Problem Statement
You are given a directed graph with $N$ vertices and $M$ edges. The $i$-th edge $(1 \leq i \leq M)$ is a directed edge from vertex $u _ i$ to vertex $v _ i$.
Initially, you are at vertex $1$. You want to repeat the following operations until you reach vertex $N$:
- Perform one of the two operations below:
- Move along a directed edge from your current vertex. This incurs a cost of $1$. More precisely, if you are at vertex $v$, choose a vertex $u$ such that there is a directed edge from $v$ to $u$, and move to vertex $u$.
- Reverse the direction of all edges. This incurs a cost of $X$. More precisely, if and only if there was a directed edge from $v$ to $u$ immediately before this operation, there is a directed edge from $u$ to $v$ immediately after this operation.
It is guaranteed that, for the given graph, you can reach vertex $N$ from vertex $1$ by repeating these operations.
Find the minimum total cost required to reach vertex $N$.
Constraints
- $2 \leq N \leq 2 \times 10^5$
- $1 \leq M \leq 2 \times 10^5$
- $1 \leq X \leq 10^9$
- $1 \leq u _ i \leq N \ (1 \leq i \leq M)$
- $1 \leq v _ i \leq N \ (1 \leq i \leq M)$
- For the given graph, it is guaranteed that you can reach vertex $N$ from vertex $1$ by the operations described.
- All input values are integers.
The input is given from Standard Input in the following format:
$N$ $M$ $X$
$u _ 1$ $v _ 1$
$u _ 2$ $v _ 2$
$\vdots$
$u _ M$ $v _ M$
Output
Print the minimum total cost required to reach vertex $N$.
Solving
求解分层图最短路，我们需要将图分解成两层，一层的边正序，一层的边倒序（反转），随后对于每层的节点，都可以与另一层的相同节点相通，其通向另一层本身节点的权值为X，随后我们在节点1处跑一边Dijkstra，比较最后通向的终点n在哪一层的路程最短，输出即可。
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#include <bits/stdc++.h>
#define int long long
#define endl '\n'
#define INF 0x3f3f3f3f
#define NINF -0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3fLL
#define Single
using namespace std;
using ll = long long;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const int N = 2e5+10;
const int M = 2e5+10;
int cnt = 0;
//两层图，双倍的节点数量
int head[N<<1];
bool vis[N<<1];
int dis[N<<1];
int n,m,x;
struct node
{
int to,w,next;
}edge[M<<2];//注意下方代码的加边次数，最多可加4*M个边(两层图的边，和连接两层之间的双向边)，故最多可有4*M条边.
void add(int u,int v,int w)
{
cnt++;
edge[cnt].w = w;
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u] = cnt;
}
void solve()
{
cin >> n >> m >> x;
for(int i=1;i<=m;i++)
{
int u,v;
cin >> u >> v;
add(u,v,1);
add(v+n,u+n,1);//第二层建反向边
}
for(int i=1;i<=n;i++)
{
//在两层图的同一节点之间建边，转换花费：X
add(i,i+n,x);
add(n+i,i,x);
}
for(int i=1;i<=n*2;i++)
{
dis[i] = LINF;//LLONG_MAX
}
priority_queue<pii,vector<pii>,greater<pii>> q;
auto dijkstra = [&](int start)
{
dis[start] = 0;
q.push({0,start});
while(!q.empty())
{
int u = q.top().second;
q.pop();
if(vis[u]) continue;
vis[u] = true;
for(int i = head[u];i;i=edge[i].next)
{
int v = edge[i].to;
int w = edge[i].w;
if(dis[v] > dis[u] + w)
{
dis[v] = dis[u] + w;
q.push({dis[v],v});
}
}
}
};
dijkstra(1);
int ans = min(dis[n],dis[n*2]);
cout << ans;
}
signed main()
{
ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
int T;
#ifdef Single
T = 1;
#else
cin >> T;
#endif
while(T--)
{
solve();
}
return 0;
}
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