2020牛客国庆集训派对day2 CHEAP DELIVERIES

https://ac.nowcoder.com/acm/contest/7818/B

很玄乎的一个状压dp

总的来说就是n个1的状态转换成n+1个1的状态，写法很奇妙，具体看代码吧。。。

#include<iostream>
#include<cstring>
#include<vector>
#include<bitset>
#include<queue>
#include<set>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 11;
ll map[50][50];
int n, m, k;
ll INF = 1e17;

set<int>sss;

struct Node {
	int p;
	ll len;
	Node(int a,ll b):p(a),len(b){}
};
vector<Node>G[maxn];

void add(int x, int y, ll len) {
	G[x].push_back(Node(y, len));
}
bitset<200>ins[100], ans[100];
int par[maxn];
int find(int x) {
	if (par[x] == -1) return x;
	return par[x] = find(par[x]);
}
bool operator<(Node a, Node b) {
	return a.len > b.len;
}
ll dis[maxn];
int vis[maxn];
int dij(int s) {
	for (int i = 0; i <= n+110 ; i++) {
		dis[i] = INF;
		vis[i] = 0;
	}
	priority_queue<Node>que;
	dis[s] = 0;
	que.push(Node(s, 0));

	while (que.size()) {
		Node ans = que.top();
		que.pop();
		if (vis[ans.p]) continue;
		vis[ans.p] = 1;
		for (int i = 0; i < G[ans.p].size(); i++) {
			int p = G[ans.p][i].p;
			if (dis[p] > dis[ans.p] + G[ans.p][i].len) {
				dis[p] = dis[ans.p] + G[ans.p][i].len;
				que.push(Node(p, dis[p]));
			}
		}
	}
	return 0;
}
int id[maxn];
struct node {
	int x, y;
}que[100];

ll dp[maxn][30];



int main() {
	scanf("%d%d%d", &n, &m, &k);
	memset(par, -1, sizeof(par));

	for (int i = 0; i < m; i++) {
		int x, y;
		ll len;
		scanf("%d%d%lld", &x, &y, &len);
		int a = find(x);
		int b = find(y);
		if (a != b) par[a] = b;
		add(x, y, len);
		add(y, x, len);
	}
	int cnt = 0;
	for (int i = 0; i < k; i++) {
		int x, y;
		scanf("%d %d", &que[i].x, &que[i].y);
		sss.insert(find(que[i].x));
		sss.insert(find(que[i].y));
	}
	if (sss.size() > 1) {
		cout << -1 << endl;
		return 0;
	}
	for (int i = 0; i < k; i++) {
		for (int j = 0; j < k; j++) {
			map[i][j] = INF;
		}
	}
	for (int i = 0; i < k; i++) {
		dij(que[i].x);
		for (int j = 0; j < k; j++) {
			map[i][j] = dis[que[j].y];
		}
	}
	for (int i = 0; i < maxn; i++) {
		for (int j = 0; j < 30; j++) {
			dp[i][j] = INF;
		}
	}
	for (int i = 0; i < k; i++) dp[1 << i][i] = map[i][i];


	ll ans = INF;
    
	for (int i = 1; i < (1 << k); i++) {
		for (int j = 0; j < k; j++) {
			if (!(i &(1 << j))) continue;//i要包括状态j才行

			for (int s = 0; s < k; s++) {
				if ((i&(1 << s))) continue;
				dp[i | (1 << s)][s] = min(dp[i | (1 << s)][s], dp[i][j] + map[j][s] + map[s][s]);
			}
		}
	}
	
	
	int len = (1 << k) - 1;
	for (int i = 0; i < k; i++) {
		ans = min(ans, dp[len][i]);
	}
	printf("%lld\n", ans);
	return 0;
}