Boring counting
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 98304/98304 K (Java/Others)
Total Submission(s): 2808 Accepted Submission(s): 826
Problem Description
In this problem we consider a rooted tree with N vertices. The vertices are numbered from 1 to N, and vertex 1 represents the root. There are integer weights on each vectice. Your task is to answer a list of queries, for each query, please tell us among all the vertices in the subtree rooted at vertice u, how many different kinds of weights appear exactly K times?
Input
The first line of the input contains an integer T( T<= 5 ), indicating the number of test cases.
For each test case, the first line contains two integers N and K, as described above. ( 1<= N <= 105, 1 <= K <= N )
Then come N integers in the second line, they are the weights of vertice 1 to N. ( 0 <= weight <= 109 )
For next N-1 lines, each line contains two vertices u and v, which is connected in the tree.
Next line is a integer Q, representing the number of queries. (1 <= Q <= 105)
For next Q lines, each with an integer u, as the root of the subtree described above.
For each test case, the first line contains two integers N and K, as described above. ( 1<= N <= 105, 1 <= K <= N )
Then come N integers in the second line, they are the weights of vertice 1 to N. ( 0 <= weight <= 109 )
For next N-1 lines, each line contains two vertices u and v, which is connected in the tree.
Next line is a integer Q, representing the number of queries. (1 <= Q <= 105)
For next Q lines, each with an integer u, as the root of the subtree described above.
Output
For each test case, output "Case #X:" first, X is the test number. Then output Q lines, each with a number -- the answer to each query.
Seperate each test case with an empty line.
Seperate each test case with an empty line.
Sample Input
1
3 1
1 2 2
1 2
1 3
3
2
1
3
3 1
1 2 2
1 2
1 3
3
2
1
3
Sample Output
Case #1:
1
1
1
Author
fish@UESTC_Oblivion
Source
题意：给你一个树 每个节点都有一个权值 q个询问 问x的子树中有多少种不同的权值出现了k次
题解：第一题莫队 莫队算法 正如莫涛本人而言这种方法是处理一类无修改的离线区间询问问题的 dfs序将子树的问题转换为区间问题
之后就是莫队的处理，只是一个神奇的算法，在我看来就是一种有技巧的暴力。另外还要注意对权值的离散化。PE orzz
1 /****************************** 2 code by drizzle 3 blog: www.cnblogs.com/hsd-/ 4 ^ ^ ^ ^ 5 O O 6 ******************************/ 7 #include<bits/stdc++.h> 8 #include<map> 9 #include<set> 10 #include<cmath> 11 #include<queue> 12 #include<bitset> 13 #include<math.h> 14 #include<vector> 15 #include<string> 16 #include<stdio.h> 17 #include<cstring> 18 #include<iostream> 19 #include<algorithm> 20 #pragma comment(linker, "/STACK:102400000,102400000") 21 using namespace std; 22 #define A first 23 #define B second 24 const int mod=1000000007; 25 const int MOD1=1000000007; 26 const int MOD2=1000000009; 27 const double EPS=0.00000001; 28 typedef __int64 ll; 29 const ll MOD=1000000007; 30 const int INF=1000000010; 31 const ll MAX=1ll<<55; 32 const double eps=1e-5; 33 const double inf=~0u>>1; 34 const double pi=acos(-1.0); 35 typedef double db; 36 typedef unsigned int uint; 37 typedef unsigned long long ull; 38 int T; 39 int n,k,w[100005],v[100005],vv[100005],x,y,q; 40 int in[100005],out[100005]; 41 int pre[100005]; 42 int pos[100005]; 43 int re[100005]; 44 int sum[100005]; 45 int dfn=0; 46 int ans;int nedge=0; 47 struct node 48 { 49 int pre,ed; 50 }N[200005]; 51 struct query 52 { 53 int id; 54 int l,r; 55 }M[100005]; 56 void add(int st,int ed ) 57 { 58 nedge++; 59 N[nedge].ed=ed; 60 N[nedge].pre=pre[st]; 61 pre[st]=nedge; 62 } 63 void getdfs(int now,int fa) 64 { 65 in[now]=++dfn; 66 for(int i=pre[now];i;i=N[i].pre) 67 { 68 if(N[i].ed!=fa) 69 { 70 getdfs(N[i].ed,now); 71 } 72 } 73 out[now]=dfn; 74 } 75 void init() 76 { 77 nedge=0; 78 memset(pre,0,sizeof(pre)); 79 dfn=0; 80 ans=0; 81 memset(sum,0,sizeof(sum)); 82 memset(N,0,sizeof(N)); 83 memset(M,0,sizeof(M)); 84 memset(re,0,sizeof(re)); 85 memset(pos,0,sizeof(pos)); 86 memset(in,0,sizeof(in)); 87 memset(out,0,sizeof(out)); 88 memset(w,0,sizeof(w)); 89 memset(v,0,sizeof(v)); 90 memset(vv,0,sizeof(vv)); 91 } 92 bool cmp(struct query aa,struct query bb) 93 { 94 if(pos[aa.id]<pos[bb.id]) 95 return true; 96 if(pos[aa.id]==pos[bb.id]) 97 { 98 if(aa.r<bb.r) 99 return true; 100 } 101 return false; 102 } 103 void ad(int value) 104 { 105 if(sum[value]==k) 106 ans--; 107 sum[value]++; 108 if(sum[value]==k) 109 ans++; 110 } 111 void del(int value) 112 { 113 if(sum[value]==k) 114 ans--; 115 sum[value]--; 116 if(sum[value]==k) 117 ans++; 118 } 119 int main() 120 { 121 scanf("%d",&T); 122 for(int i=1;i<=T;i++) 123 { 124 init(); 125 scanf("%d %d",&n,&k); 126 int block=sqrt(n); 127 for(int j=1;j<=n;j++){ 128 scanf("%d",&w[j]); 129 v[j]=w[j]; 130 } 131 sort(w+1,w+1+n); 132 for(int j=1;j<=n;j++){ 133 v[j]=lower_bound(w+1,w+n+1,v[j])-w-1; 134 } 135 for(int j=1;j<n;j++){ 136 scanf("%d %d",&x,&y); 137 add(x,y);add(y,x); 138 } 139 getdfs(1,1); 140 for(int j=1;j<=n;j++) 141 vv[in[j]]=v[j];// 将树上的权值 下放到区间当中 142 scanf("%d",&q); 143 int exm; 144 for(int j=1;j<=q;j++){ 145 scanf("%d",&exm); 146 M[j].l=in[exm]; 147 M[j].r=out[exm]; 148 M[j].id=j; 149 pos[j]=(M[j].l)/block;//分块 处理的不是特别好 150 } 151 printf("Case #%d: ",i); 152 sort(M+1,M+1+q,cmp); 153 int l=1, r=0;//初始区间 154 ad(vv[++r]); 155 for(int j=1; j<=q; j++) 156 { 157 while(r<M[j].r) ad(vv[++r]); 158 while(r>M[j].r) del(vv[r--]); 159 while(l<M[j].l) del(vv[l++]); 160 while(l>M[j].l) ad(vv[--l]); 161 re[M[j].id]=ans; 162 } 163 for(int j=1;j<=q;j++) 164 printf("%d ",re[j]); 165 if(i!=T) 166 printf(" "); 167 } 168 return 0; 169 }