报同步赛的时候出了些意外。于是仅仅能做一做“滞后赛”了2333
DAY1
T1离线+离散化搞，对于相等的部分直接并查集，不等部分查看是否在同一并查集中就可以，code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int t,n;
int father[200001];
struct hp{int kind,x,y;bool operator < (const hp &a) const{return kind>a.kind;}
}qst[200001];
int b[400001];
int find(int x)
{if (x!=father[x])father[x]=find(father[x]);return father[x];
}
int main()
{int i,st,r1,r2,size;bool f;freopen("prog.in","r",stdin);freopen("prog.out","w",stdout); scanf("%d",&t);while (t--){scanf("%d",&n);for (i=1;i<=n;++i){scanf("%d%d%d",&qst[i].x,&qst[i].y,&qst[i].kind);b[i*2-1]=qst[i].x; b[i*2]=qst[i].y; }sort(b+1,b+2*n+1);size=unique(b+1,b+2*n+1)-b-1;for (i=1;i<=n;++i){qst[i].x=upper_bound(b+1,b+size+1,qst[i].x)-b-1;qst[i].y=upper_bound(b+1,b+size+1,qst[i].y)-b-1;}sort(qst+1,qst+n+1);for (i=1;i<=size;++i)father[i]=i;st=n+1; for (i=1;i<=n;++i){if (qst[i].kind==0) {st=i; break;}r1=find(qst[i].x);r2=find(qst[i].y);if (r1!=r2) father[r1]=r2;  }f=false;for (i=st;i<=n;++i){r1=find(qst[i].x);r2=find(qst[i].y);if (r1==r2) {f=true; break;}}if (f) printf("NO\n");else printf("YES\n"); }
}

T2裸链剖啊。同【HAOI 2015】T2，对于子树问题能够在建树的时候记录一下子树的左右边界。（据说这题现场200+人AC，大天朝的数据结构啊） code：

#include<iostream>
#include<cstdio>
#include<cstring>
#define mid (l+r)/2
#define lch i<<1,l,mid
#define rch i<<1|1,mid+1,r
using namespace std;
struct hp{int size,top,wson,dep,fat;
}tree[100001];
int plc[100001],ls[100001],rs[100001];
int seg[400001],delta[400001];
int totw,e,n,m,ans;
struct hq{int u,v;
}a[200001];
int point[100001],next[200001];
void add(int u,int v)
{e++; a[e].u=u; a[e].v=v; next[e]=point[u]; point[u]=e;
}
void build_tree(int now,int last,int depth)
{int i;tree[now].dep=depth;tree[now].size=1;tree[now].wson=0;tree[now].fat=last;for (i=point[now];i;i=next[i])if (a[i].v!=last){build_tree(a[i].v,now,depth+1);tree[now].size+=tree[a[i].v].size;if (tree[tree[now].wson].size<tree[a[i].v].size)tree[now].wson=a[i].v;}
}
void build_seg(int now,int tp)
{int i;tree[now].top=tp; plc[now]=++totw;ls[now]=totw;if (tree[now].wson!=0)build_seg(tree[now].wson,tp);for (i=point[now];i;i=next[i])if (a[i].v!=tree[now].fat&&a[i].v!=tree[now].wson)build_seg(a[i].v,a[i].v);rs[now]=totw;
}
void updata(int i)
{seg[i]=seg[i<<1]+seg[i<<1|1];
}
void paint(int i,int l,int r,int a)
{seg[i]=a*(r-l+1);delta[i]=a;
}
void pushdown(int i,int l,int r)
{paint(lch,delta[i]);paint(rch,delta[i]);delta[i]=-1;
}
void insert(int i,int l,int r,int x,int y,int a)
{if (x<=l&&y>=r){paint(i,l,r,a);return;}if (delta[i]!=-1)pushdown(i,l,r);if (x<=mid) insert(lch,x,y,a);if (y>mid) insert(rch,x,y,a);updata(i);
}
void query(int i,int l,int r,int x,int y,int a)
{if (x<=l&&y>=r){if (a==1) ans=ans+seg[i];if (a==0) ans=ans+r-l+1-seg[i];return;}if (delta[i]!=-1)pushdown(i,l,r);if (x<=mid) query(lch,x,y,a);if (y>mid) query(rch,x,y,a);
}
void work(int x,int y)
{int f1=tree[x].top,f2=tree[y].top;ans=0;while (f1!=f2){if (tree[f1].dep<tree[f2].dep) {swap(x,y); swap(f1,f2);}query(1,1,n,plc[f1],plc[x],0);insert(1,1,n,plc[f1],plc[x],1);x=tree[f1].fat; f1=tree[x].top;}if (tree[x].dep>tree[y].dep) swap(x,y); query(1,1,n,plc[x],plc[y],0);insert(1,1,n,plc[x],plc[y],1);printf("%d\n",ans);
}
void build(int i,int l,int r)
{delta[i]=-1;if (l==r){seg[i]=0;return;}build(lch); build(rch);updata(i);
}
int main()
{int i,x;char s[20];scanf("%d",&n);for (i=1;i<=n-1;++i){scanf("%d",&x);x++; add(x,i+1);  }build_tree(1,0,0);build_seg(1,1);build(1,1,n);scanf("%d",&m);for (i=1;i<=m;++i){scanf("%s",&s);while (s[0]!='u'&&s[0]!='i') scanf("%s",&s);if (s[0]=='i'){scanf("%d",&x);x++;work(1,x);} if (s[0]=='u'){scanf("%d",&x);x++; ans=0;query(1,1,n,ls[x],rs[x],1);insert(1,1,n,ls[x],rs[x],0);printf("%d\n",ans);}}
} 

T3。考试没想出来，交的30分还打的表，最后10分钟跑出来n=30的数据，吓尿了= =；
UPD:这题事实上就是一个状压DP，考虑每一个数加到一个集合中，相当于增加一个其质因数，而考虑每一个数n至多一个大于其n√的因子。于是我们能够仅仅考虑小质因子的情况，最后再减去同样大质因子。详见PoPoQQQ神犇的题解 code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
long long f[301][301],p[3][301][301];
int pri[9]={0,2,3,5,7,11,13,17,19};
struct hp{int prime;int set;bool operator < (const hp &a) const{return ((prime<a.prime)||(prime==a.prime&&set<a.set));}
}num[501];
long long P;
int main()
{int i,j,k,t;long long ans;freopen("dinner.in","r",stdin);freopen("dinner.out","w",stdout);scanf("%d%I64d",&n,&P);for (i=1;i<=n;++i){num[i].set=0; t=i;for (j=1;j<=8;++j)if (t%pri[j]==0){num[i].set=num[i].set+(1<<(j-1)); while (t%pri[j]==0)t/=pri[j];}num[i].prime=t;}sort(num+2,num+n+1);f[0][0]=1;for (i=2;i<=n;++i){if (i==2||num[i].prime==1||num[i].prime!=num[i-1].prime){memcpy(p[1],f,sizeof(f));memcpy(p[2],f,sizeof(f));}for (j=255;j>=0;--j)for (k=255;k>=0;--k)if ((j&k)==0){if ((k&num[i].set)==0) p[1][j|num[i].set][k]=(p[1][j|num[i].set][k]+p[1][j][k])%P;if ((j&num[i].set)==0) p[2][j][k|num[i].set]=(p[2][j][k|num[i].set]+p[2][j][k])%P;}if (i==n||num[i].prime==1||num[i].prime!=num[i+1].prime){for (j=0;j<=255;++j)for (k=0;k<=255;++k)if ((j&k)==0){f[j][k]=((p[1][j][k]+p[2][j][k]-f[j][k])%P+P)%P;}}}ans=0;for (i=0;i<=255;++i)for (j=0;j<=255;++j)if ((i&j)==0)ans=(ans+f[i][j])%P;printf("%I64d\n",ans);
}

DAY2：
T1，考虑两个数的编码不存在一个是还有一个前缀的，在树上就相当于不存在两个点之间有父子关系。

而长度最短就相当于让权值与到根路径的乘积和最短。

这不就是k叉哈夫曼树么，对于不够k的能够先补零，然后同合并果子code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
struct hp{long long num;int dep;bool operator < (const hp &a) const{return (num>a.num)||(num==a.num&&dep>a.dep);}
};
int n,k;
priority_queue<hp> q;
int main()
{long long a,x,ans=0;int i,depth;freopen("epic.in","r",stdin);freopen("epic.out","w",stdout);hp minn;scanf("%d%d",&n,&k);for (i=1;i<=n;++i){scanf("%lld",&a);q.push((hp){a,0});}if (k!=2){while (n%(k-1)!=1){n++;q.push((hp){0,0});}}while (n!=1){x=0; depth=0;for (i=1;i<=k;++i){minn=q.top();x=x+minn.num;depth=max(depth,minn.dep);q.pop();}n=n-k+1;ans+=x;q.push((hp){x,depth+1});}minn=q.top();printf("%lld\n%d\n",ans,minn.dep);
}

T2：学后缀数组在BZOJ上就做了两道题，【AHOI 2013】差异和【JSOI 2007】字符加密。后者是裸题，而前者差点儿与此题一模一样，都能够分治的做。做法似乎被卡了常数，按点时限可能会T一到两个点。总时限的话没有问题。

code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define mid (l+r)/2
#define lch i<<1,l,mid
#define rch i<<1|1,mid+1,r
#define inf 2100000000LL
using namespace std;
char s[300001];
int n,maxm;
int sa[300001],rank[300001],height[300001];
int c[300001],x[300001],y[300001];
long long val[300001];
int t,ti;
long long txl,txr,tnl,tnr;
struct hq{long long tot,maxn;
}ans[300001];
struct hp{int sim,mini;long long minn,maxn;
}seg[1200001];
bool cmp(int *y,int i,int j,int k)
{ int a,b,c,d; a=y[i]; b=y[j]; c=i+k>=n?-1:y[i+k]; d=j+k>=n?

-1:y[j+k]; return a==b&&c==d; } void build_sa() { int i,j,k; for (i=0;i<maxm;++i) c[i]=0; for (i=0;i<n;++i) c[x[i]=s[i]]++; for (i=1;i<maxm;++i) c[i]+=c[i-1]; for (i=n-1;i>=0;--i) sa[--c[x[i]]]=i; for (k=1;k<=n;k<<=1) { int p=0; for (i=n-k;i<n;++i) y[p++]=i; for (i=0;i<n;++i) if (sa[i]>=k) y[p++]=sa[i]-k; for (i=0;i<maxm;++i) c[i]=0; for (i=0;i<n;++i) c[x[y[i]]]++; for (i=1;i<maxm;++i) c[i]+=c[i-1]; for (i=n-1;i>=0;--i) sa[--c[x[y[i]]]]=y[i]; swap(x,y); p=1; x[sa[0]]=0; for (i=1;i<n;++i) x[sa[i]]=cmp(y,sa[i],sa[i-1],k)?p-1:p++; if (p>=n) break; maxm=p; } for (i=0;i<n;++i) rank[sa[i]]=i; k=0; for (i=0;i<n;++i) { if (!rank[i]) continue; if (k) k--; j=sa[rank[i]-1]; while (s[i+k]==s[j+k]) k++; height[rank[i]]=k; } } void updata(int i) { seg[i].sim=min(seg[i<<1].sim,seg[i<<1|1].sim); if (seg[i<<1].sim<=seg[i<<1|1].sim) seg[i].mini=seg[i<<1].mini; else seg[i].mini=seg[i<<1|1].mini; seg[i].minn=min(seg[i<<1].minn,seg[i<<1|1].minn); seg[i].maxn=max(seg[i<<1].maxn,seg[i<<1|1].maxn); } void build(int i,int l,int r) { if (l==r) { seg[i].sim=height[l]; seg[i].mini=l; seg[i].minn=seg[i].maxn=val[l]; return; } build(lch); build(rch); updata(i); } void query(int i,int l,int r,int x,int y,int kind) { if (x<=l&&y>=r) { if (kind==0) if (seg[i].sim<t) {t=seg[i].sim; ti=seg[i].mini;} if (kind==1) {txl=max(txl,seg[i].maxn); tnl=min(tnl,seg[i].minn);} if (kind==2) {txr=max(txr,seg[i].maxn); tnr=min(tnr,seg[i].minn);} return; } if (x<=mid) query(lch,x,y,kind); if (y>mid) query(rch,x,y,kind); } void work(int l,int r) { int tti; if (l==r) return; t=n+1; ti=-1; query(1,0,n-1,l+1,r,0); tti=ti; ans[t+1].tot=ans[t+1].tot+(long long)((long long)(r-ti+1)*(long long)(ti-l)); txl=-inf; tnl=inf; query(1,0,n-1,l,ti-1,1); txr=-inf; tnr=inf; query(1,0,n-1,ti,r,2); ans[t+1].maxn=max(ans[t+1].maxn,max(txl*txr,tnl*tnr)); work(l,tti-1); work(tti,r); } int main() { int i; int a; freopen("savour.in","r",stdin); freopen("savour.out","w",stdout); scanf("%d",&n); scanf("%s",&s); while (s[0]<'a'||s[0]>'z') scanf("%s",&s); for (i=0;i<n;++i) maxm=max(maxm,(int)(s[i])); maxm++; build_sa(); for (i=0;i<n;++i) { scanf("%d",&a); val[rank[i]]=(long long)a; ans[i+1].tot=0; ans[i+1].maxn=-1000000000000000000LL; } build(1,0,n-1); work(0,n-1); for (i=n-1;i>=1;--i) {ans[i].tot=ans[i].tot+ans[i+1].tot; ans[i].maxn=max(ans[i].maxn,ans[i+1].maxn);} for (i=1;i<=n;++i) { if (ans[i].tot==0) ans[i].maxn=0; printf("%lld %lld\n",ans[i].tot,ans[i].maxn); } }

T3并没有看懂题解，（我会说我连题目都看得迷迷糊糊的么= =）
以后再研究吧= =