DZY 's BLOG

这道题也是分层图。k次机会让边权变0。虽然以前搞过但是好久了我也忘了。于是写死写死地写。然后纠结了下标纠结了n久之后就过了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=300000;
const int maxm=7000000;
const int inf=1061109567;
struct HN{
    int x,dis; HN(){}; HN(int _x,int _d):x(_x),dis(_d){};
    inline bool operator < (HN b)const{ return dis<b.dis;}
}H[maxn]; int pos[maxn]={0},size=0; char buf[5000000],*pt=buf; 
int h[maxn],p[maxm],w[maxm],n1[maxm],d[maxn],vis[maxn]={0},tot=0,n,m,K;
inline int getint(){    int r=0;while(*pt<'0'||*pt>'9')pt++;
    while(*pt>='0'&&*pt<='9')r=r*10+*pt++-48;return r;
}inline void sw(int a,int b){
    pos[H[a].x]=b;pos[H[b].x]=a;
    swap(H[a],H[b]);
}void up(int x){
    if(x==1) return;
    if(H[x]<H[x>>1]) sw(x,x>>1),up(x>>1);
}void down(int x){
    int son=x;   if((x<<1)<=size&&H[x<<1]<H[x]) son=x<<1;
    if((x<<1|1)<=size&&H[x<<1|1]<H[son]) son=x<<1|1;
    if(son!=x) sw(x,son),down(son);
}void del(int x){
    sw(x,size); size--; up(x); down(x);
}void ins(HN hn){
    H[++size]=hn; pos[hn.x]=size; up(size);
}inline void ae(int a,int b,int c){
    p[tot]=b; w[tot]=c; n1[tot]=h[a]; h[a]=tot++;
}int dijkstra(int s){
    HN S(s,0),o,tmp;
    size=d[s]=0; ins(S);
    while(size){
        o=H[1]; pos[H[1].x]=0; del(1); //if(o.x==e) return o.dis;
        if(vis[o.x]) continue; vis[o.x]=1;
        for(int i=h[o.x];~i;i=n1[i]) if(!vis[p[i]])
            if(o.dis+w[i]<d[p[i]]){
                d[p[i]]=o.dis+w[i];
                if(pos[p[i]]) {H[pos[p[i]]].dis=d[p[i]];up(pos[p[i]]);}
                else ins(HN(p[i],o.dis+w[i]));
            }
    }return -1;
}int main(){
    fread(pt,1,5000000,stdin); n=getint(); m=getint(); K=getint(); 
    memset(h,0xff,sizeof(h)); memset(d,63,sizeof(d));
    int a,b,c,s=getint(),e=getint();
    for(int i=1;i<=m;i++)a=getint(),b=getint(),c=getint(),ae(a,b,c),ae(b,a,c);
    for(int k=1;k<=K;k++) for(int i=0;i<n;i++) for(int j=h[i];~j;j=n1[j]){
        ae(i+n*k,p[j]+n*k,w[j]);
        ae(i+(k-1)*n,n*k+p[j],0);
    }dijkstra(s);
    printf("%d\n",d[n*K+e]);
    return 0;
}