ACM-ICPC 2018 南京赛区网络预赛 Magical Girl Haze （分层图最短路）

题意:n个点，m条单向边，可以让k条边距离为0，求1到n的最短路。

分层图最短路的裸题，拆成k+1层，表示已经消耗j次使边距离为0（0<=j<=k）, 即每个点拆成k个点

#include <bits/stdc++.h>
using namespace std;
#define MP(X, Y) make_pair(X, Y)
#define PII pair<int, int>
#define X first
#define Y second

using namespace fastIO;
#define N 4000000
#define ll long long
#define INF 0x3f3f3f3f
struct EdegeNode{
    int v;
    ll w;
    int next;

}edge[N*2+100];
struct Node {
    int v;
    ll sum;
    bool operator < (const Node &a) const {
        return sum > a.sum;
    }
}node;
int head[N+100];
bool vis[N+100];
ll dis[N+100];
int n,k,m;
int cut;
void addedge(int u,int v,ll w)
{
    edge[++cut].w=w;
    edge[cut].v=v;
    edge[cut].next=head[u];
    head[u]=cut;
}
void dijkstra()
{
    priority_queue <Node> q;
    for(int i = 0; i <= (n)*(k+1); ++i)
        dis[i] = (ll)INF*10000;
    memset(vis, 0, sizeof(vis));
    dis[1] = 0;
    q.push(Node{1, 0});
    vis[1] = 1;
    while(!q.empty()) {
        node = q.top();
        int u = node.v;
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; ~i; i = edge[i].next) {
            int v = edge[i].v;
            if(dis[u]+edge[i].w<dis[v]) {
                dis[v] = dis[u] + edge[i].w;
                if(!vis[v]) {
                    vis[v] = 1;
                    q.push(Node{v, dis[v]});
                }
            }
        }
    }
}
int main ()
{
    int T, u, v;
    ll w;
    scanf("%d", &T);
    while(T--) {
        cut = 0;
        memset(head, -1, sizeof(head));
        scanf("%d%d%d", &n, &m, &k);
        for(int i = 1; i <= m; ++i) {
            scanf("%d%d%lld", &u, &v, &w);
            for(int j = 0; j <= k; ++j) {
                addedge(u + j*n, v + j*n, w);
                if(j != k) {
                    addedge(u+j*n, v + (j+1)*n, 0);
                }
            }
        }
        dijkstra();
        ll ans = (ll)INF*100000;
        for(int i = 0; i <= k; ++i) {
            ans = min(ans, dis[i*n+n]);
        }
        printf("%lld\n", ans);
    }
    return 0;
}