P3645 [APIO2015] 雅加达的摩天楼

Description

https://www.luogu.com.cn/problem/P3645

Solution

根号分治。

跳跃能力相同且小于阈值 $B$ 的优化建边。

这样边数是 $O(nB+\dfrac{n^2}{B})$ 的。取 $B=\sqrt n$。

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#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define maxn 30005
#define put() putchar('\n')
#define Tp template<typename Ty>
#define Ts template<typename Ty,typename... Ar>
using namespace std;
inline void read(int &x){
int f=1;x=0;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
x*=f;
}
namespace Debug{
Tp void _debug(char* f,Ty t){cerr<<f<<'='<<t<<endl;}
Ts void _debug(char* f,Ty x,Ar... y){while(*f!=',') cerr<<*f++;cerr<<'='<<x<<",";_debug(f+1,y...);}
Tp ostream& operator<<(ostream& os,vector<Ty>& V){os<<"[";for(auto& vv:V) os<<vv<<",";os<<"]";return os;}
#define gdb(...) _debug((char*)#__VA_ARGS__,__VA_ARGS__)
}using namespace Debug;
int n,m,b[maxn],p[maxn];
vector<int>O[maxn];
int head=1,h[maxn];
int pre[maxn],nex[maxn],t[maxn];
struct yyy{
int to,z,w;
inline void add(int x,int y,int val) {
to=y;z=h[x];h[x]=head;w=val;
}
}a[12000005];
int dis[maxn],vis[maxn];
#define mk make_pair
#define fi first
#define se second
priority_queue<pair<int,int> >q;
inline void dj(int s) {
int i,x;
memset(dis,0x3f,sizeof(dis));
q.push(mk(0,s));dis[s]=0;
while (!q.empty()) {
x=q.top().se;q.pop();
if (vis[x]) continue;vis[x]=1;
for (i=h[x];i;i=a[i].z) if (dis[a[i].to]>dis[x]+a[i].w) {
dis[a[i].to]=dis[x]+a[i].w;
q.push(mk(-dis[a[i].to],a[i].to));
}
}
}
signed main(void){
int i,j,k,o;
read(m);read(n);
for (i=1;i<=n;i++) read(b[i]),b[i]++,read(p[i]),O[p[i]].push_back(i);
int block=sqrt(m)+1;
for (j=1;j<=block;j++) {
sort(O[j].begin(),O[j].end());
for (auto tmp:O[j]) t[b[tmp]]=1;
for (k=1;k<=j;k++) {
int las=k;
for (i=k;i<=m;i+=j) pre[i]=las,las=(t[i]?i:las);
i-=j;las=i;
for (;i>=k;i-=j) nex[i]=las,las=(t[i]?i:las);
for (i=k;i<=m;i+=j) if (t[i]) {
for (o=pre[i];o<=nex[i];o+=j) if (i^o) a[++head].add(i,o,abs(i-o)/j);
}
}
for (auto tmp:O[j]) t[b[tmp]]=0;
}
for (j=block+1;j<=m;j++) {
for (auto tmp:O[j]) {
for (i=b[tmp]-j;i>=1;i-=j) a[++head].add(b[tmp],i,abs(b[tmp]-i)/j);
for (i=b[tmp]+j;i<=m;i+=j) a[++head].add(b[tmp],i,abs(b[tmp]-i)/j);
}
}
int tmp=1;
for (i=b[tmp]-j;i>=1;i-=j) a[++head].add(b[tmp],i,abs(b[tmp]-i)/p[tmp]);
for (i=b[tmp]+j;i<=m;i+=j) a[++head].add(b[tmp],i,abs(b[tmp]-i)/p[tmp]);

dj(b[1]);
if (dis[b[2]]>1e9) puts("-1");
else printf("%d\n",dis[b[2]]);
return 0;
}