@OceanEye7年前
08/9
11:04
这是新坑……
同时开了KDT和LCT的坑【不知道的同学等下就知道了】
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@OceanEye7年前
08/8
21:55
先上一份代码……回去研究研究 @Sdchr
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Link Cut Tree 学习笔记
@OceanEye7年前
08/6
21:44
后缀数组学习笔记
@OceanEye7年前
05/26
14:55
XDDD YY了半天的点双联通忽略了最初点双的性质—-割点可能属于多个块
代码施工中
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#include <cstdio> #include <algorithm> #include <cstring> #include <vector> #include <cstdlib> #define ll long long #define p(x) ('0'<=x&&x<='9') #define For(i,l,r) for(int i=l;i<=r;i++) char cc; int C; template <class T> void read( T &x ) { x=0; cc=getchar(); C=1; while(!p(cc)) { if(cc=='-') C=-1; cc=getchar(); } while(p(cc)) { x=x*10+cc-48; cc=getchar(); } x*=C; } using namespace std; #define SIZ 1010 #define N 1000 struct node { int fa,dfn,mn,childcnt; bool iscut; vector <int> belong; void init() { belong.clear(); fa=dfn=mn=childcnt=iscut=0; } }pool[SIZ]; #define FA pool[x].fa #define MN pool[x].mn #define DFN pool[x].dfn #define CNT pool[x].childcnt #define ISCUT pool[x].iscut struct Tedge { int from,to; bool use; Tedge(int _=0,int __=0,bool ___=false) { from=_; to=__; use=false; } bool operator != (const Tedge &x) { return (from!=x.from||to!=x.to); } } edge[N<<1],STA[N]; int TOP; #define E edge[i] #define TO E.to int e_tot,fst[SIZ],cur[SIZ],NXT[N<<1]; void add_e(int a,int b) { edge[e_tot]=Tedge(a,b,false); NXT[e_tot]=fst[a]; fst[a]=e_tot++; edge[e_tot]=Tedge(b,a,false); NXT[e_tot]=fst[b]; fst[b]=e_tot++; } vector <int> BCC[SIZ]; int siz[SIZ],cnt[SIZ],tot; int sta[SIZ],top; bool use[SIZ]; unsigned ll ans1,ans2; // void init(int n) { memset(fst,-1,sizeof(fst)); memset(siz,0,sizeof(siz)); memset(cnt,0,sizeof(cnt)); e_tot=0; tot=0; ans1=0; ans2=1; for(int i=1;i<=N;i++) BCC[i].clear(); int _,__; For(i,1,n) { read(_); read(__); use[_]|=1; use[__]|=1; add_e(_,__); } } void cp(int &x,int y) { x>y?x=y:x=x; } int cd; void tarjan(int x) { use[x]=false; DFN=MN=++cd; for(int i=fst[x];~i;i=NXT[i]) { if(TO==FA) continue; if(edge[i].use) continue; STA[++TOP]=E; E.use=true; i^=1; E.use=true; i^=1; if(!use[TO]) cp(MN,pool[TO].dfn); else { CNT++; pool[TO].fa=x; tarjan(TO); MN=min(MN,pool[TO].mn); if(pool[TO].mn>=DFN) { Tedge cur; tot++; ISCUT=true; do { cur=STA[TOP--]; if(pool[cur.from].belong.size()==0||pool[cur.from].belong.back()!=tot) { pool[cur.from].belong.push_back(tot); BCC[tot].push_back(cur.from); } if(pool[cur.to].belong.size()==0||pool[cur.to].belong.back()!=tot) { pool[cur.to].belong.push_back(tot); BCC[tot].push_back(cur.to); } } while(cur!=E); } } } if(CNT<2&&!FA) ISCUT=false; } int time; void Main(int _) { time++; init(_); For(i,1,1000) cur[i]=fst[i],pool[i].init(); For(i,1,1000) if(use[i]) tarjan(i); For(i,1,1000) { if(!pool[i].dfn) continue; if(!pool[i].iscut) { siz[pool[i].belong[0]]++; continue; } for(int j=0;j<pool[i].belong.size();j++) cnt[pool[i].belong[j]]++,siz[pool[i].belong[j]]++; } For(i,1,tot) if(cnt[i]==1) { ans1++; ans2*=(unsigned ll)(siz[i]==1?1:(siz[i]-1)); } else if(cnt[i]==0) { ans1+=min(siz[i],2); ans2*=(unsigned ll)siz[i]*(siz[i]-1)/2; } printf("Case %d: %llu %llu\n",time,ans1,ans2); } int main() { freopen("BZOJ2730.in","r",stdin); memset(cur,-1,sizeof(cur)); int T; read(T); for(;T;read(T)) Main(T); return 0; } |
BZOJ2730
@OceanEye7年前
05/24
13:18
Lucas定理裸题
恩具体证明请百度 Lucas定理
刚刚发现我是RK3 :-D高兴
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//---------------OceanEye's Code----------------------// #include <cstdio> #include <algorithm> #include <cstring> #include <cstdlib> #include <cmath> #define ll long long #define INF 1000000000 #define p(x) ('0'<=x&&x<='9') #define For(i,l,r) for(int i=l;i<=r;i++) char cc; int C; template <class T> void read( T &x ) { x=0; cc=getchar(); C=1; while(!p(cc)) { if(cc=='-') C=-1; cc=getchar(); } while(p(cc)) { x=x*10+cc-48; cc=getchar(); } x*=C; } using namespace std; #define MOD 10007 #define M MOD+100 int pre[M]; void init() { pre[0]=1; for(int i=1;i<=MOD;i++) pre[i]=pre[i-1]*i%MOD; } int ksm(int a,int k) { int ret=1; for(;k;k>>=1) { if(k&1) ret=(ret*a)%MOD; a*=a; a%=MOD; } return ret; } #define calc_inv(p) (ksm(p,MOD-2)) int comb(int n,int m) { return pre[n]*calc_inv(pre[m]*pre[n-m]%MOD)%MOD; } int Lucas(int n,int m) { int ret=1,_,__; while(n&&m&&ret) { _=n%MOD; __=m%MOD; if(_<__) return 0; ret*=comb(_,__); ret%=MOD; n/=MOD; m/=MOD; } return ret; } int main() { int t,n,m; init(); for(read(t);t;t--) { read(n); read(m); printf("%d\n",Lucas(n,m)); } return 0; } |
BZOJ2982
@OceanEye7年前
05/21
13:46
先上几个好的教程
- 某不知名大佬的教程 简直良心治好了我一直没搞清楚的FFT:-D
- zky大佬的FFT教程
- kry大佬的板子:-D
[手动分割]
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#include <cstdio> #include <algorithm> #include <cstring> #include <cstdlib> #include <iostream> #include <cmath> #define ll long long #define p(x) ('0'<=x&&x<='9') #define For(i,l,r) for(int i=l;i<=r;i++) char cc; int C; template <class T> void read(T &x) { x=0; cc=getchar(); C=1; while(!p(cc)) { if(cc=='-') C=-1; cc=getchar(); } while(p(cc)) { x=x*10+cc-48; cc=getchar(); } x*=C; } using namespace std; #define N 1<<22 int n,m; int pos[N]; int k,j; const double pi=acos(-1.0); struct comp { double r,i; comp(double _r=0,double _i=0){ r=_r; i=_i; } comp operator+(const comp&x) { return comp(r+x.r,i+x.i); } comp operator-(const comp&x) { return comp(r-x.r,i-x.i); } comp operator*(const comp&x) { return comp(r*x.r-i*x.i,r*x.i+i*x.r); } comp conj() {return comp(r,-i);} }a[N],b[N]; void fft(comp a[],int n,int t) { for(int i=1;i<n;i++) if(pos[i]>i) swap(a[i],a[pos[i]]); for(int d=0;(1<<d)<n;d++) { int m=1<<d,m2=m<<1; double o=pi*2/m2*t; comp _w(cos(o),sin(o)); for(int i=0;i<n;i+=m2) { comp w(1,0); for(int j=0;j<m;j++) { comp &A=a[i+j+m],&B=a[i+j],t=w*A; A=B-t; B=B+t; w=w*_w; } } } if(t==-1) For(i,0,n-1) a[i].r/=n; } int main() { freopen("BZOJ2179FFT.in","r",stdin); read(n); read(m); for(k=1;k<=n||k<=m;k<<=1); k<<=1; j=__builtin_ctz(k)-1; For(i,0,k-1) pos[i]=pos[i>>1]>>1|((i&1)<<j); // For(i,0,k-1) printf("%d\n",pos[i]); For(i,0,n) read(j),a[i].r=j; For(i,0,m) read(j),a[i].i=j; fft(a,k,1); For(i,0,k-1) { j=(k-i)&(k-1); b[i]=(a[i]*a[i]-(a[j]*a[j]).conj())*comp(0,-0.25); } fft(b,k,-1); For(i,0,n+m) printf("%d ",(int)(b[i].r+0.5)); return 0; } |
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#include <cstdio> #include <algorithm> #include <cstdlib> #include <cstring> #include <cmath> #define ll long long #define db double #define p(x) ('0'<=x&&x<='9') #define For(i,l,r) for(int i=l;i<=r;i++) char cc; int C; template <class T> void read( T &x ) { x=0; cc=getchar(); C=1; while(!p(cc)) { if(cc=='-') C=-1; cc=getchar(); } while(p(cc)) { x=x*10+cc-48; cc=getchar(); } x*=C; } template <class T>void rd( T &x ) { cc=getchar(); while(!p(cc)) cc=getchar(); x=cc-'0'; } using namespace std; #define N 1<<17 const double pi=acos(-1.0); struct comp { double r,i; comp(double _r=0,double _i=0) { r=_r; i=_i; } comp operator+ (const comp &x) { return comp(r+x.r,i+x.i); } comp operator- (const comp &x) { return comp(r-x.r,i-x.i); } comp operator* (const comp &x) { return comp(r*x.r-i*x.i,r*x.i+i*x.r); } comp conj() { return comp(r,-i); } }a[N],b[N]; int n,k,j,pos[N]; ll c[N]; char na[N]; void fft(comp a[],int n,int t) { For(i,0,n-1) if(pos[i]>i) swap(a[i],a[pos[i]]); for(int d=0;(1<<d)<n;d++) { int m=1<<d,m2=m<<1; double o=pi*2/m2*t; comp _w(cos(o),sin(o)); for(int i=0;i<n;i+=m2) { comp w(1,0); for(j=0;j<m;j++) { comp &A=a[i+j+m],&B=a[i+j],t=A*w; A=B-t; B=B+t; w=w*_w; } } } if(t==-1) For(i,0,n-1) a[i].r/=n; } int main() { read(n); n--; for(k=1;k<=n;k<<=1); k<<=1; j=__builtin_ctz(k)-1; For(i,0,k-1) pos[i]=(pos[i>>1]>>1)|((i&1)<<j); For(i,0,n) rd(j),a[i].r=j; For(i,0,n) rd(j),a[i].i=j; fft(a,k,1); For(i,0,k-1) { j=(k-i)&(k-1); b[i]=(a[i]*a[i]-(a[j]*a[j]).conj())*comp(0,-0.25); } fft(b,k,-1); For(i,0,n*2) c[i]=(ll)(b[i].r+0.4); bool flag=false; int a=0,tot=0; for(int i=n*2;i>=0;i--) { a+=c[i]; na[tot++]=(char)(a%10); a/=10; } if(a) na[tot++]=(char)a; for(;tot;putchar(na[--tot]+'0')); return 0; } |
BZOJ2194
好像就是反过来？
挺裸的:-D
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#include <cstdio> #include <algorithm> #include <cstring> #include <cstdlib> #include <iostream> #include <cmath> #define ll long long #define INF 1000000000 #define p(x) ('0'<=x&&x<='9') #define For(i,l,r) for(int i=l;i<=r;i++) char cc; int C; template <class T> void read( T &x ) { x=0; cc=getchar(); C=1; while(!p(cc)) { if(cc=='-') C=-1; cc=getchar(); } while(p(cc)) { x=x*10+cc-48; cc=getchar(); } x*=C; } using namespace std; #define N 1<<21 const double pi=acos(-1.0); struct comp { double r,i; comp(double _r=0,double _i=0) { r=_r; i=_i; } comp operator+ (const comp &x) { return comp(r+x.r,i+x.i); } comp operator- (const comp &x) { return comp(r-x.r,i-x.i); } comp operator* (const comp &x) { return comp(r*x.r-i*x.i,r*x.i+i*x.r); } comp conj() { return comp(r,-i); } }a[N],b[N]; int k,j,n; int pos[N]; void fft(comp a[],int n,int t) { For(i,0,n-1) if(pos[i]>i) swap(a[i],a[pos[i]]); for(int k=0;(1<<k)<n;k++) { int m=1<<k,m2=m<<1; double o=2*pi/m2*t; comp _w(cos(o),sin(o)); for(int i=0;i<n;i+=m2) { comp w(1,0); for(j=0;j<m;j++) { comp &A=a[i+j+m],&B=a[i+j],t=A*w; A=B-t; B=B+t; w=w*_w; } } } if(t==-1) For(i,0,n-1) a[i].r/=n; } int main() { read(n); n--; for(k=1;k<=n;k<<=1); k<<=1; j=__builtin_ctz(k)-1; For(i,0,k-1)pos[i]=(pos[i>>1]>>1)|((i&1)<<j); For(i,0,n){ read(j); a[i].r=j; read(j); a[n-i].i=j; } fft(a,k,1); For(i,0,k-1) { j=(k-1)&(k-i); b[i]=(a[i]*a[i]-(a[j]*a[j]).conj())*comp(0,-0.25); } fft(b,k,-1); for(int i=n;i<=n*2;i++) printf("%d\n",(int)(b[i].r+0.5)); return 0; } |
[代码施工中]
FFT—-学习笔记
@OceanEye7年前
04/8
00:03
嘛……几万年没好好写过网络流了……
gap优化真的强，把我的1s的代码降成了0.05s
相信前面榜上的就是gap的吧……
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/************************************************************** Problem: 1834 User: OceanEYe Language: C++ Result: Accepted Time:56 ms Memory:1948 kb ****************************************************************/ #include <cstdio> #include <algorithm> #include <cstring> #include <cstdlib> #include <vector> #include <queue> #include <set> #define p(x) ('0'<=x&&x<='9') #define INF 100000000 char cc; int C; template <class T> void read(T &x) { cc=getchar(); C=1; x=0; while(!p(cc)) { if(cc=='-') C=-1; cc=getchar(); } while(p(cc)) x=x*10+cc-48,cc=getchar(); x*=C; } template <class T> void MN(T &x,T y) { x>y?x=y:x=x; } using namespace std; #define N 1005 #define E 10010 struct Tedge { int from,to,cap,flow; int val; }edge[E<<2]; #define TO edge[i].to #define VAL edge[i].val #define SIZ (edge[i].cap-edge[i].flow) #define CAP edge[i].cap int fst[N],NXT[E<<2],e_tot; void add_e(int _,int __,int ___,int ____) { edge[e_tot].to=__; edge[e_tot].cap=___; edge[e_tot].from=_; edge[e_tot].val=____; NXT[e_tot]=fst[_]; fst[_]=e_tot++; edge[e_tot].to=_; edge[e_tot].cap=___; edge[e_tot].flow=___; edge[e_tot].from=__; edge[e_tot].val=-____; NXT[e_tot]=fst[__]; fst[__]=e_tot++; } int S,T; struct rec { int a,b,c,d; rec(int _,int __,int ___,int ____): a(_),b(__),c(___),d(____) {} }; vector <rec> R; int n,m,k; void init1() { read(n); read(m); read(k); S=1; T=n; int _,__,___,____; memset(fst,-1,sizeof(fst)); for(int i=1;i<=m;i++) { read(_); read(__); read(___); read(____); add_e(_,__,___,0); R.push_back(rec(_,__,___,____)); } } // bool vis[N]; int d[N],cnt[N],p[N],pre[N],ans; int Augment() { int r=T,c=INF,i; while(r!=S) { i=pre[r]; MN(c,CAP-SIZ); r=TO; } r=T; while(r!=S) { i=pre[r]; edge[i].flow-=c; edge[i^1].flow+=c; r=TO; } return c; } void ISAP() { int node=S,c; for(int i=1;i<=n;i++) p[i]=fst[i]; bool flag; while(d[S]<=n) { if(node==T) { ans+=Augment(); node=S; } //solve flag=false; for(int i=p[node];~i;i=NXT[i]) { if(d[TO]!=d[node]-1||SIZ==0) continue; p[node]=i; pre[TO]=i^1; node=TO; flag=true; break; } if(flag) continue; c=n+10; for(int i=fst[node];~i;i=NXT[i]) { if(!SIZ) continue; MN(c,d[TO]+1); } if(--cnt[d[node]]==0) { printf("%d ",ans); return; } cnt[d[node]=c]++; p[node]=fst[node]; if(node!=S)node=edge[pre[node]].to; //retreat } printf("%d ",ans); } queue<int> Q; void bfs() { int x; memset(vis,0,sizeof(vis)); d[T]=1; Q.push(T); vis[T]^=1; while(!Q.empty()) { x=Q.front(); Q.pop(); cnt[d[x]]++; for(int i=fst[x];~i;i=NXT[i]) { if(vis[TO]||SIZ) continue; d[TO]=d[x]+1; Q.push(TO); vis[TO]^=1; } } } // int dis[N],cost; void init2() { int _,__,___,____; for(int i=0;i<m;i++) { _=R[i].a; __=R[i].b; ____=R[i].d; add_e(_,__,INF,____); } } int Augment2() { int r=T,c=0,i; while(r!=S) { i=pre[r]; edge[i].flow--; edge[i^1].flow++; c+=edge[i^1].val; r=TO; } return c; } void SPFA() { memset(dis,0x7f,sizeof(dis)); memset(vis,false,sizeof(vis)); dis[S]=0; Q.push(S); vis[S]=true; int x,i; while(!Q.empty()) { x=Q.front(); Q.pop(); vis[x]=false; for(i=fst[x];~i;i=NXT[i]) { if(!SIZ) continue; if(dis[TO]<=dis[x]+VAL) continue; dis[TO]=dis[x]+VAL; pre[TO]=i^1; if(vis[TO]) continue; vis[TO]=true; Q.push(TO); } } cost+=Augment2(); } // int main() { init1(); bfs(); ISAP(); init2(); for(int i=1;i<=k;i++) SPFA(); printf("%d\n",cost); return 0; } |