2.28杂题选讲
T1
曾老师做法：把限制从小到大排序，然后￥%……—*（—……%
曾老师太神仙了吊打官方题解QwQ
然而我觉得官方题解更好懂QwQ
#include<bits/stdc++.h>
//clock_t t=clock();
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define templ template<typename T>
#define sz 1010
#define mod 1000000007ll
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
templ inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
inline void print(register int x)
{
if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
while(z[++Z]=x%10+48,x/=10);
while(sr[++C]=z[Z],--Z);sr[++C]='\n';
}
// void file()
// {
// #ifndef ONLINE_JUDGE
// freopen("a.in","r",stdin);
// #endif
// }
// inline void chktime()
// {
// #ifndef ONLINE_JUDGE
// cout<<(clock()-t)/1000.0<<'\n';
// #endif
// }
// #ifdef mod
// ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
// ll inv(ll x){return ksm(x,mod-2);}
// #else
// ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
// #endif
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
int n,m;
bool visa[sz*sz],visb[sz*sz];
int main()
{
// file();
read(n,m);
int x;
rep(i,1,n)
{
read(x);
if (visa[x]) return puts("0"),0;
else visa[x]=1;
}
rep(i,1,m)
{
read(x);
if (visb[x]) return puts("0"),0;
else visb[x]=1;
}
ll ans=1;
ll A=0,B=0;
drep(i,n*m,1)
{
ll AA=A,BB=B;
if (visa[i]) AA=1,++A;
if (visb[i]) BB=1,++B;
if (visa[i]||visb[i]) ans=ans*AA%mod*BB%mod;
else ans=ans*(1ll*AA*BB-(1ll*n*m-i))%mod;
}
cout<<ans;
return 0;
}
T2
曾老师做法：发现1~n这一条链上点的编号是单调递增的，然后￥%……—（*—……
曾老师太神仙了吊打我QwQ
然而我觉得暴力分治之后最小生成树更好想QwQ
#include<bits/stdc++.h>
//clock_t t=clock();
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define templ template<typename T>
#define sz 202002
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
templ inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
inline void print(register int x)
{
if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
while(z[++Z]=x%10+48,x/=10);
while(sr[++C]=z[Z],--Z);sr[++C]='\n';
}
// void file()
// {
// #ifndef ONLINE_JUDGE
// freopen("a.in","r",stdin);
// #endif
// }
// inline void chktime()
// {
// #ifndef ONLINE_JUDGE
// cout<<(clock()-t)/1000.0<<'\n';
// #endif
// }
// #ifdef mod
// ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
// ll inv(ll x){return ksm(x,mod-2);}
// #else
// ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
// #endif
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
int n;
ll D;
ll a[sz];
struct hh{int u,v;ll w;inline const bool operator < (const hh &y) const {return w<y.w;}}edge[sz*30];
int ecnt;
int fa[sz];
int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);}
void link(int x,int y){edge[++ecnt]=(hh){x,y,D*(y-x)+a[x]+a[y]};}
int solve(int l,int r,int opt)
{
if (l==r) return l;
int mid=(l+r)>>1;
int L=solve(l,mid,-1),R=solve(mid+1,r,1);
rep(i,l,mid) link(i,R);
rep(i,mid+1,r) link(L,i);
ll mn=1e18;int ret;
rep(i,l,r) if (chkmin(mn,D*opt*i+a[i])) ret=i;
return ret;
}
ll solve()
{
rep(i,1,n) fa[i]=i;
sort(edge+1,edge+ecnt+1);
ll ret=0;
rep(i,1,ecnt)
{
int u=getfa(edge[i].u),v=getfa(edge[i].v);
if (u!=v) ret+=edge[i].w,fa[u]=v;
}
return ret;
}
int main()
{
// file();
read(n,D);
rep(i,1,n) read(a[i]);
solve(1,n,0);
cout<<solve();
return 0;
}
T3
似乎很神仙，但成功听懂了。
果然hz大佬的题更亲民QwQ
#include<bits/stdc++.h>
//clock_t t=clock();
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define templ template<typename T>
#define sz 102020
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
templ inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
inline void print(register int x)
{
if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
while(z[++Z]=x%10+48,x/=10);
while(sr[++C]=z[Z],--Z);sr[++C]='\n';
}
// void file()
// {
// #ifndef ONLINE_JUDGE
// freopen("a.in","r",stdin);
// #endif
// }
// inline void chktime()
// {
// #ifndef ONLINE_JUDGE
// cout<<(clock()-t)/1000.0<<'\n';
// #endif
// }
// #ifdef mod
// ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
// ll inv(ll x){return ksm(x,mod-2);}
// #else
// ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
// #endif
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
int n,m;
struct hh{int u,v,w;}edge[sz];
inline bool cmp(const hh &x,const hh &y){return x.w<y.w;}
int fa[sz];
ll sum[sz];
int cnt[sz];
int getfa(int x){return x==fa[x]?x:fa[x]=getfa(fa[x]);}
int main()
{
// file();
read(n,m);
rep(i,1,n) read(sum[fa[i]=i]);
int x,y,z;
rep(i,1,m) read(x,y,z),edge[i]=(hh){x,y,z};
sort(edge+1,edge+m+1,cmp);
int ans=0;
rep(i,1,m)
{
int u=edge[i].u,v=edge[i].v;
u=getfa(u),v=getfa(v);
if (u!=v) fa[u]=v,sum[v]+=sum[u],cnt[v]+=cnt[u],cnt[u]=sum[u]=0;
++cnt[v];
if (sum[v]>=edge[i].w) ans+=cnt[v],cnt[v]=0;
}
cout<<m-ans;
return 0;
}
T4
这题就是我选的啦。
你们的题都这么难，我当然要给一道SB题啦。
但是…这么简单的题你们竟然不做…你们欺人太甚……
一定是看这题太简单不想做了……
