洛谷 #3044. 搬家

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题意

无向图，选1个节点为家，使得绕完k个城镇回家的线路最短(k <= 5)

题解

因为k很小，所以预处理城镇到各个节点的距离，然后dfs找最优解

调试记录

要加上最后一个城镇回家的距离

#include <cstdio>
#include <cstring>
#include <queue>
#define maxn 50005
#define INF 0x3f3f3f3f

using namespace std;

struct node{
	int to, next, l;
}e[maxn << 1];
int head[maxn], tot = 0;
void addedge(int u, int v, int l){
	e[++tot].to = v, e[tot].next = head[u], e[tot].l = l;
	head[u] = tot;
}

int dis[6][maxn];
bool inq[maxn];
void SPFA(int id, int s){
	memset(dis[id], 0x3f, sizeof dis[id]);
	queue <int> q; while (!q.empty()) q.pop();
	q.push(s); dis[id][s] = 0; inq[s] = true;
	
	while (!q.empty()){
		int cur = q.front(); q.pop(); inq[cur] = false;
		for (int i = head[cur]; i; i = e[i].next){
			if (dis[id][e[i].to] > dis[id][cur] + e[i].l){
				dis[id][e[i].to] = dis[id][cur] + e[i].l;
				if (!inq[e[i].to]){
					q.push(e[i].to);
					inq[e[i].to] = true;
				}
			}
		}
	}
}

int n, m, k, town[6], ans = INF;
bool istown[maxn];

int res; bool vis[maxn];
void dfs(int start, int cur, int cost, int dep, int preid){
	if (dep == k + 1){
		res = min(res, cost + dis[preid][start]);
		return;
	}
	vis[cur] = true;
	for (int i = 1; i <= k; i++)
		if (!vis[town[i]]) dfs(start, town[i], cost + dis[i][cur], dep + 1, i);
	vis[cur] = false;
}

int main(){
	scanf("%d%d%d", &n, &m, &k);
	for (int i = 1; i <= k; i++) scanf("%d", town + i), istown[town[i]] = true;
	for (int u, v, l, i = 1; i <= m; i++){
		scanf("%d%d%d", &u, &v, &l);
		addedge(u, v, l);
		addedge(v, u, l);
	}
	
	for (int i = 1; i <= k; i++) SPFA(i, town[i]);
	
	for (int s = 1; s <= n; s++){
		if (istown[s]) continue;
		res = INF;
		dfs(s, s, 0, 1, 0);
		ans = min(ans, res);
	}
	printf("%d\n", ans);
	
	return 0;
}