[NOIP2017 提高组] 宝藏

考虑到这种对于某种操作顺序有一个权值。

且这个权值有一个\(O(n)\)或者更好的复杂度求出。

求最值。

那可以用模拟退火。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cmath>
#define ll long long
#define N 20

ll n,m;

ll f[N][N];

ll in[N],dis[N];

inline ll find(){
	ll ans = 0;
	for(int i = 1;i <= n;++i)
	dis[i] = 0;
	for(int i = 2;i <= n;++i){
		ll lim = 1e18;
		for(int j = 1;j <= i - 1;++j){
			if(lim > ((dis[in[j]] + 1) * f[in[j]][in[i]]))
			lim = ((dis[in[j]] + 1) * f[in[j]][in[i]]),dis[in[i]] = dis[in[j]] + 1;
		}
		ans = ans + lim;
	}
	return ans;
}

ll fans = 1e18;

inline void sa(){
	double T = 20000;
	double eps = 1e-15;
	while(T > eps){
		ll z = -find();
		int x,y;
		x = rand() % n + 1;
		y = rand() % n + 1;
		fans = std::min(fans,-z);
		std::swap(in[x],in[y]);
		z = z + find();
		if(z > 0 && exp(-z / T) * RAND_MAX < rand())
		std::swap(in[x],in[y]);
		T *= 0.996;
	}
}

int main(){
	scanf("%lld%lld",&n,&m);
	for(int i = 1;i <= N;++i)
	for(int j = 1;j <= N;++j)
	f[i][j] = 1e18;
	for(int i = 1;i <= m;++i){
		ll x,y,z;
		scanf("%lld%lld%lld",&x,&y,&z);
		f[x][y] = std::min(z,f[x][y]);
		f[y][x] = std::min(z,f[y][x]);
	}
	for(int i = 1;i <= n;++i)
	in[i] = i;
	fans = find();
	while(((double)(clock())/CLOCKS_PER_SEC)<0.5)
	sa();
	std::cout<<fans<<std::endl;
}