洛谷

BZOJ

分析

经典的分层最短路题(我不会)。

建 \(k+1\) 层图，跑一遍最短路，找到 \(dis[i]\) 最小的一个。

代码

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define il inline
#define re register
#define INF 0x3f3f3f3f
#define tie0 cin.tie(0),cout.tie(0)
#define fastio ios::sync_with_stdio(false)
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
typedef long long ll;template <typename T> inline void read(T &x) {T f = 1; x = 0; char c;for (c = getchar(); !isdigit(c); c = getchar()) if (c == '-') f = -1;for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);x *= f;
}const int N = 600005;
const int M = 2000005;struct node {int id, v;bool operator < (const node &x) const {return v > x.v;}
};struct edge {int to, nxt, val;
} e[M*2];int n, m, k, s, t, ans;
int head[M*2], cnt, dis[N];
bool vis[N];priority_queue <node> q;void insert(int u, int v, int w) {e[++cnt].to = v, e[cnt].val = w, e[cnt].nxt = head[u], head[u] = cnt;
}void Dijkstra(int s) {memset(dis, 0x3f, sizeof(dis));dis[s] = 0;q.push((node){s, 0});while (!q.empty()) {int u = q.top().id; q.pop();if (vis[u]) continue;vis[u] = 1;for (int i = head[u]; i; i = e[i].nxt) {int v = e[i].to, w = e[i].val;if (dis[v] > dis[u] + w) {dis[v] = dis[u] + w;q.push((node){v, dis[v]});}}}
}int main() {int u, v, w; ans = INF;read(n), read(m), read(k);read(s), read(t);for (int i = 1; i <= m; ++i) {read(u), read(v), read(w);insert(u, v, w); insert(v, u, w);for (int j = 0; j < k; ++j) {insert(u + j * n, v + (j + 1) * n, 0);insert(v + j * n, u + (j + 1) * n, 0);insert(u + (j + 1) * n, v + (j + 1) * n, w);insert(v + (j + 1) * n, u + (j + 1) * n, w);}}Dijkstra(s);for (int i = 0; i <= k; ++i) ans = min(ans, dis[t + i * n]);printf("%d", ans);return 0;
}nsert(v + j * n, u + (j + 1) * n, 0);insert(u + (j + 1) * n, v + (j + 1) * n, w);insert(v + (j + 1) * n, u + (j + 1) * n, w);}}Dijkstra(s);for (int i = 0; i <= k; ++i) ans = min(ans, dis[t + i * n]);printf("%d", ans);return 0;
}