题目链接:【AtCoder 2069】Snuke’s Subway Trip
题目大意:有 个节点, 条边,每条边属于一条地铁线路。坐上或换乘一次地铁需要花费 日元,求从 到 的最小花费。
// [AtCoder 2069] Snuke's Subway Trip
#include <queue>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define x first
#define y second
#define pb push_back
typedef pair<int, int> pii;
const int maxn = 1000005;
vector<pii> ver, line[maxn];
vector<int> g[maxn];
int n, m, fa[maxn], cur, was[maxn], newn, dis[maxn];
int getfa(int u) {
return fa[u] == u ? u : fa[u] = getfa(fa[u]);
}
void merge(int u, int v) {
if (was[u] != cur) {
was[u] = cur;
ver.pb(pii(0, u));
fa[u] = u;
}
if (was[v] != cur) {
was[v] = cur;
ver.pb(pii(0, v));
fa[v] = v;
}
fa[getfa(u)] = getfa(v);
}
int main() {
scanf("%d %d", &n, &m), newn = n;
for (int u, v, w, i = 1; i <= m; i++) {
scanf("%d %d %d", &u, &v, &w);
line[w].pb(pii(u, v));
}
for (int i = 1; i <= 1000000; i++) {
ver.clear(), ++cur;
for (int j = 0; j < line[i].size(); j++) {
merge(line[i][j].x, line[i][j].y);
}
for (int j = 0; j < ver.size(); j++) {
ver[j].x = getfa(ver[j].y);
}
sort(ver.begin(), ver.end());
for (int j = 0, k = 0; j < ver.size(); j = k) {
++newn;
while (k < ver.size() && ver[j].x == ver[k].x) {
g[newn].pb(ver[k].y);
g[ver[k++].y].pb(newn);
}
}
}
queue<int> que;
que.push(1);
for (int i = 2; i <= newn; i++) {
dis[i] = -2;
}
dis[1] = 0;
while (!que.empty()) {
int u = que.front();
que.pop();
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (dis[v] == -2) {
dis[v] = dis[u] + 1;
que.push(v);
}
}
}
printf("%d\n", dis[n] >> 1);
return 0;
}