Lvat_s Blog

FFT模版题

话说网上对FFT讲的神乎其神，以后我来搞一份小学生都能看懂的FFT入门吧

最好没有公式和严谨的证明，搞一些非常通俗的就比较好

因为我自己也只会模版。。。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;

const int SIZE=300005;
const double Pi=acos(-1);

int An,Bn,Cn,n,Step,Rev[SIZE],Ans[SIZE];
char s[SIZE];

struct Complex{
double a,b;
Complex(double sa=0.0,double sb=0.0){a=sa;b=sb;}
friend Complex operator+(Complex a,Complex b){return Complex(a.a+b.a,a.b+b.b);}
friend Complex operator-(Complex a,Complex b){return Complex(a.a-b.a,a.b-b.b);}
friend Complex operator*(Complex a,Complex b){return Complex(a.a*b.a-a.b*b.b,b.a*a.b+a.a*b.b);}
}A[SIZE],B[SIZE],C[SIZE];

void FFT(Complex *x,int flag){
for(int i=0;i<n;i++)if(i<Rev[i])swap(x[i],x[Rev[i]]);
for(int k=1;k<n;k<<=1){
	Complex wk=Complex(cos(Pi/k),flag*sin(Pi/k));
	for(int i=0;i<n;i+=k<<1){
		Complex wkj=Complex(1.0,0.0);
		for(int j=0;j<k;j++){
			Complex a=x[i+j],b=x[i+j+k]*wkj;
			x[i+j]=a+b;
			x[i+j+k]=a-b;
			wkj=wkj*wk;
		}
	}
}
if(flag==-1)for(int i=0;i<n;i++)x[i].a/=n;
}

void Init_Solve_Out(){
scanf("%d",&An);
Bn=An;Cn=An+Bn-1;
scanf("%s",s);
for(int i=0;i<An;i++)A[i]=Complex(s[i]-'0');
scanf("%s",s);
for(int i=0;i<Bn;i++)B[i]=Complex(s[i]-'0');
for(n=1,Step=0;n<Cn;n<<=1,Step++);
for(int i=0;i<n;i++)Rev[i]=(Rev[i>>1]>>1)|((i&1)<<(Step-1));
FFT(A,1);FFT(B,1);
for(int i=0;i<n;i++)C[i]=A[i]*B[i];
FFT(C,-1);
for(int i=0;i<Cn;i++)Ans[Cn-i]=(int)(C[i].a+0.5);
for(int i=1;i<=Cn;i++)Ans[i+1]+=Ans[i]/10,Ans[i]%=10;
if(Ans[Cn+1])Cn++;
for(int i=Cn;i>=1;i--)printf("%d",Ans[i]);
}

int main(){
freopen("2179.in","r",stdin);
freopen("2179.out","w",stdout);
Init_Solve_Out();
return 0;
}

第一道FFT

话说后缀数组我只会敲模版……这几天的比赛验证了我连暴力都能写挂

但是最近好像碰到好多FFT题目，于是就学习了一下

#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
using namespace std;

const double Pi=2*asin(1);
const int NUM_SIZE=100005;

int An,Bn,Cn,Rev[NUM_SIZE*3],Step,n;

struct Complex{
double a,b;
Complex(double as=0.0,double bs=0.0){a=as;b=bs;}
friend Complex operator+(Complex a,Complex b){return Complex(a.a+b.a,a.b+b.b);}
friend Complex operator-(Complex a,Complex b){return Complex(a.a-b.a,a.b-b.b);}
friend Complex operator*(Complex a,Complex b){return Complex(a.a*b.a-a.b*b.b,b.a*a.b+a.a*b.b);}
friend Complex operator/(Complex a,Complex b){return Complex((a.a*b.a+a.b*b.b)/(b.a*b.a+b.b*b.b),(b.a*a.b-a.a*b.b)/(b.a*b.a+b.b*b.b));}
double Mod(){return sqrt(a*a+b*b);}
}A[NUM_SIZE*3],B[NUM_SIZE*3],C[NUM_SIZE*3];

template<typename T>void Read(T &x){
int flag=1;
char ch;
while((ch=getchar())<'0' || ch>'9')if(ch=='-')flag=-1;
x=ch-'0';
while((ch=getchar())>='0' && ch<='9')x=x*10+ch-'0';
x*=flag;
}

void FFT(Complex *x,int flag){
for(int i=0;i<n;i++)if(i<Rev[i])swap(x[i],x[Rev[i]]);
for(int k=1;k<n;k<<=1){
    Complex wk=Complex(cos(Pi/k),flag*sin(Pi/k));
    for(int i=0;i<n;i+=k<<1){
		Complex wkj=Complex(1.0,0.0);
		for(int j=0;j<k;j++){
			Complex a=x[i+j],b=x[i+j+k]*wkj;
			x[i+j]=a+b;
			x[i+j+k]=a-b;
			wkj=wkj*wk;
		}
    }
}
if(flag==-1){for(int i=0;i<n;i++)x[i].a/=n;}
}

int main(){
freopen("34.in","r",stdin);
freopen("34.out","w",stdout);
Read(An);Read(Bn);
An++;Bn++;
Cn=An+Bn-1;
for(int i=0;i<An;i++)Read(A[i].a);
for(int i=0;i<Bn;i++)Read(B[i].a);
for(n=1,Step=0;n<Cn;Step++,n<<=1);
for(int i=0;i<n;i++)Rev[i]=(Rev[i>>1]>>1)|((i&1)<<(Step-1));
FFT(A,1);
FFT(B,1);
for(int i=0;i<n;i++)C[i]=A[i]*B[i];
FFT(C,-1);
for(int i=0;i<Cn;i++)printf("%d ",(int)(C[i].a+0.5));
putchar('\n');
return 0;
}

分块

每个块维护两个数组，一个是按位置排序，另一个按大小排序

修改暴力做两块，中间搞个tag

询问两块暴力，中间二分

#include<cstdio>
#include<algorithm>
using namespace std;

int n,q,cnt;
const int block_size=1000,block_num=1000;

struct Block{
int block1[block_size+5],block2[block_size+5];
int n,tag;
}block[block_num+5];

void Add(int l,int r,int v){
int first_block=(l-1)/block_size+1,last_block=(r-1)/block_size+1,first_pos=(l-1)%block_size+1,last_pos=(r-1)%block_size+1;
if(first_block==last_block){
    for(int i=first_pos;i<=last_pos;i++)block[first_block].block1[i]+=v;
    for(int i=1;i<=block[first_block].n;i++)block[first_block].block2[i]=block[first_block].block1[i];
    sort(block[first_block].block2+1,block[first_block].block2+block[first_block].n+1);
	return;
}
for(int i=first_block+1;i<last_block;i++)block[i].tag+=v;
for(int i=first_pos;i<=block[first_block].n;i++)block[first_block].block1[i]+=v;
for(int i=1;i<=block[first_block].n;i++)block[first_block].block2[i]=block[first_block].block1[i];
sort(block[first_block].block2+1,block[first_block].block2+block[first_block].n+1);
for(int i=1;i<=last_pos;i++)block[last_block].block1[i]+=v;
for(int i=1;i<=block[last_block].n;i++)block[last_block].block2[i]=block[last_block].block1[i];
sort(block[last_block].block2+1,block[last_block].block2+block[last_block].n+1);
}

int Div(int id,int k){
int l=1,r=block[id].n,ans=0;
while(l<=r){
	int mid=l+r>>1;
	if(block[id].block2[mid]+block[id].tag<k){l=mid+1;ans=mid;}
	else {r=mid-1;}
}
return block[id].n-ans;
}

int Query(int l,int r,int k){
int first_block=(l-1)/block_size+1,last_block=(r-1)/block_size+1,first_pos=(l-1)%block_size+1,last_pos=(r-1)%block_size+1,ans=0;
if(first_block==last_block){
    for(int i=first_pos;i<=last_pos;i++)ans+=(k<=(block[first_block].block1[i]+block[first_block].tag));
	return ans;
}
for(int i=first_block+1;i<last_block;i++)ans+=Div(i,k);
for(int i=first_pos;i<=block[first_block].n;i++)ans+=(k<=(block[first_block].block1[i]+block[first_block].tag));
for(int i=1;i<=last_pos;i++)ans+=(k<=(block[last_block].block1[i]+block[last_block].tag));
return ans;
}

int main(){
freopen("3343.in","r",stdin);
freopen("3343.out","w",stdout);
scanf("%d %d",&n,&q);
cnt=(n-1)/block_size+1;
for(int i=1;i<cnt;i++)block[i].n=block_size;
block[cnt].n=(n-1)%block_size+1;
for(int i=1;i<=n;i++){
	int pos_block=(i-1)/block_size+1,pos=(i-1)%block_size+1;
	scanf("%d",&block[pos_block].block1[pos]);
}
for(int i=1;i<=cnt;i++){
	for(int j=1;j<=block[i].n;j++)block[i].block2[j]=block[i].block1[j];
	sort(block[i].block2+1,block[i].block2+block[i].n+1);
}
for(int i=1;i<=q;i++){
	int l,r,v;
	char s[5];
    scanf("%s %d %d %d",s,&l,&r,&v);
    if(s[0]=='M')Add(l,r,v);
    if(s[0]=='A')printf("%d\n",Query(l,r,v));
}
return 0;
}

一模一样的点分治……

为了凑博客数又复制了一遍

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
  
int n,k,h[40005],en,mx[40005],siz[40005],ans,root,val,vis[40005],temp[40005],cnt;
  
struct Edge{
int b,v,next;
}e[80005];
  
void AddEdge(int sa,int sb,int sc){
e[++en].b=sb;
e[en].v=sc;
e[en].next=h[sa];
h[sa]=en;
}
  
void DfsSiz(int u,int fa){
siz[u]=1;
mx[u]=0;
for(int i=h[u];i;i=e[i].next){
    int v=e[i].b;
    if(v!=fa && !vis[v]){
        DfsSiz(v,u);
        siz[u]+=siz[v];
        mx[u]=max(mx[u],siz[v]);
    }
}
}
  
void DfsRoot(int point,int u,int fa){
mx[u]=max(mx[u],siz[point]-siz[u]);
if(val>mx[u]){val=mx[u];root=u;}
for(int i=h[u];i;i=e[i].next){
    int v=e[i].b;
    if(v!=fa && !vis[v])DfsRoot(point,v,u);
}
}
  
void GetDist(int u,int fa,int val){
temp[++cnt]=val;
for(int i=h[u];i;i=e[i].next){
    int v=e[i].b;
    if(v!=fa && !vis[v])GetDist(v,u,val+e[i].v);
}
}
  
int Cal(int u,int val){
cnt=0;
int now=0;
GetDist(u,u,val);
sort(temp+1,temp+cnt+1);
//for(int i=1;i<=cnt;i++)printf("Temp:%d\n",temp[i]);
int i=1,j=cnt;
while(i<j){
    while(temp[i]+temp[j]>k && i<j)j--;
    now+=j-i;
    i++;
}
//printf("Now:%d\n",now);
return now;
}
  
void Dfs(int u){
val=2100000000;
DfsSiz(u,u);
DfsRoot(u,u,0);
ans+=Cal(root,0);
//printf("Ans:%d\n",ans);
vis[root]=1;
for(int i=h[root];i;i=e[i].next){
    int v=e[i].b;
    if(!vis[v]){
        ans-=Cal(v,e[i].v);
        Dfs(v);
    }
}
}
  
int main(){
freopen("1468.in","r",stdin);
freopen("1468.out","w",stdout);
scanf("%d",&n);
//memset(h,0,sizeof(h));
//memset(vis,0,sizeof(vis));
//ans=en=0;
for(int i=1;i<n;i++){
    int sa,sb,sc;
    scanf("%d %d %d",&sa,&sb,&sc);
    AddEdge(sa,sb,sc);
    AddEdge(sb,sa,sc);
}
scanf("%d",&k);
Dfs(1);
printf("%d\n",ans);
return 0;
}

点分治第一题

这题我们考虑点分治

首先我们需要找到树的重心

然后做一遍

然后分成左右两块继续做，每一块重复上面的操作

具体的，我们每次计算每个点到重心的距离

然后放到一个数组里，再将距离和小于k的算进去

但是有可能重复统计了子树里面的数值

所以点分治的时候再减一下就好了

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

int n,k,h[10005],en,mx[10005],siz[10005],ans,root,val,vis[10005],temp[10005],cnt;

struct Edge{
int b,v,next;
}e[20005];

void AddEdge(int sa,int sb,int sc){
e[++en].b=sb;
e[en].v=sc;
e[en].next=h[sa];
h[sa]=en;
}

void DfsSiz(int u,int fa){
siz[u]=1;
mx[u]=0;
for(int i=h[u];i;i=e[i].next){
	int v=e[i].b;
	if(v!=fa && !vis[v]){
		DfsSiz(v,u);
		siz[u]+=siz[v];
		mx[u]=max(mx[u],siz[v]);
	}
}
}

void DfsRoot(int point,int u,int fa){
mx[u]=max(mx[u],siz[point]-siz[u]);
if(val>mx[u]){val=mx[u];root=u;}
for(int i=h[u];i;i=e[i].next){
	int v=e[i].b;
	if(v!=fa && !vis[v])DfsRoot(point,v,u);
}
}

void GetDist(int u,int fa,int val){
temp[++cnt]=val;
for(int i=h[u];i;i=e[i].next){
	int v=e[i].b;
	if(v!=fa && !vis[v])GetDist(v,u,val+e[i].v);
}
}

int Cal(int u,int val){
cnt=0;
int now=0;
GetDist(u,u,val);
sort(temp+1,temp+cnt+1);
int i=1,j=cnt;
while(i<j){
	while(temp[i]+temp[j]>k && i<j)j--;
    now+=j-i;
    i++;
}
return now;
}

void Dfs(int u){
val=2100000000;
DfsSiz(u,u);
DfsRoot(u,u,0);
ans+=Cal(root,0);
vis[root]=1;
for(int i=h[root];i;i=e[i].next){
	int v=e[i].b;
	if(!vis[v]){
		ans-=Cal(v,e[i].v);
		Dfs(v);
	}
}
}

int main(){
freopen("1741.in","r",stdin);
freopen("1741.out","w",stdout);
while(scanf("%d %d",&n,&k),n|k){
	memset(h,0,sizeof(h));
	memset(vis,0,sizeof(vis));
	ans=en=0;
	for(int i=1;i<n;i++){
		int sa,sb,sc;
		scanf("%d %d %d",&sa,&sb,&sc);
		AddEdge(sa,sb,sc);
		AddEdge(sb,sa,sc);
	}
	Dfs(1);
    printf("%d\n",ans);
}
return 0;
}

随便构造一下

#include<cstdio>
  
int main(){
freopen("2296.in","r",stdin);
freopen("2296.out","w",stdout);
long long a;
int t;
scanf("%d",&t);
while(t--){
scanf("%lld",&a);
if(a==0){puts("-1");continue;}
printf("%lld\n",1234567890ll*1000000ll+a-(1234567890ll*1000000ll)%a);
}
return 0;
}

拓扑排序消圈+最大权闭合子图

因为每次僵尸都是从右往左走，那么我们对于右往左依次连边

然后每个点的攻击集合就是相当于选了攻击集合中的点就必须选择这个点。

那么我们反向建边，因为有可能出现循环保护的情况，所以先拓扑排序消圈然后按照最大权闭合子图的模型建网络流图即可。

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;

int en,en1,n,m,ok[1005],cur[1005],h[1005],h1[1005],score[1005],du[1005],level[1005],S,T,now,mxval;

struct Edge{
int b,f,next,back;
}e[1000005],e1[1000005];

int Cal(int x,int y){return (x-1)*m+y;}

void AddEdge1(int sa,int sb){
e1[++en1].b=sb;
e1[en1].next=h1[sa];
h1[sa]=en1;
du[sb]++;
}

void AddEdge(int sa,int sb,int sc){
e[++en].b=sb;
e[en].back=en+1;
e[en].f=sc;
e[en].next=h[sa];
h[sa]=en;
swap(sa,sb);
e[++en].b=sb;
e[en].back=en-1;
e[en].f=0;
e[en].next=h[sa];
h[sa]=en;
}

int BFS(){
queue<int> Q;
Q.push(S);
memset(level,0,sizeof(level));
level[S]=1;
while(!Q.empty()){
	int u=Q.front();
	Q.pop();
	for(int i=h[u];i;i=e[i].next){
		int v=e[i].b;
		if(level[v] || !e[i].f)continue;
		Q.push(v);
		level[v]=level[u]+1;
	}
}
return level[T];
}

int DFS(int u,int flow){
if(u==T)return flow;
int f=flow;
for(int &i=cur[u];i;i=e[i].next){
	int v=e[i].b,fl;
	if(level[v]==level[u]+1 && e[i].f && (fl=DFS(v,min(f,e[i].f)))){
		f-=fl;
		e[i].f-=fl;
		e[e[i].back].f+=fl;
		if(!f)return flow;
	}
}
return flow-f;
}

int Dinic(){
int ans=0;
while(BFS()){
	for(int i=1;i<=T;i++)cur[i]=h[i];
	ans+=DFS(S,2100000000);
}
return ans;
}

void Toposort(){
queue<int> Q;
for(int i=1;i<=now;i++){
	if(!du[i]){ok[i]=1;Q.push(i);}
}
while(!Q.empty()){
	int u=Q.front();
	Q.pop();
	for(int i=h1[u];i;i=e1[i].next){
		int v=e1[i].b;
		du[v]--;
		if(!du[v]){ok[v]=1;Q.push(v);}
	}
}
}

int main(){
freopen("1565.in","r",stdin);
freopen("1565.out","w",stdout);
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++){
	for(int j=1;j<=m;j++){
		int can;
		now++;
		scanf("%d %d",&score[now],&can);
        for(int k=1;k<=can;k++){
			int dx,dy;
			scanf("%d %d",&dx,&dy);
			dx++;dy++;
            AddEdge1(now,Cal(dx,dy));
        }
        if(j!=m)AddEdge1(now+1,now);
	}
}
Toposort();
S=now+1;T=now+2;
for(int i=1;i<=now;i++){
	if(ok[i]){
		if(score[i]>=0){AddEdge(S,i,score[i]);mxval+=score[i];}
		else AddEdge(i,T,-score[i]);
		for(int j=h1[i];j;j=e1[j].next){
			int v=e1[j].b;
			if(ok[v])AddEdge(v,i,2100000000);
		}
	}
}
printf("%d\n",mxval-Dinic());
return 0;
}

很久以前wzf出给我们的题目。。。

考虑加边，每次并查集加完只要并查集内联通就说明实际堵塞

不联通就说明实际联通

#include<cstdio>
   
int n,k,ans=1,f[10000005];
   
int Find(int x){
return x==f[x]?f[x]:f[x]=Find(f[x]);
}
   
void Union(int sa,int sb){
if(sa>sb)f[sa]=sb;
else f[sb]=sa;
}
   
int Get(int x,int y){
if(x==n || y==n || x==0 || y==0)return 0;
return (x-1)*n+y;
}
   
int main(){
freopen("4423.in","r",stdin);
freopen("4423.out","w",stdout);
scanf("%d %d",&n,&k);
for(int i=1;i<=n*n+n;i++)f[i]=i;
while(k--){
    int sa1,sb1,sa2,sb2;
    char w1[5],w2[5];
    scanf("%d %d %s %d %d %s",&sa1,&sb1,w1,&sa2,&sb2,w2);
    if(ans==0){sa1=sa2;sb1=sb2;w1[0]=w2[0];}
    if(w1[0]=='N'){
        if(Find(Get(sa1,sb1))!=Find(Get(sa1-1,sb1))){Union(Find(Get(sa1,sb1)),Find(Get(sa1-1,sb1)));ans=1;}
        else ans=0;
    }
    else {
        if(Find(Get(sa1,sb1))!=Find(Get(sa1,sb1-1))){Union(Find(Get(sa1,sb1)),Find(Get(sa1,sb1-1)));ans=1;}
        else ans=0;
    }
    if(ans)puts("TAK");
    else puts("NIE");
}
return 0;
}

sb题写挂还有资格参加省选？

写挂10+次，莫名RE，简直崩溃

最后选择了抄hzwer的代码

这题是一个裸的欧拉函数

把原题中的式子拆一下就行了

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
long long n,ans=0;

long long phiQ(long long x)
{
    long long t=x;
    for(long long i=2;i*i<=n;i++)
        if(x%i==0)
        {
            t=t/i*(i-1);
            while(x%i==0)x/=i;
        }
    if(x>1)t=t/x*(x-1);
    return t;
}

int main(){
freopen("2705.in","r",stdin);
freopen("2705.out","w",stdout);
scanf("%lld",&n);
for(long long i=1;i*i<=n;i++){
	if(n%i==0){
		ans+=(long long)i*phiQ(n/i);
		if(i*i<n)ans+=(long long)(n/i)*phiQ(i);
	}
}
printf("%lld\n",ans);
return 0;
}

二分判定,使用网络流

暴力重构图即可

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;

int n,m,k,en,h[705],level[705],S,T,cur[705],scr[255][255];

struct Edge{
int b,next,f,back;
}e[1000005];

void AddEdge(int sa,int sb,int sc){
e[++en].b=sb;
e[en].f=sc;
e[en].back=en+1;
e[en].next=h[sa];
h[sa]=en;
swap(sa,sb);
e[++en].b=sb;
e[en].f=0;
e[en].back=en-1;
e[en].next=h[sa];
h[sa]=en;
}

int Bfs(){
queue<int> Q;
memset(level,0,sizeof(level));
level[S]=1;
Q.push(S);
while(!Q.empty()){
	int u=Q.front();
	Q.pop();
	for(int i=h[u];i;i=e[i].next){
		int v=e[i].b;
		if(!e[i].f || level[v])continue;
		level[v]=level[u]+1;
		Q.push(v);
	}
}
return level[T];
}

int Dfs(int u,int flow){
if(u==T)return flow;
int f=flow;
for(int &i=cur[u];i;i=e[i].next){
	int v=e[i].b,fl;
	if(e[i].f && level[v]==level[u]+1 && (fl=Dfs(v,min(f,e[i].f)))){
		f-=fl;
		e[i].f-=fl;
		e[e[i].back].f+=fl;
		if(f==0)return flow;
	}
}
return flow-f;
}

int Dinic(){
int ans=0;
while(Bfs()){for(int i=1;i<=T;i++)cur[i]=h[i];ans+=Dfs(S,2100000000);}
return ans;
}

int Check(int score){
memset(h,0,sizeof(h));
en=0;
for(int i=1;i<=n;i++)AddEdge(S,i,1);
for(int i=1;i<=m;i++)AddEdge(i+n,T,1);
for(int i=1;i<=n;i++){
	for(int j=1;j<=m;j++){
		if(scr[i][j]<=score)AddEdge(i,j+n,1);
	}
}
return Dinic()>n-k;
}

int main(){
freopen("4443.in","r",stdin);
freopen("4443.out","w",stdout);
scanf("%d %d %d",&n,&m,&k);
for(int i=1;i<=n;i++){
	for(int j=1;j<=m;j++){
		scanf("%d",&scr[i][j]);
	}
}
S=n+m+1;
T=n+m+2;
int l=1,r=100000000,ans=0;
while(l<=r){
	int mid=l+r>>1;
	if(Check(mid)){ans=mid;r=mid-1;}
	else {l=mid+1;}
}
printf("%d\n",ans);
return 0;
}