Decreasing Debts
Contest: Codeforces Global Round 6
Difficulty rating: 2000
Tags: constructive algorithms, data structures, graphs, greedy, implementation, math, two pointers

There are $$$n$$$ people in this world, conveniently numbered $$$1$$$ through $$$n$$$. They are using burles to buy goods and services. Occasionally, a person might not have enough currency to buy what he wants or needs, so he borrows money from someone else, with the idea that he will repay the loan later with interest. Let $$$d(a,b)$$$ denote the debt of $$$a$$$ towards $$$b$$$, or $$$0$$$ if there is no such debt.

Sometimes, this becomes very complex, as the person lending money can run into financial troubles before his debtor is able to repay his debt, and finds himself in the need of borrowing money.

When this process runs for a long enough time, it might happen that there are so many debts that they can be consolidated. There are two ways this can be done:

1. Let $$$d(a,b) > 0$$$ and $$$d(c,d) > 0$$$ such that $$$a \neq c$$$ or $$$b \neq d$$$. We can decrease the $$$d(a,b)$$$ and $$$d(c,d)$$$ by $$$z$$$ and increase $$$d(c,b)$$$ and $$$d(a,d)$$$ by $$$z$$$, where $$$0 < z \leq \min(d(a,b),d(c,d))$$$.
2. Let $$$d(a,a) > 0$$$. We can set $$$d(a,a)$$$ to $$$0$$$.

The total debt is defined as the sum of all debts:

$$$$$$\Sigma_d = \sum_{a,b} d(a,b)$$$$$$

Your goal is to use the above rules in any order any number of times, to make the total debt as small as possible. Note that you don't have to minimise the number of non-zero debts, only the total debt.

Input format:
The first line contains two space separated integers $$$n$$$ ($$$1 \leq n \leq 10^5$$$) and $$$m$$$ ($$$0 \leq m \leq 3\cdot 10^5$$$), representing the number of people and the number of debts, respectively.

$$$m$$$ lines follow, each of which contains three space separated integers $$$u_i$$$, $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n, u_i \neq v_i$$$), $$$d_i$$$ ($$$1 \leq d_i \leq 10^9$$$), meaning that the person $$$u_i$$$ borrowed $$$d_i$$$ burles from person $$$v_i$$$.

Output format:
On the first line print an integer $$$m'$$$ ($$$0 \leq m' \leq 3\cdot 10^5$$$), representing the number of debts after the consolidation. It can be shown that an answer always exists with this additional constraint.

After that print $$$m'$$$ lines, $$$i$$$-th of which contains three space separated integers $$$u_i, v_i, d_i$$$, meaning that the person $$$u_i$$$ owes the person $$$v_i$$$ exactly $$$d_i$$$ burles. The output must satisfy $$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$ and $$$0 < d_i \leq 10^{18}$$$.

For each pair $$$i \neq j$$$, it should hold that $$$u_i \neq u_j$$$ or $$$v_i \neq v_j$$$. In other words, each pair of people can be included at most once in the output.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
In the first example the optimal sequence of operations can be the following:

1. Perform an operation of the first type with $$$a = 1$$$, $$$b = 2$$$, $$$c = 2$$$, $$$d = 3$$$ and $$$z = 5$$$. The resulting debts are: $$$d(1, 2) = 5$$$, $$$d(2, 2) = 5$$$, $$$d(1, 3) = 5$$$, all other debts are $$$0$$$;
2. Perform an operation of the second type with $$$a = 2$$$. The resulting debts are: $$$d(1, 2) = 5$$$, $$$d(1, 3) = 5$$$, all other debts are $$$0$$$.

In the second example the optimal sequence of operations can be the following:

1. Perform an operation of the first type with $$$a = 1$$$, $$$b = 2$$$, $$$c = 3$$$, $$$d = 1$$$ and $$$z = 10$$$. The resulting debts are: $$$d(3, 2) = 10$$$, $$$d(2, 3) = 15$$$, $$$d(1, 1) = 10$$$, all other debts are $$$0$$$;
2. Perform an operation of the first type with $$$a = 2$$$, $$$b = 3$$$, $$$c = 3$$$, $$$d = 2$$$ and $$$z = 10$$$. The resulting debts are: $$$d(2, 2) = 10$$$, $$$d(3, 3) = 10$$$, $$$d(2, 3) = 5$$$, $$$d(1, 1) = 10$$$, all other debts are $$$0$$$;
3. Perform an operation of the second type with $$$a = 2$$$. The resulting debts are: $$$d(3, 3) = 10$$$, $$$d(2, 3) = 5$$$, $$$d(1, 1) = 10$$$, all other debts are $$$0$$$;
4. Perform an operation of the second type with $$$a = 3$$$. The resulting debts are: $$$d(2, 3) = 5$$$, $$$d(1, 1) = 10$$$, all other debts are $$$0$$$;
5. Perform an operation of the second type with $$$a = 1$$$. The resulting debts are: $$$d(2, 3) = 5$$$, all other debts are $$$0$$$.

Example input:
3 2
1 2 10
2 3 5

Example output:
2
1 2 5
1 3 5

Example input:
3 3
1 2 10
2 3 15
3 1 10

Example output:
1
2 3 5

Editorial:
Consider a solution which minimises the total debt.
Assume for contradiction that there is a triple of vertices $$$u \neq v \neq w$$$, such that $$$d(u,v) > 0$$$ and $$$d(v,w) > 0$$$. We can use the first rule with $$$a = u$$$, $$$b = c = v$$$, $$$d = w$$$ and $$$z = min(d(u,v), d(v,w))$$$ and then the second rule with $$$a = v$$$ and the same $$$z$$$. We have just reduced the total debt by $$$z$$$, which is a contradiction. So, there cannot be such a triple, in particular there cannot be a vertex $$$v$$$ that has both incoming and outgoing debt. Hence, every vertex has only outgoing edges or only incoming ones.
Define $$$bal(u) = \sum_v d(u,v) - \sum_v d(v,u)$$$ to be the balance of $$$u$$$. Any application of either rule preserves balance of all vertices. It follows that any solution in which every vertex has either outgoing or incoming edges is constructible using finite number of applications of rules. This means that we can just find balances, and greedily match vertices with positive balance to vertices with negative balance.
The total debt is then $$$\Sigma_d = \frac{\sum_v |bal(v)|}{2}$$$ and it is clear that we cannot do better than that.