CF733F Drivers Dissatisfaction【链剖】【最小生成树应用】

F. Drivers Dissatisfaction

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In one kingdom there are n cities and m two-way roads. Each road connects a pair of cities, and for each road we know the level of drivers dissatisfaction — the value wi.

For each road we know the value ci — how many lamziks we should spend to reduce the level of dissatisfaction with this road by one. Thus, to reduce the dissatisfaction with the i-th road by k, we should spend k·ci lamziks. And it is allowed for the dissatisfaction to become zero or even negative.

In accordance with the king's order, we need to choose n - 1 roads and make them the main roads. An important condition must hold: it should be possible to travel from any city to any other by the main roads.

The road ministry has a budget of S lamziks for the reform. The ministry is going to spend this budget for repair of some roads (to reduce the dissatisfaction with them), and then to choose the n - 1 main roads.

Help to spend the budget in such a way and then to choose the main roads so that the total dissatisfaction with the main roads will be as small as possible. The dissatisfaction with some roads can become negative. It is not necessary to spend whole budget S.

It is guaranteed that it is possible to travel from any city to any other using existing roads. Each road in the kingdom is a two-way road.

Input

The first line contains two integers n and m (2 ≤ n ≤ 2·105, n - 1 ≤ m ≤ 2·105) — the number of cities and the number of roads in the kingdom, respectively.

The second line contains m integers w1, w2, ..., wm (1 ≤ wi ≤ 109), where wi is the drivers dissatisfaction with the i-th road.

The third line contains m integers c1, c2, ..., cm (1 ≤ ci ≤ 109), where ci is the cost (in lamziks) of reducing the dissatisfaction with the i-th road by one.

The next m lines contain the description of the roads. The i-th of this lines contain a pair of integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) which mean that the i-th road connects cities ai and bi. All roads are two-way oriented so it is possible to move by the i-th road from aito bi, and vice versa. It is allowed that a pair of cities is connected by more than one road.

The last line contains one integer S (0 ≤ S ≤ 109) — the number of lamziks which we can spend for reforms.

Output

In the first line print K — the minimum possible total dissatisfaction with main roads.

In each of the next n - 1 lines print two integers x, vx, which mean that the road x is among main roads and the road x, after the reform, has the level of dissatisfaction vx.

Consider that roads are numbered from 1 to m in the order as they are given in the input data. The edges can be printed in arbitrary order. If there are several answers, print any of them.

Examples

input

6 9
1 3 1 1 3 1 2 2 2
4 1 4 2 2 5 3 1 6
1 2
1 3
2 3
2 4
2 5
3 5
3 6
4 5
5 6
7

output

0
1 1
3 1
6 1
7 2
8 -5

input

3 3
9 5 1
7 7 2
2 1
3 1
3 2
2

output

5
3 0
2 5

---

Solution

题意：给一个无向图，每条边有一个边权$w$和一个费用$c$，你现在有$s$元，对于每条边可以选择花费$c$将这条边边权减少1（允许负边权），询问这种操作过后最小生成树的最小总权值以及树上的所有边和它们的权值。

思路很简单，明显可以把所有的花费全部砸在一条边上，可以首先建一棵最小生成树，如果把钱砸在树边，那么选择的一定是树边中$c$最小的那条，往死里减就可以了。

如果要砸在非树边上，那么就是在这条非树边两端点在树上的链中找到最长的边删除，用这条边代替即可。

主要是代码实现太复杂了！！！！（虽然我一次a掉嘻嘻嘻嘻嘻

捋清楚每条边的标号是最复杂的？？？链剖+线段树随便搞搞就好。

Code

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#define LL long long
using namespace std;

LL n, m, s;

struct tree {
    LL w, id;
    tree operator + (const tree &a) const {
        tree c;
        if(a.w > w)    c.w = a.w, c.id = a.id;
        else        c.w = w, c.id = id;
        return c;
    }
} TR[];

struct Node {
    LL u, v, w, id, tag, nex;
} Edge[], Edge_inv[];
bool cmp(Node a, Node b) { return a.w < b.w; }

LL h[], stot;
void add(LL u, LL v, LL w, LL id) {
    Edge[++stot] = (Node) {u, v, w, id, , h[u]};
    h[u] = stot;
}

LL stot_inv;
void add_inv(LL u, LL v, LL w, LL id) {
    Edge_inv[++stot_inv] = (Node) {u, v, w, id, , };
}

LL fa[], dep[], siz[], son[], sw[], sid[];
void dfs1(LL u, LL f) {
    fa[u] = f;    dep[u] = dep[f] + ;  siz[u] = ;
    for(LL i = h[u]; i; i = Edge[i].nex) {
        LL v = Edge[i].v;
        if(v == f)    continue;
        dfs1(v, u);
        siz[u] += siz[v];
        if(siz[v] > siz[son[u]])    son[u] = v, sw[u] = Edge[i].w, sid[u] = Edge[i].id;
    }
}

LL top[], seq[], seq1[], in[], idc;
void dfs2(LL u, LL t, LL w, LL id) {
    top[u] = t;    seq[++idc] = w; seq1[idc] = id, in[u] = idc;
    if(son[u])    dfs2(son[u], t, sw[u], sid[u]);
    for(LL i = h[u]; i; i = Edge[i].nex) {
        LL v = Edge[i].v;
        if(v == fa[u] || v == son[u])    continue;
        dfs2(v, v, Edge[i].w, Edge[i].id);
    }
}

void update(LL nd) {
    TR[nd] = TR[nd << ] + TR[nd <<  | ];
}

void build(LL nd, LL l, LL r) {
    if(l == r) {
        TR[nd].w = seq[l];
        TR[nd].id = seq1[l];
        return ;
    }
    LL mid = (l + r) >> ;
    build(nd << , l, mid);    build(nd <<  | , mid + , r);
    update(nd);
}

tree query(LL nd, LL l, LL r, LL L, LL R) {
    if(l >= L && r <= R)    return TR[nd];
    LL mid = (l + r) >> ; tree ans;    ans.w = -0x3f3f3f3f, ans.id = ;
    if(L <= mid)    ans = ans + query(nd << , l, mid, L, R);
    if(R > mid)        ans = ans + query(nd <<  | , mid + , r, L, R);
    return ans;
}

tree query(LL u, LL v) {
    tree ans; ans.w = -0x3f3f3f3f, ans.id = ;
    while(top[u] != top[v]) {
        if(dep[top[u]] < dep[top[v]])    swap(u, v);
        ans = ans + query(, , n, in[top[u]], in[u]);
        u = fa[top[u]];
    }
    if(dep[u] < dep[v])    swap(u, v);
    ans = ans + query(, , n, in[v] + , in[u]);
    return ans;
}

LL f[];
LL find(LL x) {
    if(x != f[x])    f[x] = find(f[x]);
    return f[x];
}

LL w[], c[], tot, ans1, ans2;
void Kruskal() {
    sort(Edge_inv + , Edge_inv +  + m, cmp);
    for(LL i = ; i <= n; i ++)    f[i] = i;
    for(LL i = ; i <= m; i ++) {
        LL u = Edge_inv[i].u, v = Edge_inv[i].v, id = Edge_inv[i].id;
        LL uu = find(u), vv = find(v);
        if(uu != vv) {
            Edge_inv[i].tag = ;
            f[uu] = vv;
            add(u, v, w[id], id);    add(v, u, w[id], id);
            tot += w[id];
            if(c[id] < c[ans1])    ans1 = id;
        }
    }
}

int main() {
    scanf("%lld%lld", &n, &m);
    for(LL i = ; i <= m; i ++)    scanf("%lld", &w[i]);
    for(LL i = ; i <= m; i ++)    scanf("%lld", &c[i]);    c[] = 0x3f3f3f3f;
    for(LL i = ; i <= m; i ++) {
        LL u, v;
        scanf("%lld%lld", &u, &v);
        add_inv(u, v, w[i], i);
    }
    scanf("%lld", &s);
    Kruskal();    ans2 = tot - s / c[ans1];
    dfs1(, );    dfs2(, , -0x3f3f3f3f, );    build(, , n);
    LL flag = ;
    for(LL i = ; i <= m; i ++) {
        if(!Edge_inv[i].tag) {
            LL u = Edge_inv[i].u, v = Edge_inv[i].v, id = Edge_inv[i].id;
            tree a = query(u, v);
            LL tmp = tot - a.w + w[id] - s / c[id];
            if(tmp < ans2)    ans1 = id, ans2 = tmp, flag = a.id;
        }
    }
    printf("%lld\n", ans2);
    for(LL i = ; i <= m; i ++) {
        LL id = Edge_inv[i].id;
        if(ans1 == id) {
            printf("%lld %lld\n", id, w[id] - s / c[id]);
        } else if(Edge_inv[i].tag) {
            if(flag != id) {
                printf("%lld %lld\n", id, w[id]);
            }
        }
    }
    return ;
}