Successor
Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on HDU. Original ID: 436664-bit integer IO format: %I64d Java class name: Main
Sean owns a company and he is the BOSS.The other Staff has one Superior.every staff has a loyalty and ability.Some times Sean will fire one staff.Then one of the fired man’s Subordinates will replace him whose ability is higher than him and has the highest loyalty for company.Sean want to know who will replace the fired man.
Input
In the first line a number T indicate the number of test cases. Then for each case the first line contain 2 numbers n,m (2<=n,m<=50000),indicate the company has n person include Sean ,m is the times of Sean’s query.Staffs are numbered from 1 to n-1,Sean’s number is 0.Follow n-1 lines,the i-th(1<=i<=n-1) line contains 3 integers a,b,c(0<=a<=n-1,0<=b,c<=1000000),indicate the i-th staff’s superior Serial number,i-th staff’s loyalty and ability.Every staff ‘s Serial number is bigger than his superior,Each staff has different loyalty.then follows m lines of queries.Each line only a number indicate the Serial number of whom should be fired.
Output
For every query print a number:the Serial number of whom would replace the losing job man,If there has no one to replace him,print -1.
Sample Input
1 3 2 0 100 99 1 101 100 1 2
Sample Output
2 -1
Source
解题：线段树，dfs序列得出区间，然后将员工按能力降序排序，然后利用线段树查询即可！不过G++爆栈，选择C++交
1 #include <iostream> 2 #include <vector> 3 #include <cstdio> 4 #include <algorithm> 5 #include <cstring> 6 #pragma comment(linker, "/STACK:102400000,102400000") 7 using namespace std; 8 const int maxn = 200010; 9 struct node { 10 int maxv,id; 11 } tree[maxn<<2]; 12 void update(int L,int R,int lt,int rt,int val,int id,int v) { 13 if(lt <= L && rt >= R) { 14 tree[v].maxv = val; 15 tree[v].id = id; 16 return ; 17 } 18 int mid = (L + R)>>1; 19 if(lt <= mid) update(L,mid,lt,rt,val,id,v<<1); 20 if(rt > mid) update(mid+1,R,lt,rt,val,id,v<<1|1); 21 if(tree[v<<1].maxv > tree[v<<1|1].maxv) 22 tree[v] = tree[v<<1]; 23 else tree[v] = tree[v<<1|1]; 24 } 25 node query(int L,int R,int lt,int rt,int v) { 26 if(lt <= L && rt >= R) return tree[v]; 27 int mid = (L + R)>>1; 28 node a,b; 29 a.maxv = -1,b.maxv = -1; 30 a.id = -1,b.id = -1; 31 if(lt <= mid) a = query(L,mid,lt,rt,v<<1); 32 if(rt > mid) b = query(mid+1,R,lt,rt,v<<1|1); 33 if(a.maxv > b.maxv) return a; 34 return b; 35 } 36 vector<int>g[maxn]; 37 int L[maxn],R[maxn],ret[maxn],tt; 38 void dfs(int u) { 39 L[u] = tt++; 40 for(int i = g[u].size()-1; i >= 0; --i) 41 dfs(g[u][i]); 42 R[u] = tt-1; 43 } 44 struct STAFF { 45 int id,ability,loyalty; 46 bool operator<(const STAFF &t)const { 47 if(ability == t.ability) return id < t.id; 48 return ability > t.ability; 49 } 50 } staff[maxn]; 51 int main() { 52 int kase,n,m,fa; 53 scanf("%d",&kase); 54 while(kase--) { 55 scanf("%d%d",&n,&m); 56 for(int i = tt = 0; i <= n; ++i) g[i].clear(); 57 for(int i = 1; i < n; ++i) { 58 scanf("%d%d%d",&fa,&staff[i-1].loyalty,&staff[i-1].ability); 59 g[fa].push_back(i); 60 staff[i-1].id = i; 61 } 62 dfs(0); 63 sort(staff,staff+n-1); 64 memset(tree,-1,sizeof tree); 65 for(int i = 0; i < n-1; ++i) { 66 node now = query(0,R[0],L[staff[i].id],R[staff[i].id],1); 67 ret[staff[i].id] = now.id; 68 update(0,R[0],L[staff[i].id],L[staff[i].id],staff[i].loyalty,staff[i].id,1); 69 } 70 while(m--) { 71 scanf("%d",&fa); 72 printf("%d ",ret[fa]); 73 } 74 } 75 return 0; 76 }