#10078. 「一本通 3.2 练习 4」新年好

题目链接

loj

思路

亲戚很少，可以每个点都算一遍单源最短路
然后dfs

错误原因

算错复杂度

#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxm = 2e5 + 7;
const int maxn = 5e4 + 7;
struct edge {
	int v, q, nxt;
} e[maxm];
struct node {
	int x, y;
	node(int a, int b) {
		x = a, y = b;
	}
	bool operator < (const node &b) const {
		return y > b.y;
	}
};
int head[maxm], tot;
void add_edge(int u, int v, int q) {
	e[++tot].v = v;
	e[tot].q = q;
	e[tot].nxt = head[u];
	head[u] = tot;
}
int n, m;
int qinqi[7];
int a[10][10];
int dis[6][maxn];
int read() {
	int x = 0, f = 1; char s = getchar();
	for (; s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
	for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
	return x * f;
}
void dijstra(int s) {
	priority_queue<node> q;
	q.push(node(qinqi[s], 0));
	dis[s][qinqi[s]] = 0;
	while (!q.empty()) {
		node u = q.top();
		q.pop();
		if (dis[s][u.x] != u.y) continue;
		for (int i = head[u.x]; i; i = e[i].nxt) {
			int v = e[i].v;
			if (dis[s][v] > dis[s][u.x] + e[i].q) {
				dis[s][v] = dis[s][u.x] + e[i].q;
				q.push(node(v, dis[s][v]));
			}
		}
	}
}
int ans=inf;
int vis[10];
void dfs(int last,int js,int tot)
{
	if(js==5) {
		ans=min(ans,tot);
		return;
	}
	if(tot > ans) {
		return;
	}
	for(int i=1;i<=5;++i)
	{
		if(!vis[i]) {
			vis[i]=1;
			dfs(i,js+1,tot+a[last][i]);	
			vis[i]=0;
		}
	}
}

int main() {
	freopen("a.in", "r", stdin);
	memset(dis, inf, sizeof(dis));
	n = read(), m = read();
	qinqi[0] = 1;
	for (int i = 1; i <= 5; ++i)
		qinqi[i] = read();
	for (int i = 1; i <= m; ++i) {
		int x = read(), y = read(), z = read();
		add_edge(x, y, z);
		add_edge(y, x, z);
	}
	for (int i = 0; i <= 5; ++i) {
		dijstra(i);
	}
	for (int i = 0; i <= 5; ++i) {
		for (int j = 0; j <= 5; ++j) {
			a[i][j] = dis[i][qinqi[j]];
		}
	}
	dfs(0,0,0);
	cout<<ans;
	return 0;
}