初三奥赛模拟测试4
初三奥赛模拟测试4
崩三回
$ T1 $ 最后一课
无意义水题，全场切（除了 HANGRY_Sol 大聪明快读写挂了)
小学数学秒了
CODE
#include<bits/stdc++.h>
using namespace std;
long long k,a,b,x,y,ans;
int main()
{
#ifndef ONLINE_JUDGE
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
scanf("%lld%lld%lld%lld%lld",&k,&a,&b,&x,&y);
if((b>=k&&y>=k)||(b<=k&&y<=k))
y=2*k-y,ans=(a-x)*(a-x)+(b-y)*(b-y);
else
ans=(a-x)*(a-x)+(b-y)*(b-y);
printf("%lld\n",ans);
return 0;
}
$ T2 $ 日常
依旧是水题，把区间排序之后对于每个询问二分得到它所在区间，然后直接判断即可，复杂度 \(O(klogn)\)
CODE
#include<bits/stdc++.h>
using namespace std;
long long n,m,k,d,st,ed,mid;
struct ask
{long long l,r,x,y;bool vis=0;}a[100100];
bool cmp(ask u,ask v){return u.l<v.l;}
int main()
{
#ifndef ONLINE_JUDGE
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
scanf("%lld%lld%lld",&n,&m,&k);
for(int i=1;i<=n;i++)
scanf("%lld%lld%lld%lld",&a[i].l,&a[i].x,&a[i].y,&a[i].r);
stable_sort(a+1,a+1+n,cmp);
for(int i=1;i<=k;i++)
{
scanf("%lld",&d);
st=1,ed=n;
while(st<ed)
{
mid=(st+ed)>>1;
if(d>a[mid].r) st=mid+1;
else ed=mid;
}
if(a[st].l>d||a[ed].r<d)
puts("Failed");
else if(a[st].vis==1)
puts("Again");
else if(a[st].x>d||a[ed].y<d)
puts("Normal"),a[st].vis=1;
else
puts("Perfect"),a[st].vis=1;
}
return 0;
}
$ T3 $ 渡尘
子段问题显而易见的可以转化成前缀和问题，所求就是两个段头，段尾的前缀和的差的最大绝对值
设段头一段是 $ s_{a} $ ,段尾一段是 $ s_{b} $ , $ \left\vert s_{b}-s_{a} \right\vert $ 显然就等于 $ \max (s_{a}-s_{b},s_{b}-s_{a}) $ ，直接用 $ \max_{l-1 \le a \le r } s_{a} - \min_{l-1 \le a \le r } s_{a} $ \(ST\)表维护一下两项即可，时间复杂度 $ O(nlogn) $ ,赛时因为巨大常数和逆天评测机还有1s时限被卡了，另外线段树做法是 $ O(mlogn) $ 的,代码使用了 Vsinger_LouTianyi 的 $ Fastio $
CODE
#include<bits/stdc++.h>
using namespace std;
#define endl "\n"
namespace Fread{const int SIZE=(1<<18);char buf[SIZE],*S,*T;inline char getchar(){if(S==T){T=(S=buf)+fread(buf,1,SIZE,stdin);if(S==T)return '\n';}return *S++;}}
namespace Fwrite{const int SIZE=(1<<18);char buf[SIZE],*S=buf,*T=buf+SIZE;inline void flush(){fwrite(buf,1,S-buf,stdout),S=buf;}inline void putchar(char c){*S++=c;if(S==T)flush();}struct NTR{ ~NTR(){flush();}}ztr;}
#ifndef ONLINE_JUDGE
#define getchar_ getchar
#define putchar_ putchar
#else
#define getchar_ Fread::getchar
#define putchar_ Fwrite::putchar
#endif
namespace Fastio{struct Reader{template<typename T>Reader&operator>>(T&x){char c=getchar_();bool f=false;while(c<'0'||c>'9'){if(c=='-')f=true;c=getchar_();}x=0;while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+(c^48);c=getchar_();}if(f)x=-x;return *this;}Reader&operator>>(double & x){char c=getchar_();short f=1,s=0;x=0;double t=0;while((c<'0'||c>'9')&&c!='.'){if(c=='-')f*=-1;c=getchar_();}while(c>='0'&&c<='9'&&c!='.')x=x*10+(c^48),c=getchar_();if(c=='.')c=getchar_();else return x*=f,*this;while(c>='0'&&c<='9')t=t*10+(c^48),s++,c=getchar_();while(s--)t/=10.0;x=(x+t)*f;return*this;}Reader&operator>>(long double&x){char c=getchar_();short f=1,s=0;x=0;long double t=0;while((c<'0'||c>'9')&&c!='.'){if(c=='-')f*=-1;c=getchar_();}while(c>='0'&&c<='9'&&c!='.')x=x*10+(c^48),c=getchar_();if(c=='.')c=getchar_();else return x*=f,*this;while(c>='0'&&c<='9')t=t*10+(c^48),s++,c=getchar_();while(s--)t/=10.0;x=(x+t)*f;return*this;}Reader&operator>>(__float128&x){char c=getchar_();short f=1,s=0;x=0;__float128 t=0;while((c<'0'||c>'9')&&c!='.'){if(c=='-')f*=-1;c=getchar_();}while(c>='0'&&c<='9'&&c!='.')x=x*10+(c^48),c=getchar_();if(c=='.') c = getchar_();else return x*=f,*this;while(c>='0'&&c<='9')t=t*10+(c^48),s++,c=getchar_();while(s--)t/=10.0;x=(x+t)*f;return *this;}Reader&operator>>(char&c){c=getchar_();while(c=='\n'||c==' '||c=='\r')c=getchar_();return *this;}Reader&operator>>(char*str){int len=0;char c=getchar_();while(c=='\n'||c==' '||c=='\r')c=getchar_();while(c!='\n'&&c!=' '&&c!='\r')str[len++]=c,c=getchar_();str[len]='\0';return *this;}Reader&operator>>(string&str){char c=getchar_();str.clear();while(c=='\n'||c==' '||c=='\r')c=getchar_();while(c!='\n'&&c!=' '&&c!='\r')str.push_back(c),c=getchar_();return *this;}template<class _Tp>Reader&operator>>(vector<_Tp>&vec){for(unsigned i=0;i<vec.size();i++)cin>>vec[i];return *this;}template<class _Tp,class _tp>Reader&operator>>(pair<_Tp,_tp>&a){cin>>a.first>>a.second;return *this;}Reader(){}}cin;struct Writer{static const int set_precision = 6;typedef int mxdouble;template<typename T>Writer&operator<<(T x){if(x==0)return putchar_('0'),*this;if(x<0)putchar_('-'),x=-x;static int sta[45];int top=0;while(x)sta[++top]=x%10,x/=10;while(top)putchar_(sta[top]+'0'),--top;return*this;}Writer&operator<<(double x){if(x<0)putchar_('-'),x=-x;mxdouble _=x;x-=(double)_;static int sta[45];int top=0;while(_)sta[++top]=_%10,_/=10;if(!top)putchar_('0');while(top)putchar_(sta[top]+'0'),--top;putchar_('.');for(int i=0;i<set_precision;i++)x*=10;_=x;while(_)sta[++top]=_%10,_/=10;for(int i=0;i<set_precision-top;i++)putchar_('0');while(top)putchar_(sta[top]+'0'),--top;return*this;}Writer&operator<<(long double x){if(x<0)putchar_('-'),x=-x;mxdouble _=x;x-=(long double)_;static int sta[45];int top=0;while(_)sta[++top]=_%10,_/=10;if(!top)putchar_('0');while(top)putchar_(sta[top]+'0'),--top;putchar_('.');for(int i=0;i<set_precision;i++)x*=10;_=x;while(_)sta[++top]=_%10,_/=10;for(int i=0;i<set_precision-top;i++)putchar_('0');while(top)putchar_(sta[top]+'0'),--top;return*this;}Writer&operator<<(__float128 x){if(x<0)putchar_('-'),x=-x;mxdouble _=x;x-=(__float128)_;static int sta[45];int top=0;while(_)sta[++top]=_%10,_/=10;if(!top)putchar_('0');while(top)putchar_(sta[top]+'0'),--top;putchar_('.');for(int i=0;i<set_precision;i++)x*=10;_=x;while(_)sta[++top]=_%10,_/=10;for(int i=0;i<set_precision-top;i++)putchar_('0');while(top)putchar_(sta[top]+'0'),--top;return*this;}Writer&operator<<(char c){putchar_(c);return*this;}Writer&operator<<(char*str){int cur=0;while(str[cur])putchar_(str[cur++]);return *this;}Writer&operator<<(const char*str){int cur=0;while(str[cur])putchar_(str[cur++]);return *this;}Writer&operator<<(string str){int fir=0,ed=str.size();while(fir<ed) putchar_(str[fir++]);return *this;}template<class _Tp>Writer&operator<<(vector<_Tp>vec){for(unsigned i=0;i<vec.size()-1;i++)cout<<vec[i]<<" ";cout<<vec[vec.size()-1];return *this;}template<class _Tp,class _tp>Writer&operator<<(pair<_Tp,_tp>a){cout<<a.first<<" "<<a.second;return *this;}Writer(){}}cout;}
#define cin Fastio::cin
#define cout Fastio::cout
long long n,m,a[201000],s[201000],ST[201000][40][2],lb[201000],x,y;
inline long long check_max(int st,int ed)
{
long long dis=lb[ed-st+1];
return max(ST[st][dis][0],ST[ed-(1<<dis)+1][dis][0]);
}
inline long long check_min(int st,int ed)
{
long long dis=lb[ed-st+1];
return min(ST[st][dis][1],ST[ed-(1<<dis)+1][dis][1]);
}
int main()
{
cin>>n>>m;
for(register int i=1;i<=n;i++)
cin>>a[i];
for(register int i=1;i<=n;i++)
s[i]=s[i-1]+a[i],ST[i][0][0]=ST[i][0][1]=s[i];
for(int i=1,mid=0;i<=n+2;i++)
{
if(i==(1<<(mid+1))) mid++;
lb[i]=mid;
}
for(register int i=n;i>=0;i--)
for(register int j=1;j<=lb[n-i+1];j++)
ST[i][j][0]=max(ST[i][j-1][0],ST[i+(1<<(j-1))][j-1][0]),
ST[i][j][1]=min(ST[i][j-1][1],ST[i+(1<<(j-1))][j-1][1]);
for(register int i=1;i<=m;i++)
{
cin>>x>>y;
cout<<check_max(x-1,y)-check_min(x-1,y)<<endl;
}
return 0;
}
$ T4 $ 罪人挽歌
蚌，赛时欧拉路写挂了，爆零了
首先，根据题意，相当于如果前两个二元组是 \(first\) 一样的话，第三个和第二个一定是 \(second\) 一样，类似这样交替
我们发现可以根据每个二元组建图，如果这次走正边（就是说从左边相同到右边相同），下次就要走反边，之后遍历每条边
容易发现这玩意其实是个强大欧拉路，对于每个点在正图（有向）里如果入度是奇数，那么它一定是 起点/终点（因为这样的话它总有一条边走不了），入度同理，所以处理出 入度为奇数的点个数加上出度为奇数的点的个数，设为 \(k\) ，如果为 \(2\) ，这两个点分别是起点和终点（两个点重合也行），如果是 \(0\) 则从任意一点出发总有一条合法的欧拉回路，上述结论可以归纳证明
直接 \(dfs\) 一遍贪心即可，需要保证走完每一步之后图依然联通，回溯的时候把后面整个插入到这条边后面，我直接链表维护了，注意删边，否则就会导致无用边重复遍历退化成 $ O(n^2) $ 对于 \(k=2\) 枚举两个起点,然后比较字典序即可，对于 \(k=0\) 可以删去任何一条边，所以第一步一定走编号是一的边（除非它是桥）,复杂度 $ O(n) $ 的
CODE
#include<bits/stdc++.h>
using namespace std;
#define N 501000
long long n,a[N],b[N],head1[N],head2[N],in[N],out[N],c,num1[N],num2[N],us1[N],us2[N],siz;
bool vis[N],jud;
vector<int> vec1,vec2;
struct NODE{long long nxt,ele;}ans1[N],ans[N];
struct EDGE{long long to,nxt;}e1[N],e2[N];
void insert(NODE& now,int d){ans[++siz].nxt=now.nxt,now.nxt=siz,ans[siz].ele=d;}
void add(int fr,int ed,int r)
{
e1[r].nxt=head1[fr];
head1[fr]=r;
e1[r].to=ed;
e2[r].nxt=head2[ed];
head2[ed]=r;
e2[r].to=fr;
}
void clear()
{
memcpy(us1,head1,sizeof(head1));
memcpy(us2,head2,sizeof(head2));
memset(vis,0,sizeof(vis));
siz=0;
memset(ans,0,sizeof(ans));
}
void dfs(int now,int state,NODE& p)
{
long long mid=0;
if(state==1)
{
while(1)
{
mid=us1[now];
while(vis[mid]==1)
mid=e1[mid].nxt;
us1[now]=mid;
if(mid==0)
return;
vis[mid]=1;
insert(p,mid);
dfs(e1[mid].to,state^1,ans[p.nxt]);
}
}
if(state==0)
{
while(1)
{
mid=us2[now];
while(vis[mid]==1)
mid=e2[mid].nxt;
us2[now]=mid;
if(mid==0)
return;
vis[mid]=1;
insert(p,mid);
dfs(e2[mid].to,state^1,ans[p.nxt]);
}
}
}
void rr(){memcpy(ans1,ans,sizeof(ans1));}
void change(NODE* Q)
{
vec1.clear();
for(int i=Q[0].nxt;i>0;i=Q[i].nxt)
vec1.push_back(Q[i].ele);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
scanf("%lld",&n);
for(int i=1;i<=n;i++)
{
scanf("%lld%lld",&a[i],&b[i]);
if(out[a[i]]==0)
num1[a[i]]=i;
if(in[b[i]]==0)
num2[b[i]]=i;
out[a[i]]++,in[b[i]]++;
}
for(int i=n;i>=1;i--)
add(a[i],b[i],i);
for(int i=1;i<=n;i++)
{
if(in[i]&1==1) c++;
if(out[i]&1==1) c++;
}
if(c!=0&&c!=2){puts("No");return 0;}
puts("Yes");
if(c==0)
{
clear();
dfs(b[1],0,ans[0]),rr();
clear();
dfs(a[1],1,ans[0]);
}
else
{
for(int i=1;i<=n;i++)
{
if(in[i]&1==1)
{
clear(),dfs(i,0,ans[0]);
if(jud==0)
rr(),jud=1;
else
break;
}
if(out[i]&1==1)
{
clear(),dfs(i,1,ans[0]);
if(jud==0)
rr(),jud=1;
else
break;
}
}
}
change(ans),vec2=vec1;change(ans1);
if(vec2<vec1)
for(int i=0;i<n;i++)
printf("%d ",vec2[i]);
else
for(int i=0;i<n;i++)
printf("%d ",vec1[i]);
puts("");
return 0;
}
