NOI 2015 滞后赛解题报告
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
#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= =
