BZOJ 2763 [JLOI2011]飞行路线（分层最短路）

题意：给你一张图，有s和t，然后有k次免费的机会，问你最小花费是多少

思路：之前其实网络赛考过一道这样的题，其实就是原题，记得那次被spfa卡死了，后来也没太看分层图，今天补下

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int maxn=1200100;
const int maxm=1200100;
int cnt;
struct node
{
    int from,to,val,next;
}edge[maxm<<1];
int head[maxm];
LL dis[maxn];
int vis[maxn];
struct Node
{
    int val,id;
    bool operator <(const Node &b)const
    {
        if(val==b.val)return id<b.id;
        return val>b.val;
    }
};
void init()
{
    memset(head,-1,sizeof(head));
    cnt=1;
}
void addedge(int from,int to,int val)
{
    edge[cnt].from=from;
    edge[cnt].to=to;
    edge[cnt].val=val;
    edge[cnt].next=head[from];
    head[from]=cnt++;
}

void dijkstra(int s,int e)
{
    memset(vis,0,sizeof(vis));
    memset(dis,0x7f,sizeof(dis));
    Node now;
    priority_queue<Node>q;
    while(q.size())q.pop();
    now.val=0,now.id=s;
    dis[s]=0;
    q.push(now);
    while(!q.empty()){
        Node u=q.top();
        q.pop();
        if(vis[u.id])continue;
        vis[u.id]=1;
        for(int i=head[u.id];i!=-1;i=edge[i].next){
            int to=edge[i].to;
            if(dis[u.id]+edge[i].val<dis[to]){
                dis[to]=dis[u.id]+edge[i].val;
                Node pus;
                pus.id=to,pus.val=dis[to];
                q.push(pus);
            }
        }
    }
    return ;
}

int main()
{
    int n=read(),m=read(),k=read();
    int s=read(),t=read();
    s++;t++;
    init();
    for(int i=1;i<=m;i++){
        int u=read(),v=read(),w=read();
        u++;v++;
        for(int j=0;j<=k;j++){
            addedge(j*n+u,j*n+v,w);
            addedge(j*n+v,j*n+u,w);
            if(j!=k){
                addedge(j*n+u,(j+1)*n+v,0);
                addedge(j*n+v,(j+1)*n+u,0);
            }
        }
    }
    dijkstra(s,t);
    LL ans=0x3f3f3f3f3f3f3f3fLL;
    for(int i=0;i<=k;i++){
        ans=min(ans,dis[i*n+t]);
    }
    printf("%lld\n",ans);
    return 0;
}