D. Decreasing Debts
tags: cf brush questions
Topic 1
Title:
gives the debt relationship of n individuals, and asks to simplify the debt relationship. For example, a owes b10 yuan and b owes c10 yuan, so you can directly let a owe c10 yuan.
analysis:
We think about the debt relationship after the final simplification, for everyone, either owes money to others or others owe him money. Because if it exists at the same time, you can offset the borrowed money with the money you owe. So we only need to pay attention to everyone's final money and match.
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
typedef long long ll;
ll debt[100005];
vector<ll> ans1,ans2,ans3;
struct node{
ll id,val;
node(ll a,ll b)
{
id = a;
val = b;
}
};
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
queue<node> q1,q2;
int n,m;
cin >> n >> m;
for (int i = 1; i <= m; i++)
{
ll x,y,v;
cin >> x >> y >> v;
debt[x] += v;
debt[y] -= v;
}
for (int i = 1; i <= n; i++)
{
if( debt[i] > 0 ) q1.push(node(i,debt[i]));
else if( debt[i] < 0 ) q2.push(node(i,-debt[i]));
}
ll t1 = 0,t2 = 0;
int id1,id2;
while( !q1.empty() || t1 != 0 )
{
if( t1 == 0 )
{
t1 = q1.front().val;
id1 = q1.front().id;
q1.pop();
}
if( t2 == 0 )
{
t2 = q2.front().val;
id2 = q2.front().id;
q2.pop();
}
ans1.push_back(id1);
ans2.push_back(id2);
ll z = min(t1,t2);
ans3.push_back(z);
t1 -= z;
t2 -= z;
}
cout << ans1.size() << '\n';
for (int i = 0; i < ans1.size(); i++)
{
cout << ans1[i] << ' ' << ans2[i] << ' ' << ans3[i] << '\n';
}
return 0;
}
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