数据结构（主席树）：HDU 4729 An Easy Problem for Elfness

An Easy Problem for Elfness

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1148    Accepted Submission(s): 234

Problem Description

Pfctgeorge
is totally a tall rich and handsome guy. He plans to build a huge water
transmission network that covers the whole southwest China. To save the
fund, there will be exactly one path between two cities.

Since
the water every city provides and costs every day is different, he needs
to transfer water from one particular city to another as much as
possible in the next few days. However the pipes which connect the
cities have a limited capacity for transmission. (Which means the water
that transfer though the pipe should not exceed a particular amount) So
he has to know the maximum water that the network can transfer in the
next few days.

He thought it's a maximum flow problem, so he
invites an expert in this field, Elfness (Also known as Xinhang senior
sister) to help him figure it out.

Unlike Pfctgeorge, Elfness
quickly finds that this problem is much easier than a normal maximum
flow problem, and is willing to help Pfctgeorge.

"Oh well, this problem is not a tough one. We can ..."

Abruptly, Pfctgeorge's iPhone rings, and ... the ringtone is Mo Di Da Biao Ke.

"You can make that? Excellent! "Pfctgeorge hangs up his iPhone, and turns to Elfness.

"Here's
good news for you. A construction team told me that every pipe's
capacity can be extended for one day. And the price for extending one
unit capacity varies from day to day. "

"Eh well, that's a good
news for you, not me. Now it's rather like a minimum cost ow problem,
right? But it's still not a tough one, let me have a think. "

After a few seconds' thought, Elfness comes up with a simple solution.

"Ok, we can solve it like... "

Abruptly, here comes Mo Di Da Biao Ke again.

"Seriously? You can build new pipes? Thank you very much. "

"OK,
my dear Elfness, we got more good news. Another construction team said
they can build one or more pipes between any two cities and their pipes
are exactly like the original ones except that they only work for one
day. And the capacity of the new pipes is only one, but they can be
extended, too. Of course, their price to build a single pipe also varies
in days. "

"You mean the new pipes can be extended too? Wow, things are getting more interesting. Give me a few minutes. "

Elfness takes out his new ultrabook which is awarded in VK cup and does some basic calculation.

"I get it. The problem can be solved ..."

Mo Di Da Biao Ke again, but this time it's from Elfness's phone.

"As you see, I have to go out. But I know someone else who can also solve this; I'll recommend this guy for you. "

And
of course, that poor guy is YOU. Help Pfctgeorge solve his problem, and
then the favorability about you from Elfness will raise a lot.

 

Input

The first line has a number T (T <= 10) , indicating the number of test cases.

The
first line of each test case is two integers N (1 <= N <= 100000)
and M (1 <= M <= 100000), indicating the number of the city that
the original network connects and the number of days when Pfctgeorge
needs to know about the maximum water transmissions. Then next N - 1
lines each describe a pipe that connects two cities. The format will be
like U, V , cap (1 <= U, V <= N and 0 <= cap < 10000), which
means the ids of the two cities the pipe connects and the transmission
limit of the pipe. As is said in description, the network that the
cities and pipes form is a tree (an undirected acyclic graph).

Then next M lines of the test case describe the information about the next few days. The format is like S, T, K, A, B(0 <= K <= 2^31 - 1, 1 <= A, B <= 2^31 - 1).
S means the source of the water while T means the sink. K means the
total budget in the day. A means the cost for a construction team to
build a new pipe and B means the cost for a construction team to extend
the capacity of a pipe.

I am glad to list the information of building a new pipe and extending the capacity.

1.
Pfctgeorge can build a new pipe between any two cities, no matter they
have been directly connected or not. Pfctgeorge can build more than one
new pipe between any two cities.
2. The capacity of the pipe that was newly built is one.
3. Pfctgeorge can extend the capacity of any existed pipe including the newly built one and the original one.
4. Each time you extend the capacity of one pipe, the capacity of that pipe increases one.
5. The cost of building a new pipe is A and the cost of extending a pipe is B.
6. You can take any constructions in any times and the only limit is to make sure the total costs not exceed the budget.
7. All the work that construction team does only lasts one single day.

 

Output

For every case, you should output "Case #t:" at first, without quotes. The t is the case number starting from 1.
Then for each day, output the maximum water Pfctgeorge can transfer from S and T with a budget of K.

 

Sample Input

2
5 1
1 2 2
1 3 5
2 4 1
4 5 2
1 5 3 3 2
5 5
1 2 10
2 3 2
3 4 7
2 5 7
1 5 0 1 3
1 3 0 2 3
1 5 3 2 3
1 2 7 3 1
1 3 2 3 1

 

Sample Output

Case #1:
2
Case #2:
7
2
8
17
4

Hint

In the first sample case, you can extend the capacity of the pipe which connects city2 and city4 by one, or just build a new pipe between city2 and city4.

　　感觉这道题就是一个笑话。

　　本地对拍会爆，但HDU上AC了……

　　考虑按深度建主席树，儿子承接父亲的信息，这样就可以在线段树上查询一条链的信息了。

　　三种决策：不停地新加边；先加一条边，再不停扩容；不停扩容。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int maxn=;
 #pragma comment(linker,"/STACK:1024000000,1024000000")
 int cntE,fir[maxn],to[maxn*],nxt[maxn*],cap[maxn*];
 int rt[maxn],ch[maxn*][],sum[maxn*],cal[maxn*],cnt;
 int dep[maxn];
 int max(int a,int b){return a>b?a:b;}
 void addedge(int a,int b,int c){
     nxt[++cntE]=fir[a];
     fir[a]=cntE;
     cap[cntE]=c;
     to[cntE]=b;
 }

 void Insert(int pre,int &rt,int l,int r,int g){
     rt=++cnt;
     ch[rt][]=ch[pre][];
     ch[rt][]=ch[pre][];
     sum[rt]=sum[pre]+;
     cal[rt]=cal[pre]+r-g+;
     if(l==r)return;
     int mid=(l+r)>>;
     if(mid>=g)Insert(ch[pre][],ch[rt][],l,mid,g);
     else Insert(ch[pre][],ch[rt][],mid+,r,g);
 }

 int ID[maxn],tot;
 int mm[maxn*],Min[maxn*][];

 void DFS(int x,int fa){
     Min[ID[x]=++tot][]=x;
     for(int i=fir[x];i;i=nxt[i])
         if(to[i]!=fa){
             dep[to[i]]=dep[x]+;
             Insert(rt[x],rt[to[i]],,,cap[i]);
             DFS(to[i],x);Min[++tot][]=x;
         }
 }

 void Prepare(){
     for(int i=;i<=tot;i++){
         if((i&(i-))==)mm[i]=mm[i-]+;
         else mm[i]=mm[i-];
     }
     for(int k=;k<=mm[tot];k++)
         for(int i=;i+(<<k)-<=tot;i++){
             if(dep[Min[i][k-]]<dep[Min[i+(<<k-)][k-]])
                 Min[i][k]=Min[i][k-];
             else
                 Min[i][k]=Min[i+(<<k-)][k-];
         }
 }

 int Lca(int x,int y){
     if(ID[x]>ID[y])swap(x,y);
     int k=mm[ID[y]-ID[x]+];
     if(dep[Min[ID[x]][k]]<dep[Min[ID[y]-(<<k)+][k]])
         return Min[ID[x]][k];
     return Min[ID[y]-(<<k)+][k];
 }

 #define num() sum[ch[r1][0]]+sum[ch[r2][0]]-2*sum[ch[lca][0]]
 #define key() cal[ch[r1][0]]+cal[ch[r2][0]]-2*cal[ch[lca][0]]

 int Query1(int lca,int r1,int r2,int l,int r){
     if(l==r)return l;
     int mid=(l+r)>>;
     if(num()>)
     return Query1(ch[lca][],ch[r1][],ch[r2][],l,mid);
     return Query1(ch[lca][],ch[r1][],ch[r2][],mid+,r);
 }

 int Query2(int lca,int r1,int r2,int l,int r,int k,int t){
     if(l==r)return l;
     int mid=(l+r)>>,tmp;
     tmp=key()+t*(mid-l+);
     if(tmp>k)return Query2(ch[lca][],ch[r1][],ch[r2][],l,mid,k,t);
     return Query2(ch[lca][],ch[r1][],ch[r2][],mid+,r,k-tmp,t+num());
 }

 int T,n,Q,cas;
 int s,t,k,a,b,ans;
 void Init(){
     printf("Case #%d:\n",++cas);
     memset(sum,,sizeof(sum));
     memset(cal,,sizeof(cal));
     memset(fir,,sizeof(fir));
     memset(Min,,sizeof(Min));
     mm[]=-;cntE=cnt=tot=;
 }

 int main(){
     scanf("%d",&T);
     while(T--){
         Init();
         scanf("%d%d",&n,&Q);
         for(int i=,c;i<n;i++){
             scanf("%d%d%d",&a,&b,&c);
             addedge(a,b,c);
             addedge(b,a,c);
         }

         DFS(,);
         Prepare();

         while(Q--){
             scanf("%d%d",&s,&t);
             scanf("%d%d%d",&k,&a,&b);
             int lca=Lca(s,t),tmp;
             tmp=Query1(rt[lca],rt[s],rt[t],,);
             if(a<=b)
                 ans=tmp+k/a;
             else{
                 ans=tmp;
                 if(k>=a)ans+=+(k-a)/b;
                 ans=max(ans,Query2(rt[lca],rt[s],rt[t],,,k/b,));
             }
             printf("%d\n",ans);
         }
     }
     return ;
 }