he settled on the second approach: N network cards and a hub.

Luke lives in a loft and so is prepared to run the cables and place the hub anywhere. But he won't move his computers. He wants to minimize the total length of cable he must buy.

The first line of input contains the number of case that you need to test followed by a blank line. Each test case start with a positive integer N <= 100, the number of computers. N lines follow; each gives the (x,y) coordinates (in mm.) of a computer within the room. All coordinates are integers between 0 and 10,000.There is a blank line between each consecutive test case.

Output consists of one number, the total length of the cable segments, rounded to the nearest mm. Print a blank line between 2 consecutive test case.

Sample Input

1 4 0 0 0 10000 10000 10000 10000 0 

Output for Sample Input

28284 

目解法：

模退火法－冶金技，不地分子分散再聚合，，使得品越越好。

而退火就是冷，之後再加，如同一般。

用在此，退火意指往梯度(也就是好的位置)小的地方跑，至於加部分就有使用。

原本加是根某率函容忍跳到不好的地方，但是有可能跳出去又回原本的域最佳解。

其方法跟多撒後，再分退火的效用差不多，但是撒有，充性不。

由於少，有很，所以撒和多跑次，每次跳距逐小即可。

#include <stdio.h> 
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
double distForAllPoints(double x, double y, 
 vector< pair<int, int> > &D) {
 double sum = 0;
 for(int i = D.size()-1; i >= 0; i--) {
 sum += hypot(D[i].first - x, D[i].second - y);
 }
 return sum;
}
double randDouble() {
 return (rand() % 32767) / 32767.0;
}
double annealing(vector< pair<int, int> > &D) {
#define S_MUL 0.6f
#define S_LEN 1000
#define T_CNT 10
#define E_CNT 10
 double step = S_LEN;
 double x[E_CNT], y[E_CNT], val[E_CNT];
 double Lx, Ly, Rx, Ry, tx, ty, tcost;
 Lx = Rx = D[0].first;
 Ly = Ry = D[0].second;
 for(int i = 0; i < D.size(); i++) {
 Lx = min(Lx, (double)D[i].first);
 Rx = max(Rx, (double)D[i].first);
 Ly = min(Ly, (double)D[i].second);
 Ry = max(Ry, (double)D[i].second);
 }
 for(int i = 0; i < E_CNT; i++) {
 x[i] = randDouble() * (Rx - Lx) + Lx;
 y[i] = randDouble() * (Ry - Ly) + Ly;
 val[i] = distForAllPoints(x[i], y[i], D);
 }
 while(step > 0.1) {
 for(int i = 0; i < E_CNT; i++) {
 for(int j = 0; j < T_CNT; j++) {
 tx = x[i] + randDouble() * 2 *step - step;
 ty = y[i] + randDouble() * 2 * step - step;
 tcost = distForAllPoints(tx, ty, D);
 if(tcost < val[i]) {
 val[i] = tcost, x[i] = tx, y[i] = ty;
 }
 }
 }
 step *= S_MUL;
 }
 double ret = val[0];
 for(int i = 0; i < E_CNT; i++) {
 ret = min(ret, val[i]);
 }
 printf("%.0lf\n", ret);
}
int main() {
 int testcase, N;
 scanf("%d", &testcase);
 while(testcase--) {
 scanf("%d", &N);
 vector< pair<int, int> > D;
 int x, y;
 for(int i = 0; i < N; i++) {
 scanf("%d %d", &x, &y);
 D.push_back(make_pair(x, y));
 }
 annealing(D);
 if(testcase)
 puts("");
 }
 return 0;
}

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