HDU 5927 Auxiliary Set 【DFS+树】(2016CCPC东北地区大学生程序设计竞赛)

Auxiliary Set

Time Limit: 9000/4500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 873 Accepted Submission(s): 271


[align=left]Problem Description[/align]
Given a rooted tree with n vertices, some of the vertices are important.

An auxiliary set is a set containing vertices satisfying at least one of the two conditions：

∙It is an important vertex
∙It is the least common ancestor of two different important vertices.

You are given a tree with n vertices (1 is the root) and q queries.

Each query is a set of nodes which indicates the unimportant vertices in the tree. Answer the size (i.e. number of vertices) of the auxiliary set for each query.

[align=left]Input[/align]
The first line contains only one integer T (T≤1000), which indicates the number of test cases.

For each test case, the first line contains two integers n (1≤n≤100000), q (0≤q≤100000).

In the following n -1 lines, the i-th line contains two integers ui,vi(1≤ui,vi≤n) indicating there is an edge between uii and vi in the tree.

In the next q lines, the i-th line first comes with an integer mi(1≤mi≤100000) indicating the number of vertices in the query set.Then comes with mi different integers, indicating the nodes in the query set.

It is guaranteed that ∑qi=1mi≤100000.

It is also guaranteed that the number of test cases in which n≥1000 or ∑qi=1mi≥1000 is no more than 10.

[align=left]Output[/align]
For each test case, first output one line "Case #x:", where x is the case number (starting from 1).

Then q lines follow, i-th line contains an integer indicating the size of the auxiliary set for each query.

[align=left]Sample Input[/align]

1
6 3
6 4
2 5
5 4
1 5
5 3
3 1 2 3
1 5
3 3 1 4

[align=left]Sample Output[/align]

Case #1:
3
6
3

Hint

1 //
2 //by coolxxx
3 //#include<bits/stdc++.h>
4 #include<iostream>
5 #include<algorithm>
6 #include<string>
7 #include<iomanip>
8 #include<map>
9 #include<stack>
10 #include<queue>
11 #include<set>
12 #include<bitset>
13 #include<memory.h>
14 #include<time.h>
15 #include<stdio.h>
16 #include<stdlib.h>
17 #include<string.h>
18 //#include<stdbool.h>
19 #include<math.h>
20 #pragma comment(linker,"/STACK:1024000000,1024000000")
21 #define min(a,b) ((a)<(b)?(a):(b))
22 #define max(a,b) ((a)>(b)?(a):(b))
23 #define abs(a) ((a)>0?(a):(-(a)))
24 #define lowbit(a) (a&(-a))
25 #define sqr(a) ((a)*(a))
26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
27 #define mem(a,b) memset(a,b,sizeof(a))
28 #define eps (1e-10)
29 #define J 10000
30 #define mod 1000000007
31 #define MAX 0x7f7f7f7f
32 #define PI 3.14159265358979323
33 #define N 100004
34 using namespace std;
35 typedef long long LL;
36 double anss;
37 LL aans;
38 int cas,cass;
39 int n,m,lll,ans;
40 int s
,b
,c
,fa
,last
,d
;
41 struct xxx
42 {
43     int to,next;
44 }a[N+N];
45 void add(int x,int y)
46 {
47     a[++lll].next=last[x];
48     last[x]=lll;
49     a[lll].to=y;
50 }
51 bool cmp(int a,int b)
52 {
53     return d[a]>d[b];
54 }
55 void dfs(int now,int ffa)
56 {
57     int i,to;
58     s[now]=1;c[now]=0;d[now]=d[ffa]+1;fa[now]=ffa;
59     for(i=last[now];i;i=a[i].next)
60     {
61         to=a[i].to;
62         if(to==ffa)continue;
63         dfs(to,now);
64         s[now]++;
65     }
66 }
67 void work()
68 {
69     int i,x,mm;
70     ans=n;
71     scanf("%d",&mm);
72     for(i=1;i<=mm;i++)scanf("%d",&b[i]);
73     sort(b+1,b+1+mm,cmp);
74     for(i=1;i<=mm;i++)
75     {
76         x=b[i];
77         c[x]++;
78         if(s[x]==c[x])c[fa[x]]++;
79         if(s[x]-c[x]<2)ans--;
80     }
81     for(i=1;i<=mm;i++)c[b[i]]=c[fa[b[i]]]=0;
82     printf("%d\n",ans);
83 }
84 int main()
85 {
86     #ifndef ONLINE_JUDGEW
87 //    freopen("1.txt","r",stdin);
88 //    freopen("2.txt","w",stdout);
89     #endif
90     int i,j,k;
91     int x,y,z;
92 //    init();
93 //    for(scanf("%d",&cass);cass;cass--)
94     for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
95 //    while(~scanf("%s",s))
96 //    while(~scanf("%d%d",&n,&m))
97     {
98         printf("Case #%d:\n",cass);
99         lll=0;mem(last,0);
100         scanf("%d%d",&n,&m);
101         for(i=1;i<n;i++)
102         {
103             scanf("%d%d",&x,&y);
104             add(x,y),add(y,x);
105         }
106         dfs(1,0);
107         for(i=1;i<=m;i++)
108             work();
109     }
110     return 0;
111 }
112 /*
113 //
114
115 //
116 */

View Code