标签:iat std bug media air queue tar ios 单点
InputThe first line contains a single positive integer T( T <= 10 ), indicates the number of test cases.
For each test case:
The first line contains an integer N (N ≤ 50,000) , which is the number of the employees.
The following N - 1 lines each contain two integers u and v, which means the employee v is the immediate boss of employee u(1<=u,v<=N).
The next line contains an integer M (M ≤ 50,000).
The following M lines each contain a message which is either
"C x" which means an inquiry for the current task of employee x
or
"T x y"which means the company assign task y to employee x.
(1<=x<=N,0<=y<=10^9)OutputFor each test case, print the test case number (beginning with 1) in the first line and then for every inquiry, output the correspond answer per line.Sample Input
1 5 4 3 3 2 1 3 5 2 5 C 3 T 2 1 C 3 T 3 2 C 3
Sample Output
Case #1: -1 1 2
读题之后发现建树是个问题,然后就学习了dfs建树。有一个结论,如果v是u的祖先,那么dfs序st[v]<st[u]&&ed[v]>ed[u]
于是就可以建树了。然后就是打标记查标记
1 #include <iostream> 2 #include <stdio.h> 3 #include <math.h> 4 #include <string.h> 5 #include <stdlib.h> 6 #include <string> 7 #include <vector> 8 #include <set> 9 #include <map> 10 #include <queue> 11 #include <algorithm> 12 #include <sstream> 13 #include <stack> 14 using namespace std; 15 #define FO freopen("in.txt","r",stdin); 16 #define rep(i,a,n) for (int i=a;i<n;i++) 17 #define per(i,a,n) for (int i=n-1;i>=a;i--) 18 #define pb push_back 19 #define mp make_pair 20 #define all(x) (x).begin(),(x).end() 21 #define fi first 22 #define se second 23 #define SZ(x) ((int)(x).size()) 24 #define debug(x) cout << "&&" << x << "&&" << endl; 25 #define lowbit(x) (x&-x) 26 #define mem(a,b) memset(a, b, sizeof(a)); 27 typedef vector<int> VI; 28 typedef long long ll; 29 typedef pair<int,int> PII; 30 const ll mod=1000000007; 31 const int inf = 0x3f3f3f3f; 32 ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} 33 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 34 //head 35 36 const int maxn=50010; 37 int _,lazy[maxn<<3],st[maxn],ed[maxn],cur,m,vis[maxn],n; 38 vector<int> boss[maxn]; 39 40 void dfs(int rt) {//建树 41 st[rt]=++cur; 42 for(int i=0;i<boss[rt].size();i++) { 43 dfs(boss[rt][i]); 44 } 45 ed[rt]=cur; 46 } 47 48 void pushdown(int rt) { 49 if(lazy[rt]!=-1) { 50 lazy[rt<<1]=lazy[rt]; 51 lazy[rt<<1|1]=lazy[rt]; 52 lazy[rt]=-1; 53 } 54 } 55 56 void build(int rt,int L,int R) { 57 lazy[rt]=-1; 58 if(L==R) return; 59 int mid=(L+R)>>1; 60 build(rt<<1,L,mid); 61 build(rt<<1|1,mid+1,R); 62 } 63 64 void updata(int rt,int L,int R,int l,int r,int zhi) { 65 if(L>=l&&R<=r) { 66 lazy[rt]=zhi; 67 return; 68 } 69 pushdown(rt); 70 int mid=(L+R)>>1; 71 if(l<=mid) updata(rt<<1,L,mid,l,r,zhi); 72 if(r>mid) updata(rt<<1|1,mid+1,R,l,r,zhi); 73 } 74 75 int query(int rt,int L,int R,int pos) { 76 if(L==R) return lazy[rt];//单点查 77 pushdown(rt); 78 int mid=(L+R)>>1; 79 if(pos<=mid) query(rt<<1,L,mid,pos); 80 else query(rt<<1|1,mid+1,R,pos); 81 } 82 83 int curr=1; 84 int main() { 85 for(scanf("%d",&_);_;_--) { 86 printf("Case #%d:\n",curr++); 87 cur=0; 88 mem(boss,0); 89 mem(vis,0); 90 scanf("%d",&n); 91 int u,v; 92 rep(i,1,n) {//存关系 93 scanf("%d%d",&u,&v); 94 boss[v].push_back(u); 95 vis[u]=1; 96 } 97 rep(i,1,n+1) {//找到根 98 if(!vis[i]) { 99 dfs(i); 100 break; 101 } 102 } 103 build(1,1,cur);//建树 104 scanf("%d",&m); 105 char s[2]; 106 int pos,zhi; 107 while(m--) { 108 scanf("%s",s); 109 if(s[0]==‘T‘) { 110 scanf("%d%d",&pos,&zhi); 111 updata(1,1,cur,st[pos],ed[pos],zhi);//区间st[pos]-ed[pos]是pos的员工 112 } else { 113 scanf("%d",&pos); 114 printf("%d\n",query(1,1,cur,st[pos]));//查pos的任务(ed[pos]就不是了) 115 } 116 } 117 } 118 }
kuangbin专题七 HDU3974 Assign the task (dfs时间戳建树)
标签:iat std bug media air queue tar ios 单点
原文地址:https://www.cnblogs.com/ACMerszl/p/9895421.html