【分层最短路】Magical Girl Haze
https://nanti.jisuanke.com/t/31001
有K次机会可以让一条边的权值变为0,求最短路。
在存储单源最短路的数组上多开一维状态,d[i][k]表示走到序号i的点,且让k条边权值为0时的最短路。
对于每个待更新的点,尝试不置零此边的状态和置零此边的状态,分别压入优先队列去更新其他状态。
另外,此题由于有重边,需要先去重,保留同起始点最短的边。
代码:
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstring>
#include <vector>
#include <cstdio>
using namespace std;
typedef long long ll;
const ll INF = 0x3f3f3f3f3f3f3f3fLL;
const int MAX_V = 100005;
const int MAX_E = 200005;
int N, M, K;
struct Dijkstra {
struct Edge {
int to, cost, next;
} es[MAX_E];
struct Node {
int u, k;
ll d;
Node(int u, ll d, int k) : u(u), d(d), k(k) {}
bool operator< (const Node& n) const {
return d > n.d;
}
};
int head[MAX_V];
int V, E;
ll d[MAX_V][15];
bool vis[MAX_V][15];
void init(int V) {
this->V = V;
this->E = 0;
memset(head, -1, sizeof head);
}
void addEdge(int u, int v, int w) {
es[E].to = v;
es[E].cost = w;
es[E].next = head[u];
head[u] = E++;
}
void dijkstra(int s) {
priority_queue <Node> Q;
memset(d, 0x3f, sizeof d);
memset(vis, 0, sizeof(vis));
d[s][0] = 0;
Q.push(Node(s, 0, 0));
while (!Q.empty()) {
int u = Q.top().u, k = Q.top().k;
Q.pop();
if (vis[u][k])
continue;
vis[u][k] = true;
for (int i = head[u]; i != -1; i = es[i].next) {
int v = es[i].to, w = es[i].cost;
if (d[v][k] > d[u][k] + w) {
d[v][k] = d[u][k] + w;
Q.push(Node(v, d[v][k], k));
}
if (k + 1 <= K) {
if (d[v][k + 1] > d[u][k]) {
d[v][k + 1] = d[u][k];
Q.push(Node(v, d[v][k + 1], k + 1));
}
}
}
}
}
} dijk;
struct Elem {
int u, v, w;
bool operator< (const Elem& e) const {
if (u == e.u && v == e.v) {
return w < e.w;
}
if (u == e.u) {
return v < e.v;
}
return u < e.u;
}
} e[MAX_E];
int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d%d%d", &N, &M, &K);
dijk.init(N);
for (int i = 0; i < M; i++) {
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
}
sort(e, e + M);
int preu = 0, prev = 0;
for (int i = 0; i < M; i++) {
if (preu != e[i].u || prev != e[i].v) {
dijk.addEdge(e[i].u, e[i].v, e[i].w);
preu = e[i].u, prev = e[i].v;
}
}
dijk.dijkstra(1);
ll ans = INF;
for (int i = 0; i <= K; i++) {
ans = min(ans, dijk.d[N][i]);
}
printf("%lld\n", ans);
}
}