csps-模拟7980题解

题面：https://www.cnblogs.com/Juve/articles/11712702.html

树：

我太sb了不知道DROT是1，还在sb找根

记录一个fa[]数组，表示x的祖先中第一个智商比x大的，用栈维护

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define re register
 using namespace std;
 const int MAXN=1e5+;
 int n,q,w[MAXN],sta[MAXN],top=,rb[MAXN],topp=,fa[MAXN],deep[MAXN];
 int to[MAXN<<],nxt[MAXN<<],pre[MAXN],cnt=;
 inline void add(re int u,re int v){
     ++cnt,to[cnt]=v,nxt[cnt]=pre[u],pre[u]=cnt;
 }
 void dfs(int x,int f){
     deep[x]=deep[f]+;
     int res=;
     while(w[x]>w[sta[top]]){
         rb[++topp]=sta[top];
         --top,++res;
     }
     fa[x]=sta[top];
     sta[++top]=x;
     for(int i=pre[x];i;i=nxt[i]){
         int y=to[i];
         if(y==f) continue;
         dfs(y,x);
     }
     --top;
     while(res){
         sta[++top]=rb[topp--];
         --res;
     }
 }
 signed main(){
     scanf("%d%d",&n,&q);
     for(re int i=;i<=n;++i) scanf("%d",&w[i]);
     for(re int i=,u,v;i<n;++i){
         scanf("%d%d",&u,&v);
         add(u,v),add(v,u);
     }
     w[]=0x7fffffff;sta[++top]=;
     dfs(,);
     for(re int i=,u,v,c,res;i<=q;++i){
         scanf("%d%d%d",&u,&v,&c);
         res=;
         if(c<w[u]) c=w[u],++res;
         while(deep[u]>=deep[v]){
             if(c<w[u]) c=w[u],++res;
             u=fa[u];
         }
         if(c<w[v]) ++res;
         printf("%d\n",res);
     }
     return ;
 }

环：

我没脸抄的方程

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define int long long
 using namespace std;
 const int MAXN=1e5+,mod=1e9+;
 int n,e,p[MAXN],q[MAXN],ans;
 int q_pow(int a,int b,int p){
     int res=;
     while(b){
         if(b&) res=res*a%p;
         a=a*a%p;
         b>>=;
     }
     return res;
 }
 signed main(){
     scanf("%lld%lld",&n,&e);
     for(int i=;i<=n;++i) q[i]=n-;
     for(int i=,u,v;i<=e;++i){
         scanf("%lld%lld",&u,&v);
         ++p[u],--q[u],--q[v];
     }
     ans=n*(n-)%mod*(n-)%mod*q_pow(,mod-,mod)%mod;
     for(int i=;i<=n;++i){
         int res=p[i]*(p[i]-)%mod*q_pow(,mod-,mod)%mod+p[i]*q[i]%mod*q_pow(,mod-,mod)%mod+q[i]*(q[i]-)%mod*q_pow(,mod-,mod)%mod;
         ans=(ans-res+mod)%mod;
     }
     printf("%lld\n",ans);
     return ;
 }

礼物：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define int long long
 #define re register
 using namespace std;
 const int mod=1e9+,MAXN=1e7+,MAXM=1e7+;
 int n,a[MAXN],b[MAXN],ans;
 inline int q_pow(re int a,re int b,re int p){
     int res=;
     while(b){
         if(b&) res=res*a%p;
         a=a*a%p;
         b>>=;
     }
     return res;
 }
 int fac[MAXM<<],inv[MAXM<<];
 inline void get_c(re int N){
     fac[]=fac[]=inv[]=;
     for(int i=;i<=N;++i) fac[i]=fac[i-]*i%mod;
     inv[N]=q_pow(fac[N],mod-,mod);
     for(int i=N-;i>=;--i) inv[i]=inv[i+]*(i+)%mod;
 }
 inline int C(re int n,re int m){
     if(m>n) return ;
     if(n==m) return ;
     return fac[n]*inv[m]%mod*inv[n-m]%mod;
 }
 int maxx=,tong[MAXM<<],bas=1e7+;
 signed main(){
     scanf("%lld",&n);
     for(re int i=;i<=n;++i){
         scanf("%lld%lld",&a[i],&b[i]);
         maxx=max(maxx,a[i]+b[i]);
     }
     get_c(maxx*+);
     for(int i=;i<=n;++i){
         for(int t=-a[i];t<=b[i];++t) ans=(ans+tong[t+bas]*C(a[i]+b[i],a[i]+t)%mod)%mod;
         for(int t=-b[i];t<=a[i];++t) tong[t+bas]=(tong[t+bas]+C(a[i]+b[i],a[i]-t))%mod;
     }
     printf("%lld\n",*ans%mod);
     return ;
 }

最长不下降子序列：

n有1e12，但值域150，发现它是循环的，处理循环节即可

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define int long long
 using namespace std;
 const int MAXN=1e6+;
 int n,t,a,b,c,d,p[MAXN],ans=,maxx=,cc[MAXN];
 int lowbit(int x){
     return x&(-x);
 }
 void update(int pos,int val){
     for(int i=pos;i<=maxx;i+=lowbit(i)){
         cc[i]=max(cc[i],val);
     }
 }
 int query(int pos){
     int res=;
     for(int i=pos;i>;i-=lowbit(i)){
         res=max(res,cc[i]);
     }
     return res;
 }
 int vis[MAXN];
 signed main(){
     scanf("%lld%lld%lld%lld%lld%lld",&n,&t,&a,&b,&c,&d);
     if(n<=){
         p[]=maxx=t;
         for(int i=;i<=n;++i){
             p[i]=(p[i-]*p[i-]%d*a%d+b*p[i-]%d+c%d)%d;
             maxx=max(p[i],maxx);
         }
         ++maxx;
         for(int i=;i<=n;++i){
             int res=query(p[i]+)+;
             update(p[i]+,res);
             ans=max(ans,res);
         }
         printf("%lld\n",ans);
         return ;
     }
     p[]=maxx=t;
     vis[t]=;
     int pos=,q=;//q:xunhuanchangdu
     for(int i=;i<=n;++i){
         p[i]=(p[i-]*p[i-]%d*a%d+b*p[i-]%d+c%d)%d;
         maxx=max(p[i],maxx);
         if(vis[p[i]]){
             pos=i;
             q=i-vis[p[i]];
             break;
         }
         vis[p[i]]=i;
     }
     ++maxx;
     if(!pos){
         //cout<<pos<<endl;
         for(int i=;i<=n;++i){
             int res=query(p[i]+)+;
             update(p[i]+,res);
             ans=max(ans,res);
         }
         printf("%lld\n",ans);
         return ;
     }else{
         int len=n-vis[p[pos]]+;
         int xh=len/q;//xunhuanjici
         int ys=len%q;//yuji
         for(int i=;i<=pos-;++i){
             int res=query(p[i]+)+;
             update(p[i]+,res);
             ans=max(ans,res);
         }
         for(int k=;k<=q;++k){
             for(int i=pos;i<=min(n,pos+q-);++i){
                 p[i]=p[i-q];
                 int res=query(p[i]+)+;
                 update(p[i]+,res);
                 int tmp=i-pos+;
                 res=res+xh-k-+(tmp<=ys);
                 ans=max(ans,res);
             }
             pos=pos+q;
         }
         printf("%lld\n",ans);
         return ;
     }
     return ;
 }

完全背包问题：

同余系最短路，咕咕

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #define int long long
 using namespace std;
 const int MAXN=;
 int n,m,l,c,v[MAXN],pos,minn=0x3f3f3f3f3f3f3f3f;
 int dis[],cnt[];
 queue<int>q;
 bool in[];
 void spfa(){
     memset(dis,0x3f,sizeof(dis));
     memset(cnt,0x3f,sizeof(cnt));
     dis[]=cnt[]=;
     in[]=;
     q.push();
     while(!q.empty()){
         int x=q.front();
         q.pop();
         in[x]=;
         for(int i=;i<=n;++i){
             int y=(x+v[i])%minn;
             if(v[i]>=l&&cnt[x]>=c) break;
             if(dis[y]>dis[x]+v[i]){
                 dis[y]=dis[x]+v[i];
                 if(v[i]>=l) cnt[y]=cnt[x]+;
                 if(!in[y]) q.push(y),in[y]=;
             }else if(dis[y]==dis[x]+v[i]){
                 if(cnt[y]>cnt[x]+(v[i]>l)){
                     cnt[y]=cnt[x]+;
                     if(!in[y]) q.push(y),in[y]=;
                 }
             }
         }
     }
 }
 signed main(){
     scanf("%lld%lld",&n,&m);
     for(int i=;i<=n;++i){
         scanf("%lld",&v[i]);
         minn=min(minn,v[i]);
     }
     scanf("%lld%lld",&l,&c);
     sort(v+,v+n+);
     spfa();
     for(int i=,x;i<=m;++i){
         scanf("%lld",&x);
         if(dis[x%minn]<=x) puts("Yes");
         else puts("No");
     }
     return ;
 }

最近公共祖先：

想到了正解，但不会维护subtree(fa[x])-subtree(x)的信息

考虑dfs序，更新时减去子树x即可

重复染的点可以直接break

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define int long long
 using namespace std;
 const int MAXN=1e5+;
 int n,m,w[MAXN];
 bool vis[MAXN];
 int to[MAXN<<],nxt[MAXN<<],pre[MAXN],cnt=;
 void add(int u,int v){
     ++cnt,to[cnt]=v,nxt[cnt]=pre[u],pre[u]=cnt;
 }
 int in[MAXN],out[MAXN],fa[MAXN],dfn=;
 void dfs(int x){
     in[x]=++dfn;
     for(int i=pre[x];i;i=nxt[i]){
         int y=to[i];
         if(y==fa[x]) continue;
         fa[y]=x;
         dfs(y);
     }
     out[x]=dfn;
 }
 int tr[MAXN<<],laz[MAXN<<];
 void down(int k){
     tr[k<<]=max(tr[k<<],laz[k]);
     tr[k<<|]=max(tr[k<<|],laz[k]);
     laz[k<<]=max(laz[k<<],laz[k]);
     laz[k<<|]=max(laz[k<<|],laz[k]);
     laz[k]=;
 }
 void update(int k,int l,int r,int opl,int opr,int val){
     if(opl<=l&&r<=opr){
         tr[k]=max(tr[k],val);
         laz[k]=max(laz[k],val);
         return ;
     }
     if(laz[k]) down(k);
     int mid=(l+r)>>;
     if(opl<=mid) update(k<<,l,mid,opl,opr,val);
     if(opr>mid) update(k<<|,mid+,r,opl,opr,val);
     tr[k]=max(tr[k<<],tr[k<<|]);
 }
 int query(int k,int l,int r,int opt){
     if(l==r) return tr[k];
     if(laz[k]) down(k);
     int mid=(l+r)>>;
     if(opt<=mid) return query(k<<,l,mid,opt);
     else return query(k<<|,mid+,r,opt);
 }
 signed main(){
     scanf("%lld%lld",&n,&m);
     for(int i=;i<=n;++i){
         scanf("%lld",&w[i]);
     }
     for(int i=,u,v;i<n;++i){
         scanf("%lld%lld",&u,&v);
         add(u,v),add(v,u);
     }
     dfs(),vis[]=;
     for(int i=;i<=m;++i){
         char opt[];
         int v,ans;
         scanf("%s%lld",opt,&v);
         if(opt[]=='M'){
             update(,,n,in[v],out[v],w[v]);
             vis[v]=;
             while(vis[fa[v]]!=){
                 update(,,n,in[fa[v]],in[v]-,w[fa[v]]);
                 update(,,n,out[v]+,out[fa[v]],w[fa[v]]);
                 v=fa[v];
             }
         }else{
             ans=query(,,n,in[v]);
             if(!ans) ans=-;
             printf("%lld\n",ans);
         }
     }
     return ;
 }