hdu4729 树链剖分+二分

An Easy Problem for Elfness

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1235    Accepted Submission(s): 257

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.

/*
hdu4729 树链剖分+二分

problem:
给你n个点,然你求两个点之间的最大流.而且你有一定的钱,可以进行两种操作
1.在任意连个点之间建立一个单位1的流,费用A
2.将原先的流扩大1个单位,费用B

solve:
在最初始的图上面u，v之间的流flow即u->v边上的最小值
①如果A<=B,那么我们可以直接在两个目标点u,v之间建边.所以答案为 flow+k/A
②如果A>B
 可以先建一条边然后不停地扩展 (flow-A)/B+1
 or不停地给最小的边扩展,使u->v的最小值变大. 可以二分找到这个值
最开始是二分(1,10000)结果超时QAQ. 后来发现可以先求出(flow-A)/B+1，那么二分的时候就是((flow-A)/B+1,10000)

hhh-2016-08-18 11:39:23
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#define lson  i<<1
#define rson  i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define key_val ch[ch[root][1]][0]
using namespace std;
const int maxn = 200100;
const int inf = 0x3f3f3f3f;
int head[maxn],tot,pos,son[maxn];
int top[maxn],fp[maxn],fa[maxn],dep[maxn],num[maxn],p[maxn];
int val[maxn];
int n;
struct Edge
{
    int to,next,w;
} edge[maxn<<1];

void ini()
{
    tot = 0,pos = 1;
    clr(head,-1),clr(son,-1);
//    clr(val,0);
}

void add_edge(int u,int v,int w)
{
    edge[tot].w = w,edge[tot].to = v,edge[tot].next = head[u],head[u] = tot++;
}

void dfs1(int u,int pre,int d)
{
    dep[u] = d;
    fa[u] = pre,num[u] = 1;
//    cout << "node:" << u<<endl;
    for(int i = head[u]; ~i; i = edge[i].next)
    {
        int v = edge[i].to;
        if(v != pre)
        {
            val[v] = edge[i].w;
            dfs1(v,u,d+1);
            num[u] += num[v];
            if(son[u] == -1 || num[v] > num[son[u]])
                son[u] = v;
        }
    }
}

void getpos(int u,int sp)
{
    top[u] = sp;
    p[u] = pos++;
    fp[p[u]] = u;
    if(son[u] == -1)return ;
    getpos(son[u],sp);
    for(int i = head[u]; ~i ; i = edge[i].next)
    {
        int v = edge[i].to;
        if(v != son[u] && v != fa[u])
            getpos(v,v);
    }
}

struct node
{
    int l,r,mid;
    ll Min;
} tree[maxn << 2];
void push_up(int i)
{
    tree[i].Min = min(tree[lson].Min,tree[rson].Min);
}
void build(int i,int l,int r)
{
    tree[i].l = l,tree[i].r = r;
    tree[i].Min = inf;
    tree[i].mid=(l+r) >>1;
    if(l == r)
    {
        tree[i].Min = val[fp[l]];
//        cout << fp[l] <<" " <<val[fp[l]]<<endl;
        return;
    }
    build(lson,l,tree[i].mid);
    build(rson,tree[i].mid+1,r);
    push_up(i);
}

void update(int i,int k,int val)
{
    if(tree[i].l == k && tree[i].r == k)
    {
        tree[i].Min = val;
        return;
    }
    int mid = tree[i].mid;
    if(k <= mid) update(lson,k,val);
    else update(rson,mid,val);
    push_up(i);
}
ll query(int i,int l,int r)
{
//    cout <<"l:"<< l <<" r:"<<r <<" min:"<< tree[i].Min<<endl;
    if(tree[i].l >= l && tree[i].r <= r)
        return tree[i].Min;
    int mid = tree[i].mid;
    if(r <= mid)
        return query(lson,l,r);
    else if(l > mid)
        return query(rson,l,r);
    else
    {
        return min(query(lson,l,mid),query(rson,mid+1,r));
    }
}
ll find_flow(int u,int v)
{
    int f1 = top[u],f2 = top[v];
    ll tmp = inf;
//    cout <<f1 <<" " <<f2 <<endl;
    while(f1 != f2)
    {
        if(dep[f1] < dep[f2])
        {
            swap(f1,f2),swap(u,v);
        }
        tmp = min(tmp,query(1,p[f1],p[u]));
        u = fa[f1],f1 = top[u];
    }
    if(u == v) return tmp;
    if(dep[u] > dep[v]) swap(u,v);
//    cout << son[u] << " " <<v <<endl;
    return min(tmp,query(1,p[son[u]],p[v]));
}
int allnum = 0;
bool can_do(int i,int l,int r,int mid)
{
    if(tree[i].l >= l && tree[i].r <= r && tree[i].Min >= mid)
    {
        return true;
    }
    if(tree[i].l == tree[i].r)
    {
//        cout << tree[i].Min <<" " <<mid<<endl;
        if(tree[i].Min >= mid)
            return true;
        allnum -= (mid-tree[i].Min);
        if(allnum>= 0) return true;
        return false;
    }
    if(r <= tree[i].mid)
        return can_do(lson,l,r,mid);
    else if(l > tree[i].mid)
        return can_do(rson,l,r,mid);
    else
        return can_do(lson,l,tree[i].mid,mid)&&can_do(rson,tree[i].mid+1,r,mid);
}

bool find_flag(int u,int v,int mid)
{
    int f1 = top[u],f2 = top[v];
    while(f1 != f2)
    {
        if(dep[f1] < dep[f2])
        {
            swap(f1,f2),swap(u,v);
        }
        if(!can_do(1,p[f1],p[u],mid))
            return false;
        u = fa[f1],f1 = top[u];
    }
    if(u == v) return true;
    if(dep[u] > dep[v]) swap(u,v);
    return can_do(1,p[son[u]],p[v],mid);
}

int a[maxn];

int main()
{
//    freopen("in.txt","r",stdin);
    int T,cas = 1;
    int x,y,k,a,b;
    int m,u,v,w;
   scanf("%d",&T);
    while(T--)
    {
        ini();
        scanf("%d%d",&n,&m);
//        cout << n <<" " <<m <<endl;
        for(int i =1; i < n; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add_edge(u,v,w);
            add_edge(v,u,w);
        }
        dfs1(1,0,0);
        getpos(1,1);
        build(1,0,pos-1);
//        char op[10];
        printf("Case #%d:\n",cas++);
        for(int i = 1;i <= m;i++)
        {
            scanf("%d%d%d%d%d",&x,&y,&k,&a,&b);
            ll flow = find_flow(x,y);
//           cout <<"flow:"<<flow <<endl;
            if(k < min(a,b))
            {
                printf("%I64d\n",flow);
            }
            else if(a <= b)
            {
                printf("%I64d\n",flow+(ll)k/a);
            }
            else
            {
                ll ans = flow;
                if(k > a)
                    ans += (k-a)/b+1;
                int l = ans,r = 10000;
                while(l <= r)
                {
                    int mid = (l+r)>>1;
                    allnum = k/b;
                    if(find_flag(x,y,mid))
                    {
                        ans = mid,l = mid + 1;
                    }
                    else
                        r = mid - 1;
                }
                printf("%I64d\n",ans);
            }
        }
    }
    return 0;
}