Topic: Given $n$n Personal debt relationship, $u,v,w$u,v,w Just $v$v Owe $u$u $w$wFor a dollar, we know that the debt relationship can be transferred, so find all the debt relationships with the least amount of debt.

If we want to minimize the debt relationship at the end, we first count everyone's net debt, then we will save the people who make money, and then use the money of those people who are in debt to pay back the money.

AC code:

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///Fast power m^k%mod
ll qpow(int m, int k, int mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}

// Quick power inversion
int Fermat(int a, int p) //Ferma asks for the inverse of a with respect to b
{
    return qpow(a, p - 2, p);
}

///Extended Euclidean
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///Use ecgcd to find the inverse x of a
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///Chinese Remainder Theorem Template
ll china(int a[], int b[], int n) //a[] is the divisor, b[] is the remainder
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //Calculate the result of their multiplication
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //Calculate inverse element
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}
#define pll pair<int, ll>
const int M = 1e5 + 10;
const int N = 2e6 + 10;
struct node
{
    int u, v;
    ll w;
} b[N];
int cnt, pos;
int n, m;
ll a[M];
int u, v;
ll w;
ll sum, res;
queue<pll> q;

int main()
{
    sdd(n, m);
    rep(i, 1, m)
    {
        sdd(u, v);
        sld(w);
        a[u] -= w;
        a[v] += w;
    }
    rep(i, 1, n)
    {
        if (a[i] > 0)
            q.push(make_pair(i, a[i]));//If the person's net debt is to make money
    }
    sum = 0;
    rep(i, 1, n)
    {
        if (a[i] >= 0)
            continue;
        while (a[i])
        {
            pos = q.front().first;
            res = q.front().second;
            q.pop();
            if (res + a[i] > 0)
            {
                res += a[i];
                q.push(make_pair(pos, res));
                b[++cnt].u = i;
                b[cnt].v = pos;
                b[cnt].w = -a[i];
                a[i] = 0;
            }
            else if (res + a[i] == 0)
            {
                b[++cnt].u = i;
                b[cnt].v = pos;
                b[cnt].w = res;
                a[i] = 0;
            }
            else
            {
                a[i] += res;
                b[++cnt].u = i;
                b[cnt].v = pos;
                b[cnt].w = res;
            }
        }
    }
    pd(cnt);
    rep(i, 1, cnt)
        printf("%d %d %lld\n", b[i].u, b[i].v, b[i].w);
    return 0;
}