题意: 给出
个人的债务关系,
就是
欠
块钱,我们知道债务关系是可以转移的,所以求出所有的欠债钱数最少的债务关系。
要想使得最后得债务关系最小,我们先统计每个人的净债务,那么我们就把赚钱的人给存起来,然后用负债的这些人的钱还给赚钱的就好了。
AC代码:
#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);
}
///快速幂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;
}
// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
return qpow(a, p - 2, p);
}
///扩展欧几里得
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;
}
///使用ecgcd求a的逆元x
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;
}
///中国剩余定理模板
ll china(int a[], int b[], int n) //a[]为除数,b[]为余数
{
int M = 1, y, x = 0;
for (int i = 0; i < n; ++i) //算出它们累乘的结果
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); //计算逆元
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]));//如果这个人的净债务是赚钱
}
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;
}