【LCA】CodeForce #326 Div.2 E:Duff in the Army
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Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.
There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is iand he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.
Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.
To answer a query:
Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order.
If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).
Duff is very busy at the moment, so she asked you to help her and answer the queries.
The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).
The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).
Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).
Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).
For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.
5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1
1 3
2 2 3
0
3 1 2 4
1 2
Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):
大约题目是给一棵树给m个人在哪个点上的信息
然后给q个询问,每次问u到v上的路径有的点上编号最小的k个人,k<=10(很关键)
u到v上路径的询问很容易想到lca
但是前k个答案很不好搞?
直接在lca数组里面开个Num[11]记录前10个在该点上的编号
码了1个半小时结果wa成狗- -
最后发现lca打挂了。。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 #include<stack> 6 #include<vector> 7 8 #define maxn 100001 9 10 using namespace std; 11 12 inline int in() 13 { 14 int x=0,f=1;char ch=getchar(); 15 while((ch<'0'||ch>'9')&&ch!='-')ch=getchar(); 16 if(ch=='-')f=-1; 17 while(ch<='9'&&ch>='0')x=x*10+ch-'0',ch=getchar(); 18 return f*x; 19 } 20 21 struct ed{ 22 int to,last; 23 }edge[maxn*2]; 24 25 struct lc{ 26 int father,num[11]; 27 }f[19][maxn]; 28 29 int last[maxn],tot=0,dep[maxn],n; 30 31 void add(int u,int v) 32 { 33 edge[++tot].to=v,edge[tot].last=last[u],last[u]=tot; 34 edge[++tot].to=u,edge[tot].last=last[v],last[v]=tot; 35 } 36 37 void dfs(int poi,int lastt,int de) 38 { 39 dep[poi]=de; 40 if(lastt!=-1)f[0][poi].father=lastt; 41 for(int i=last[poi];i;i=edge[i].last) 42 if(edge[i].to!=lastt)dfs(edge[i].to,poi,de+1); 43 } 44 45 void update(int pos,int cen) 46 { 47 int i=1,j=1; 48 while(i<=f[cen-1][f[cen-1][pos].father].num[0]&&j<=f[cen-1][pos].num[0]&&f[cen][pos].num[0]<10) 49 { 50 if(i<=f[cen-1][f[cen-1][pos].father].num[0]&&f[cen-1][f[cen-1][pos].father].num[i]<f[cen-1][pos].num[j])f[cen][pos].num[++f[cen][pos].num[0]]=f[cen-1][f[cen-1][pos].father].num[i++]; 51 else if(j<=f[cen-1][pos].num[0])f[cen][pos].num[++f[cen][pos].num[0]]=f[cen-1][pos].num[j++]; 52 } 53 while(i<=f[cen-1][f[cen-1][pos].father].num[0]&&f[cen][pos].num[0]<10)f[cen][pos].num[++f[cen][pos].num[0]]=f[cen-1][f[cen-1][pos].father].num[i++]; 54 while(j<=f[cen-1][pos].num[0]&&f[cen][pos].num[0]<10)f[cen][pos].num[++f[cen][pos].num[0]]=f[cen-1][pos].num[j++]; 55 } 56 57 void pre() 58 { 59 for(int i=1;(1<<i)<=n;i++) 60 for(int j=1;j<=n;j++) 61 if(f[i-1][f[i-1][j].father].father)f[i][j].father=f[i-1][f[i-1][j].father].father,update(j,i); 62 } 63 64 int ANS[11],ANS_B[11]; 65 66 void Up(int pos,int cen,int kk) 67 { 68 ANS_B[0]=0; 69 int i=1,j=1; 70 while(i<=ANS[0]&&j<=f[cen][pos].num[0]&&ANS_B[0]<kk) 71 { 72 if(i<=ANS[0]&&ANS[i]<f[cen][pos].num[j])ANS_B[++ANS_B[0]]=ANS[i++]; 73 else if(j<=f[cen][pos].num[0])ANS_B[++ANS_B[0]]=f[cen][pos].num[j++]; 74 } 75 while(i<=ANS[0]&&ANS_B[0]<kk)ANS_B[++ANS_B[0]]=ANS[i++]; 76 while(j<=f[cen][pos].num[0]&&ANS_B[0]<kk)ANS_B[++ANS_B[0]]=f[cen][pos].num[j++]; 77 for(int i=0;i<=ANS_B[0];i++)ANS[i]=ANS_B[i]; 78 } 79 80 void print() 81 { 82 printf("%d",ANS[0]); 83 for(int i=1;i<=ANS[0];i++)printf(" %d",ANS[i]); 84 printf("\n"); 85 } 86 87 void lca(int u,int v,int kk) 88 { 89 ANS[0]=0; 90 int nu; 91 if(dep[u]>dep[v])swap(u,v); 92 if(dep[v]!=dep[u]) 93 { 94 nu=log2(dep[v]-dep[u]); 95 for(int i=nu;i>=0;i--) 96 if((1<<i) & (dep[v]-dep[u]))Up(v,i,kk),v=f[i][v].father; 97 } 98 if(u==v) 99 {100 Up(u,0,kk);101 print();102 return;103 }104 nu=log2(dep[v]);105 while(nu!=-1)106 {107 if(f[nu][u].father==f[nu][v].father){nu--;continue;}108 Up(v,nu,kk);109 Up(u,nu,kk);110 u=f[nu][u].father;111 v=f[nu][v].father;112 nu--;113 }114 Up(v,0,kk);115 Up(u,0,kk);116 Up(f[0][u].father,0,kk);117 print();118 }119 120 int main()121 {122 freopen("t.in","r",stdin);123 int m,q,u,v,kk;124 n=in(),m=in(),q=in();125 for(int i=1;i<n;i++)126 u=in(),v=in(),add(u,v);127 for(int i=1;i<=m;i++)128 {129 u=in();130 if(f[0][u].num[0]<10)f[0][u].num[++f[0][u].num[0]]=i;131 }132 dfs(1,-1,0);133 pre();134 for(int i=1;i<=q;i++)135 {136 u=in(),v=in(),kk=in();137 lca(u,v,kk);138 }139 return 0;140 }
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