题解 P2829 大逃离

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直接根据题意把度数 \(<k\) 的点标记成不能走，然后求次短路即可。

注意重边和自环，代码如下：

#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <iostream>
#include <queue>
#include <map>
typedef long long LL;
using namespace std;
inline int read() {
	int num = 0 ,f = 1; char c = getchar();
	while (!isdigit(c)) f = c == '-' ? -1 : f ,c = getchar();
	while (isdigit(c)) num = (num << 1) + (num << 3) + (c ^ 48) ,c = getchar();
	return num * f;
}
const int N = 5e3 + 5 ,M = 2e5 + 5;
const LL INF = 0x3f3f3f3f3f3f3f3f;
struct Edge {
	int to ,w ,next;
	Edge (int to = 0 ,int w = 0 ,int next = 0) : to(to) ,w(w) ,next(next) {}
}G[M]; int head[N] ,idx;
inline void add(int u ,int v ,int w) {
	G[++idx] = Edge(v ,w ,head[u]); head[u] = idx;
	G[++idx] = Edge(u ,w ,head[v]); head[v] = idx;
}
struct node {
	int id; LL dis;
	node (int id = 0 ,LL dis = 0) : id(id) ,dis(dis) {}
	friend bool operator < (const node &a ,const node &b) {
		return a.dis > b.dis;
	}
};
LL dis[N] ,cdis[N]; int done[N]; bool used[N];
inline void dijkstra(int n ,int s) {
	for (int i = 1; i <= n; i++)
		dis[i] = cdis[i] = INF ,done[i] = 0;
	priority_queue <node> q;
	q.push(node(s ,0)); dis[s] = 0;
	while (!q.empty()) {
		int now = q.top().id ,dist = q.top().dis; q.pop();
		if (done[now] > 1) continue;
		done[now]++;
		for (int i = head[now]; i ; i = G[i].next) {
			int v = G[i].to ,w = G[i].w;
			if (done[v] > 1 || used[v]) continue;
			if (dis[v] > dist + w) {
				cdis[v] = dis[v] ,dis[v] = dist + w;
				q.push(node(v ,dis[v]));
			}
			else if (cdis[v] > dist + w) {
				cdis[v] = dist + w;
				q.push(node(v ,cdis[v])); 
			}
		}
	}
}
int n ,m ,k ,rd[N];
map <pair <int ,int> ,int> mp;
signed main() {
	n = read() ,m = read() ,k = read();
	for (int i = 1; i <= m; i++) {
		int u = read() ,v = read() ,w = read();
		add(u ,v ,w);
		if (u != v && !mp[make_pair(u ,v)])
			rd[u]++ ,rd[v]++ ,mp[make_pair(u ,v)] = true ,mp[make_pair(v ,u)] = true;
	}
	for (int i = 1; i <= n; i++) if (rd[i] < k) used[i] = true;
	used[1] = used[n] = false;
	dijkstra(n ,1);
	printf("%lld\n" ,cdis[n] == INF ? -1 : cdis[n]);
	return 0;
}