[NOIP2017 TG D2T2]宝藏（模拟退火）

题目大意：$NOIPD2T2$宝藏

题解：正常做法：状压DP 。这次模拟退火，随机一个排列，$O(n^2)$贪心按排列的顺序加入生成树

卡点：没开$long;long$，接受较劣解时判断打错，没判$n=1​$的情况

C++ Code：

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <cstdlib>
#define maxn 14
const int inf = 0x3f3f3f3f;
const int Tim = 20;
const double ST = 500, DelT = 0.9, eps = 1e-5;

inline int min(int a, int b) {return a < b ? a : b;}
inline double rand_d() {return static_cast<double> (rand()) / RAND_MAX;}

int n, m;
int e[maxn][maxn];
int dep[maxn];

struct node {
	int s[maxn];
	long long ans;
	inline long long calc() {
		ans = 0;
		dep[s[1]] = 1;
		for (register int i = 2; i <= n; i++) {
			long long MIN = inf;
			for (register int j = 1; j < i; j++) if (e[s[i]][s[j]] != inf) {
				long long tmp = static_cast<long long> (dep[s[j]]) * e[s[i]][s[j]];
				if (tmp < MIN) MIN = tmp, dep[s[i]] = dep[s[j]] + 1;
			}
			ans += MIN;
		}
		return ans;
	}
} ans, now, nxt;

void SA() {
	double T = ST;
	long long del;
	now = ans;
	while (T > eps) {
		int x = rand() % n + 1, y = rand() % n + 1;
		while (x == y) x = rand() % n + 1, y = rand() % n + 1;
		nxt = now;
		std::swap(nxt.s[x], nxt.s[y]);
		del = nxt.calc();
		if (del < now.ans || exp((now.ans - del) / T) > rand_d()) now = nxt;
		if (del < ans.ans) ans = nxt;
		T *= DelT;
	}
}

int main() {
	srand(20040826);
	scanf("%d%d", &n, &m);
	if (n == 1) {
		puts("0");
		return 0;
	}
	for (int i = 1; i < n; i++) {
		for (int j = i + 1; j <= n; j++) e[i][j] = e[j][i] = inf;
	}
	for (int i = 1, a, b, c; i <= m; i++) {
		scanf("%d%d%d", &a, &b, &c);
		e[b][a] = e[a][b] = min(e[a][b], c);
	}
	int __Tim = Tim;
	for (int i = 1; i <= n; i++) ans.s[i] = i;
	ans.calc();
	while (__Tim --> 0) SA();
	printf("%lld
", ans.ans);
	return 0;
}