给出n和点，m条边，每条边有各自的标号，进入第一个标号需要消耗1的费用，此后转换标号需要1费用，在同一个标号上走不需要费用。问你从1到n最少需要多少费用。
最短路变形，把第一个点看成不存在的标号，然后从第一个点开始走，然后就是dijkstra了...当时没好好学dij的优化，当场的时候没加堆优化，乱T....
#include<map> #include<set> #include<ctime> #include<cmath> #include<stack> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define lowbit(x) (x & (-x)) typedef unsigned long long int ull; typedef long long int ll; const double pi = 4.0*atan(1.0); const int inf = 0x3f3f3f3f; const int maxn = 200010; const int maxm = 400010; const int mod = 998244353; using namespace std; int n, m; int T, tol; struct Node { int v, f, w; int next; bool operator < (Node a) const { return w > a.w; } }; Node node[maxm]; int head[maxn]; bool vis[maxn]; int dis[maxn]; void init() { tol = 0; memset(vis, 0, sizeof vis); memset(dis, inf, sizeof dis); memset(head, -1, sizeof head); } void addnode(int u, int v, int f) { node[tol].v = v; node[tol].f = f; node[tol].next = head[u]; head[u] = tol++; } void dijkstra() { dis[1] = 0; priority_queue<Node > q; while(!q.empty()) q.pop(); Node now; now.v = 1; now.f = 0; now.w = 0; q.push(now); while(!q.empty()) { now = q.top(); q.pop(); int u = now.v; if(vis[u]) continue; vis[u] = true; dis[u] = now.w; for(int i=head[u]; ~i; i=node[i].next) { int v = node[i].v; if(!vis[v]) { Node nex; nex.v = v; nex.f = node[i].f; nex.w = dis[u] + (node[i].f != now.f); q.push(nex); } } } } int main() { while(~scanf("%d%d", &n, &m)) { init(); int u, v, w; for(int i=1; i<=m; i++) { scanf("%d%d%d", &u, &v, &w); addnode(u, v, w); addnode(v, u, w); } dijkstra(); if(dis[n] == inf) printf("-1\n"); else printf("%d\n", dis[n]); } return 0; }