2022.11.11
今天，全校都去秋游了，只有 Oier 还在训练。
其实就是前一天那道P5059 中国象棋 的加强版。
状态转移方程也就略有变动： \(f_i = \sum_{j=1}^{i-m-1} f_j = f_{i-1}+f_{i-m-2}\)
注意到 \(m\) 很小，不难想到，直接把目前 \(m\) 个元素全都加入矩阵，最后把当前矩阵的数值累加即可。
code
// I forgot all the tragedies and all I saw were miracles.
/************************************
|* Author: A.I.skeleton
|* Problem: P5004 专心OI - 跳房子
|* Contest: Luogu
|* URL: https://www.luogu.com.cn/problem/P5004
|* When: 2022-11-10 20:59:34
|*
|* Memory: 125 MB
|* Time: 1000 ms
************************************/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define lc p<<1
#define rc p<<1|1
#define lb(x) (x&-x)
#define pb push_back
#define ch(i) (i-'a')
#define vi vector<int>
#define F(i) (i).first
#define S(i) (i).second
#define X(i) (i>n?i-n:i)
#define pi pair<int,int>
#define id(x,y) ((x-1)*m+y)
#define cl(x) ((x).clear());
#define si(i) ((int)i.size())
#define kw(x,k) (((x)>>(k))&1)
#define all(x) (x).begin(),(x).end()
#define me(t,x) memset(t,x,sizeof(t))
#define L(i,j,k) for(int (i)=(j);i<=(k);(i)++)
#define R(i,j,k) for(int (i)=(j);i>=(k);(i)--)
#define ll(i,j,k,l) for(int (i)=(j);i<=(k);(i)+=(l))
#define rr(i,j,k,l) for(int (i)=(j);i>=(k);(i)-=(l))
#define dout(a,x) cout<<fixed<<setprecision(x)<<a<<'\n';
#define FST ios::sync_with_stdio(false);cin.tie(nullptr);
#define DE
#pragma GCC optimize(2)
namespace DG{
#ifdef DE
#define got cout<<"get here"<<'\n';
#define cut cout<<"-------------------------\n";
#define time cerr<<1e3*clock()/CLOCKS_PER_SEC<<" ms\n",0;
#define dgvi(v) cout<<#v": size= "<<si(v)<<"\nelement: ";for(auto p:v) cout<<p<<" ";cout<<'\n';
#define dgar(x,s) cout<<#x": len= "<<s+1<<"\nelement: ";L(i,0,s) cout<<x[i]<<" \n"[i==s];
#define dge(x,n) int _=0;L(i,0,n) _+=si(x[i]);cout<<"There are "<<_<<" edges in "<<#x<<'\n';\
L(u,0,n) for(int v:x[u]) cout<<u<<' '<<v<<'\n';
#define dg(...) W(#__VA_ARGS__,__VA_ARGS__)
template<typename ...Args>
void W(string X,Args&&...V){cout<<X<<" = ";string D=""; (...,(cout<<D<<V,D=", "));cout<<'\n';}
#else
#define got
#define time 0
#define cut
#define dgvi(v)
#define dgar(x,s)
#define dge(x,n)
#define dg(...)
#endif
}using namespace DG;namespace IO{
template<typename T>void rd(T &x){cin>>x;}template<typename T>void wr(T x){cout<<x;}
template<typename T,typename ...Args>void rd(T &x,Args &...args){rd(x),rd(args...);}
template<typename T,typename ...Args>void wr(T x,Args ...args){wr(x),wr(args...);}
}using namespace IO;
int mod=1e9+7;
#define P ((int)mod)
int O(int x){return x<0?(x+=P):x<P?x:(x-=P);}
template<class T>
T ksm(T a,ll b){T s=1;for(;b;b>>=1,a*=a) if(b&1) s*=a;return s;}
struct Z{
int x;Z(int x=0):x(O(x)){}Z(ll x):x(O(x%P)){}
bool operator<(const Z&b)const{return x<b.x;}
bool operator>(const Z&b)const{return x>b.x;}
bool operator<=(const Z&b)const{return x<=b.x;}
bool operator>=(const Z&b)const{return x>=b.x;}
bool operator==(const Z&b)const{return x==b.x;}
bool operator!=(const Z&b)const{return x!=b.x;}
int val()const{return x;}bool operator!(){return !x;}
Z operator-()const{return Z(O(P-x));}
Z inv()const{assert(x!=0);return ksm(*this,P-2);}
Z&operator++(){return x=O(x+1),*this;}
Z&operator--(){return x=O(x-1),*this;}
Z&operator*=(const Z&r){x=(ll)(x)*r.x%P;return*this;}
Z&operator+=(const Z&r){x=O(x+r.x);return*this;}
Z&operator-=(const Z&r){x=O(x-r.x);return*this;}
Z&operator/=(const Z&r){return*this*=r.inv();}
Z&operator+=(const int &r){x=O(x+r);return *this;}
Z&operator-=(const int &r){x=O(x-r);return *this;}
Z&operator*=(const int &r){x=(ll)(x)*r%P;return *this;}
Z&operator/=(const int &r){Z x=r;return *this*=x.inv();}
friend Z operator*(const Z&l,const Z&r){Z s=l;s*=r;return s;}
friend Z operator+(const Z&l,const Z&r){Z s=l;s+=r;return s;}
friend Z operator-(const Z&l,const Z&r){Z s=l;s-=r;return s;}
friend Z operator/(const Z&l,const Z&r){Z s=l;s/=r;return s;}
friend Z operator*(const Z&l,const int&r){Z s=l;s*=r;return s;}
friend Z operator+(const Z&l,const int&r){Z s=l;s+=r;return s;}
friend Z operator-(const Z&l,const int&r){Z s=l;s-=r;return s;}
friend Z operator/(const Z&l,const int&r){Z s=l;s/=r;return s;}
Z &operator^=(int b){
Z a=*this,c=1;for(;b;b>>=1,a*=a) if(b&1) c*=a;
return x=c.x,*this;
}friend Z operator^(Z a,int b){return a^=b;}
friend istream&operator>>(istream&is,Z&a){ll v;is>>v;a=Z(v);return is;}
friend ostream&operator<<(ostream&os,const Z&a){return os<<a.val();}
};
#define int long long
const int N=2e6+100,INF=1e18;
int n,m;Z ans;
struct mt{
Z a[20][20];mt(){me(a,0);}
friend mt operator*(mt a,mt b){
mt c;L(i,0,16) L(j,0,16) L(k,0,16)
c.a[i][j]+=a.a[i][k]*b.a[k][j];return c;
}
}b,s;
void power(int k){for(;k;k>>=1,b=b*b) if(k&1) s=s*b;}
void init(){
b.a[1][1]=b.a[m+1][1]=1;
L(i,1,m) b.a[i][i+1]=1;
}signed main(){
FST;rd(n,m);init();
L(i,1,m+1) s.a[1][i]=1;
if(n<=m){wr(2,'\n');return time;}
power(n-m);L(i,1,m+1) ans+=s.a[1][i];
wr(ans,'\n');return time;
}
边权最大为 \(3\)，显然直接拆点建图。
一定要注意可能有重边，邻接矩阵一定是 ++ 而不是 =1
因为是全局的路径，所以引入一个超级源点 \(0\)，所有实点(不是拆点拆出的点)向源点连边，矩阵乘统计。
显然直接乘会 T，考虑来个倍增优化下。
注意要特判，如果超过 \(64\) 次倍增仍达不到，就直接判掉。
运算过程中可能会爆 long long，所以要时刻对 \(k\) 取 \(\min\)，或者直接用 \(__int128\)。
code
// I forgot all the tragedies and all I saw were miracles.
/************************************
|* Author: A.I.skeleton
|* Problem: P3597 [POI2015] WYC
|* Contest: Luogu
|* URL: https://www.luogu.com.cn/problem/P3597
|* When: 2022-11-11 10:10:10
|*
|* Memory: 128 MB
|* Time: 1000 ms
************************************/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define lc p<<1
#define rc p<<1|1
#define lb(x) (x&-x)
#define pb push_back
#define ch(i) (i-'a')
#define vi vector<int>
#define F(i) (i).first
#define S(i) (i).second
#define X(i) (i>n?i-n:i)
#define pi pair<int,int>
#define id(x,y) ((x-1)*m+y)
#define cl(x) ((x).clear());
#define si(i) ((int)i.size())
#define kw(x,k) (((x)>>(k))&1)
#define all(x) (x).begin(),(x).end()
#define me(t,x) memset(t,x,sizeof(t))
#define L(i,j,k) for(int (i)=(j);i<=(k);(i)++)
#define R(i,j,k) for(int (i)=(j);i>=(k);(i)--)
#define ll(i,j,k,l) for(int (i)=(j);i<=(k);(i)+=(l))
#define rr(i,j,k,l) for(int (i)=(j);i>=(k);(i)-=(l))
#define dout(a,x) cout<<fixed<<setprecision(x)<<a<<'\n';
#define FST ios::sync_with_stdio(false);cin.tie(nullptr);
// #define DE
// #pragma GCC optimize(2)
namespace DG{
#ifdef DE
#define got cout<<"get here"<<'\n';
#define cut cout<<"-------------------------\n";
#define time cerr<<1e3*clock()/CLOCKS_PER_SEC<<" ms\n",0;
#define dgvi(v) cout<<#v": size= "<<si(v)<<"\nelement: ";for(auto p:v) cout<<p<<" ";cout<<'\n';
#define dgar(x,s) cout<<#x": len= "<<s+1<<"\nelement: ";L(i,0,s) cout<<x[i]<<" \n"[i==s];
#define dge(x,n) int _=0;L(i,0,n) _+=si(x[i]);cout<<"There are "<<_<<" edges in "<<#x<<'\n';\
L(u,0,n) for(int v:x[u]) cout<<u<<' '<<v<<'\n';
#define dg(...) W(#__VA_ARGS__,__VA_ARGS__)
template<typename ...Args>
void W(string X,Args&&...V){cout<<X<<" = ";string D=""; (...,(cout<<D<<V,D=", "));cout<<'\n';}
#else
#define got
#define time 0
#define cut
#define dgvi(v)
#define dgar(x,s)
#define dge(x,n)
#define dg(...)
#endif
}using namespace DG;
const int N=2e6+100,INF=1e9;
#define int __int128
void wr(int x){
if(x<0)x=-x,putchar('-');
if(x>9)wr(x/10);
putchar(x%10+48);
}void rd(int &x){
x=0;char ch=getchar();
for(;!isdigit(ch);ch=getchar());
for(;isdigit(ch);ch=getchar())x=(x<<1)+(x<<3)+(ch^48);
}int n,m,k,u,v,w,ans;
struct mt{
int a[150][150];mt(){me(a,0);}
friend mt operator*(mt a,mt b){
mt c;L(k,0,n*3) L(i,0,n*3) L(j,0,n*3)
c.a[i][j]+=a.a[i][k]*b.a[k][j];return c;
}
}b[80],s,g;
signed main(){//(x-1)*3+1 to unique
rd(n);rd(m);rd(k);
b[0].a[0][0]=1;
L(i,1,n){
s.a[0][(i-1)*3+1]=1;b[0].a[(i-1)*3+1][0]=1;
b[0].a[(i-1)*3+2][(i-1)*3+1]=1;
b[0].a[(i-1)*3+3][(i-1)*3+2]=1;
}L(i,1,m) rd(u),rd(v),rd(w),b[0].a[(u-1)*3+1][(v-1)*3+w]++;
int x;L(i,1,INF){
x=i;b[i]=b[i-1]*b[i-1];g=s*b[i];
if(g.a[0][0]-n>=k) break;
if(i>=64) return puts("-1"),time;
}R(i,x,0){
g=s*b[i];
if(g.a[0][0]-n<k)
s=g,ans+=(1ll<<i);
}wr(ans);putchar(10);
return time;
}
GSS6 - Can you answer these queries VI
其实就是 P2042 [NOI2005] 维护数列 的弱化版本，直接 FHQ-Treap 干上去即可。
但码力不及当年，\(\color{Red}{WA}\) 了好几发才过去。
剩余时间都在摆。
