Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem E: A Star not a Tree?

Luke wants to upgrade his home computer network from 10mbs to 100mbs. His existing network uses 10base2 (coaxial) cables that allow you to connect any number of computers together in a linear arrangement. Luke is particulary proud that he solved a nasty NP-complete problem in order to minimize the total cable length.

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

---

題目描述：

找一點到所有點距離總和最小。四捨五入到整數輸出距離總和。

題目解法：

模擬退火法－冶金技術，不斷地讓分子分散再聚合，讓雜質脫離，使得品質越來越好。

而退火就是冷卻，之後再加熱，如同鍛劍一般。

運用在此題時，退火意指往梯度(也就是較好的位置)較小的地方跑，至於加熱部分就沒有使用。

原本加熱是根據某個機率函數來容忍跳躍到不好的地方，但是有可能跳出去又滾回原本的區域最佳解。

其實這種方法跟隨機多撒幾個點後，再分別退火的效用差不多，但是隨機撒點有壞處，擴充性不強。

這題由於點數較少，測資沒有很強，所以隨機撒點和多跑幾次，每次將跳躍距離逐漸縮小即可。

#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;
}