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2020-2021:teams:legal_string:contest4 [2020/07/07 10:38] jxm2001 |
2020-2021:teams:legal_string:contest4 [2020/07/07 11:07] (当前版本) jxm2001 |
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</hidden> | </hidden> | ||
+ | ===== I. Running Away From the Barn ===== | ||
+ | === 题解 === | ||
+ | |||
+ | 这题思路还挺多的。有跑树上差分 $+$ 倍增的;还有按 $\text{depth}$ 降序询问,然后权值线段树/树状数组维护答案的;还有左偏树的。 | ||
+ | |||
+ | 比赛时没想到调整询问顺序,直接写了个主席树 $+$ 权值线段树,感觉又写麻烦了。 | ||
+ | |||
+ | <hidden 比赛代码> | ||
+ | <code cpp> | ||
+ | #include <iostream> | ||
+ | #include <cstdio> | ||
+ | #include <cstdlib> | ||
+ | #include <algorithm> | ||
+ | #include <string> | ||
+ | #include <sstream> | ||
+ | #include <cstring> | ||
+ | #include <cctype> | ||
+ | #include <cmath> | ||
+ | #include <vector> | ||
+ | #include <set> | ||
+ | #include <map> | ||
+ | #include <stack> | ||
+ | #include <queue> | ||
+ | #include <ctime> | ||
+ | #include <cassert> | ||
+ | #define _for(i,a,b) for(int i=(a);i<(b);++i) | ||
+ | #define _rep(i,a,b) for(int i=(a);i<=(b);++i) | ||
+ | #define mem(a,b) memset(a,b,sizeof(a)) | ||
+ | using namespace std; | ||
+ | typedef long long LL; | ||
+ | inline int read_int(){ | ||
+ | int t=0;bool sign=false;char c=getchar(); | ||
+ | while(!isdigit(c)){sign|=c=='-';c=getchar();} | ||
+ | while(isdigit(c)){t=(t<<1)+(t<<3)+(c&15);c=getchar();} | ||
+ | return sign?-t:t; | ||
+ | } | ||
+ | inline LL read_LL(){ | ||
+ | LL t=0;bool sign=false;char c=getchar(); | ||
+ | while(!isdigit(c)){sign|=c=='-';c=getchar();} | ||
+ | while(isdigit(c)){t=(t<<1)+(t<<3)+(c&15);c=getchar();} | ||
+ | return sign?-t:t; | ||
+ | } | ||
+ | inline char get_char(){ | ||
+ | char c=getchar(); | ||
+ | while(c==' '||c=='\n'||c=='\r')c=getchar(); | ||
+ | return c; | ||
+ | } | ||
+ | inline void write(LL x){ | ||
+ | register char c[21],len=0; | ||
+ | if(!x)return putchar('0'),void(); | ||
+ | if(x<0)x=-x,putchar('-'); | ||
+ | while(x)c[++len]=x%10,x/=10; | ||
+ | while(len)putchar(c[len--]+48); | ||
+ | } | ||
+ | inline void space(LL x){write(x),putchar(' ');} | ||
+ | inline void enter(LL x){write(x),putchar('\n');} | ||
+ | const int MAXN=2e5+5; | ||
+ | struct Node{ | ||
+ | int lef,rig,val; | ||
+ | }; | ||
+ | Node node[MAXN*40]; | ||
+ | int cnt,root[MAXN]; | ||
+ | int nodecopy(int k){ | ||
+ | node[++cnt]=node[k]; | ||
+ | return cnt; | ||
+ | } | ||
+ | void build(int &k,int lef,int rig){ | ||
+ | k=++cnt; | ||
+ | int mid=lef+rig>>1; | ||
+ | if(lef==rig) | ||
+ | return; | ||
+ | build(node[k].lef,lef,mid); | ||
+ | build(node[k].rig,mid+1,rig); | ||
+ | } | ||
+ | void update(int &k,int p,int lef,int rig,int pos){ | ||
+ | k=nodecopy(p); | ||
+ | node[k].val++; | ||
+ | if(lef==rig) | ||
+ | return; | ||
+ | int mid=lef+rig>>1; | ||
+ | if(mid<pos) | ||
+ | update(node[k].rig,node[p].rig,mid+1,rig,pos); | ||
+ | else | ||
+ | update(node[k].lef,node[p].lef,lef,mid,pos); | ||
+ | } | ||
+ | int query(int k1,int k2,int lef,int rig,int L,int R){ | ||
+ | if(L<=lef&&rig<=R) | ||
+ | return node[k1].val-node[k2].val; | ||
+ | int mid=lef+rig>>1; | ||
+ | if(mid>=R) | ||
+ | return query(node[k1].lef,node[k2].lef,lef,mid,L,R); | ||
+ | else if(mid<L) | ||
+ | return query(node[k1].rig,node[k2].rig,mid+1,rig,L,R); | ||
+ | else | ||
+ | return query(node[k1].lef,node[k2].lef,lef,mid,L,R)+query(node[k1].rig,node[k2].rig,mid+1,rig,L,R); | ||
+ | } | ||
+ | int head[MAXN],edge_cnt; | ||
+ | struct Edge{ | ||
+ | int to,next; | ||
+ | LL w; | ||
+ | }edge[MAXN]; | ||
+ | void Insert(int u,int v,LL w){ | ||
+ | edge[++edge_cnt].next=head[u]; | ||
+ | edge[edge_cnt].to=v; | ||
+ | edge[edge_cnt].w=w; | ||
+ | head[u]=edge_cnt; | ||
+ | } | ||
+ | LL dep[MAXN],dw[MAXN],b[MAXN]; | ||
+ | int dfs_t,lf[MAXN],rf[MAXN]; | ||
+ | void dfs(int u,LL d){ | ||
+ | lf[u]=++dfs_t;dep[u]=d;dw[dfs_t]=d; | ||
+ | for(int i=head[u];i;i=edge[i].next){ | ||
+ | int v=edge[i].to; | ||
+ | dfs(v,d+edge[i].w); | ||
+ | } | ||
+ | rf[u]=dfs_t; | ||
+ | } | ||
+ | int main() | ||
+ | { | ||
+ | int n=read_int(),u; | ||
+ | LL L=read_LL(),w; | ||
+ | _rep(i,2,n){ | ||
+ | u=read_int(),w=read_LL(); | ||
+ | Insert(u,i,w); | ||
+ | } | ||
+ | dfs(1,0); | ||
+ | memcpy(b,dep,sizeof(dep)); | ||
+ | sort(b+1,b+n+1); | ||
+ | int m=unique(b+1,b+n+1)-b; | ||
+ | build(root[0],1,m); | ||
+ | _rep(i,1,n) | ||
+ | update(root[i],root[i-1],1,m,lower_bound(b+1,b+m,dw[i])-b); | ||
+ | _rep(i,1,n) | ||
+ | enter(query(root[rf[i]],root[lf[i]-1],1,m,1,upper_bound(b+1,b+m,dep[i]+L)-b-1)); | ||
+ | return 0; | ||
+ | } | ||
+ | </code> | ||
+ | </hidden> | ||
+ | |||
+ | <hidden 正解代码> | ||
+ | <code cpp> | ||
+ | #include<bits/stdc++.h> | ||
+ | using namespace std; | ||
+ | const int N=200005; | ||
+ | int n,id[N],r[N],now,ans[N]; | ||
+ | pair<long long,int> a[N]; | ||
+ | struct BIT | ||
+ | { | ||
+ | int tree[N]; | ||
+ | void modify(int x,int val) | ||
+ | { | ||
+ | for(;x<=n;x+=x&-x) | ||
+ | tree[x]+=val; | ||
+ | } | ||
+ | int query(int x) | ||
+ | { | ||
+ | int ans=0; | ||
+ | for(;x;x-=x&-x) | ||
+ | ans+=tree[x]; | ||
+ | return ans; | ||
+ | } | ||
+ | }T; | ||
+ | //树状数组模板 | ||
+ | vector<pair<int,long long>> mat[N]; | ||
+ | void dfs(int k) | ||
+ | { | ||
+ | id[k]=++now; | ||
+ | //dfs序 | ||
+ | for(auto e:mat[k]) | ||
+ | { | ||
+ | a[e.first]=make_pair(a[k].first+e.second,e.first); | ||
+ | dfs(e.first); | ||
+ | } | ||
+ | r[k]=now; | ||
+ | //结束时间戳 | ||
+ | } | ||
+ | int main() | ||
+ | { | ||
+ | long long l; | ||
+ | cin>>n>>l; | ||
+ | for(int i=2;i<=n;i++) | ||
+ | { | ||
+ | int p; | ||
+ | long long w; | ||
+ | cin>>p>>w; | ||
+ | mat[p].push_back(make_pair(i,w)); | ||
+ | } | ||
+ | a[1]=make_pair(0ll,1); | ||
+ | dfs(1); | ||
+ | sort(a+1,a+n+1); | ||
+ | int j=n; | ||
+ | for(int i=n;i;i--) | ||
+ | { | ||
+ | for(;a[j].first-a[i].first>l;j--) | ||
+ | T.modify(id[a[j].second],-1); | ||
+ | //删除超过距离超过L的点 | ||
+ | T.modify(id[a[i].second],1); | ||
+ | //插入当前点 | ||
+ | ans[a[i].second]=T.query(r[a[i].second])-T.query(id[a[i].second]-1); | ||
+ | //统计子树答案 | ||
+ | } | ||
+ | for(int i=1;i<=n;i++) | ||
+ | cout<<ans[i]<<endl; | ||
+ | return 0; | ||
+ | } | ||
+ | </code> | ||
+ | </hidden> | ||
+ | |||
+ | ===== J. First! ===== | ||
+ | |||
+ | === 题解 === | ||
+ | |||
+ | $\text{Trie}$ 建树,然后询问每个单词结点,如果他是某个单词结点的子结点,一定不合法。 | ||
+ | |||
+ | 沿路把 $\text{Trie}$ 的其他分支代表的字母向该路径代表字母连单向边,最后跑一下 $26$ 个字母的拓扑,看看合不合法。 | ||
+ | |||
+ | <hidden 赛后代码> | ||
+ | <code cpp> | ||
+ | #include <iostream> | ||
+ | #include <cstdio> | ||
+ | #include <cstdlib> | ||
+ | #include <algorithm> | ||
+ | #include <string> | ||
+ | #include <sstream> | ||
+ | #include <cstring> | ||
+ | #include <cctype> | ||
+ | #include <cmath> | ||
+ | #include <vector> | ||
+ | #include <set> | ||
+ | #include <map> | ||
+ | #include <stack> | ||
+ | #include <queue> | ||
+ | #include <ctime> | ||
+ | #include <cassert> | ||
+ | #define _for(i,a,b) for(int i=(a);i<(b);++i) | ||
+ | #define _rep(i,a,b) for(int i=(a);i<=(b);++i) | ||
+ | #define mem(a,b) memset(a,b,sizeof(a)) | ||
+ | using namespace std; | ||
+ | typedef long long LL; | ||
+ | inline int read_int(){ | ||
+ | int t=0;bool sign=false;char c=getchar(); | ||
+ | while(!isdigit(c)){sign|=c=='-';c=getchar();} | ||
+ | while(isdigit(c)){t=(t<<1)+(t<<3)+(c&15);c=getchar();} | ||
+ | return sign?-t:t; | ||
+ | } | ||
+ | inline LL read_LL(){ | ||
+ | LL t=0;bool sign=false;char c=getchar(); | ||
+ | while(!isdigit(c)){sign|=c=='-';c=getchar();} | ||
+ | while(isdigit(c)){t=(t<<1)+(t<<3)+(c&15);c=getchar();} | ||
+ | return sign?-t:t; | ||
+ | } | ||
+ | inline char get_char(){ | ||
+ | char c=getchar(); | ||
+ | while(c==' '||c=='\n'||c=='\r')c=getchar(); | ||
+ | return c; | ||
+ | } | ||
+ | inline void write(LL x){ | ||
+ | register char c[21],len=0; | ||
+ | if(!x)return putchar('0'),void(); | ||
+ | if(x<0)x=-x,putchar('-'); | ||
+ | while(x)c[++len]=x%10,x/=10; | ||
+ | while(len)putchar(c[len--]+48); | ||
+ | } | ||
+ | inline void space(LL x){write(x),putchar(' ');} | ||
+ | inline void enter(LL x){write(x),putchar('\n');} | ||
+ | const int MAXS=3e5+5,MAXN=3e5+5; | ||
+ | int ch[MAXS][26],val[MAXS],cnt; | ||
+ | char temp[MAXN],*buf[MAXN]; | ||
+ | void Insert(char *s){ | ||
+ | int len=strlen(s),pos=0; | ||
+ | _for(i,0,len){ | ||
+ | int c=s[i]-'a'; | ||
+ | if(!ch[pos][c]) | ||
+ | ch[pos][c]=++cnt; | ||
+ | pos=ch[pos][c]; | ||
+ | } | ||
+ | val[pos]++; | ||
+ | } | ||
+ | int a[26][26],in[26],vis[26]; | ||
+ | bool topu(){ | ||
+ | mem(in,0); | ||
+ | _for(i,0,26) | ||
+ | _for(j,0,26){ | ||
+ | if(a[i][j]) | ||
+ | in[j]++; | ||
+ | } | ||
+ | queue<int> q; | ||
+ | _for(i,0,26){ | ||
+ | if(!in[i]) | ||
+ | q.push(i); | ||
+ | } | ||
+ | while(!q.empty()){ | ||
+ | int u=q.front();q.pop(); | ||
+ | _for(i,0,26){ | ||
+ | if(a[u][i]){ | ||
+ | in[i]--; | ||
+ | if(!in[i]) | ||
+ | q.push(i); | ||
+ | } | ||
+ | } | ||
+ | } | ||
+ | _for(i,0,26) | ||
+ | if(in[i]) | ||
+ | return false; | ||
+ | return true; | ||
+ | } | ||
+ | bool query(char *s){ | ||
+ | mem(a,0); | ||
+ | int len=strlen(s),pos=0; | ||
+ | _for(i,0,len){ | ||
+ | if(val[pos]) | ||
+ | return false; | ||
+ | int c=s[i]-'a'; | ||
+ | _for(j,0,26){ | ||
+ | if(j==c||ch[pos][j]==0) | ||
+ | continue; | ||
+ | a[j][c]=1; | ||
+ | } | ||
+ | pos=ch[pos][c]; | ||
+ | } | ||
+ | return topu(); | ||
+ | } | ||
+ | bool ok[MAXN]; | ||
+ | int main() | ||
+ | { | ||
+ | int n=read_int(); | ||
+ | _for(i,0,n){ | ||
+ | gets(temp); | ||
+ | buf[i]=(char*)malloc(strlen(temp)+1); | ||
+ | strcpy(buf[i],temp); | ||
+ | } | ||
+ | _for(i,0,n) | ||
+ | Insert(buf[i]); | ||
+ | int tot=0; | ||
+ | _for(i,0,n) | ||
+ | ok[i]=query(buf[i]),tot+=ok[i]; | ||
+ | enter(tot); | ||
+ | _for(i,0,n) | ||
+ | if(ok[i]) | ||
+ | puts(buf[i]); | ||
+ | return 0; | ||
+ | } | ||
+ | </code> | ||
+ | </hidden> |