[CF1266D] Decreasing Debts

Existing \ (n-\) individuals with \ (m \) of debt relationship: \ (D (A, B) \) represents \ (A \) under \ (b \ d (a, b) \) Element . Now we are given a final debt relationship, such that \ (\ sum d \) minimum.

Solution

Only needs to record the total amount of each of the input and output points, each into a negative, positive and negative portion toward the portion to do a similar operation can be matched

#include <bits/stdc++.h>
using namespace std;

#define int long long
const int N = 1000005;

struct item {int a,b,c;};

int n,m,u,v,w,d[N];

signed main() {
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1;i<=m;i++) {
        cin>>u>>v>>w;
        d[u]-=w;
        d[v]+=w;
    }
    int pos=1;
    vector <item> vec;
    for(int i=1;i<=n;i++) {
        while(d[i]>0) {
            while(d[pos]>=0&&pos<=n) ++pos;
            int tmp=min(-d[pos],d[i]);
            d[pos]+=tmp;
            d[i]-=tmp;
            vec.push_back({pos,i,tmp});
        }
    }
    cout<<vec.size()<<endl;
    for(item i:vec) cout<<i.a<<" "<<i.b<<" "<<i.c<<endl;
}