(Analysis by Benjamin Qi)

For each constraint $(a,b,x)$ draw a directed edge from $a$ to $b$ with weight $x$. Note that there cannot be a cycle in this graph or else no solution would exist. Thus, we'll process the sessions in order of the topological sort.

Without loss of generality suppose that the topological sort is $1,2,\ldots,N,$ meaning that all edges satisfy $a<b$. Then for each directed edge in increasing order of $a$, it suffices to set $S_b=\max(S_b,S_a+x)$. After all of these edges are processed, the resulting $S_1,S_2,\ldots,S_N$ are the lowest possible values that satisfy all the edge conditions (assuming all of them are less than or equal to $M$). This can be implemented in $O(N+M)$ time.

#include "bits/stdc++.h"

using namespace std;

void setIO(string s) {
ios_base::sync_with_stdio(0); cin.tie(0);
freopen((s+".in").c_str(),"r",stdin);
freopen((s+".out").c_str(),"w",stdout);
}

#define f first
#define s second

const int MX = 1e5+5;

int N,M,C,S[MX],in[MX];
bool vis[MX];
vector<pair<int,int>> adj[MX];
queue<int> q;

int main() {
setIO("timeline");
cin >> N >> M >> C;
for (int i = 1; i <= N; ++i) cin >> S[i];
for (int i = 0; i < C; ++i) {
int a,b,x; cin >> a >> b >> x;
adj[a].push_back({b,x}); in[b] ++;
}
for (int i = 1; i <= N; ++i) if (!in[i]) q.push(i);
while (q.size()) {
int x = q.front(); q.pop(); // process x in order of topological sort
vis[x] = 1; assert(S[x] <= M);
for (auto& t: adj[x]) {
S[t.f] = max(S[t.f],S[x]+t.s);
if (!(--in[t.f])) q.push(t.f);
}
}
for (int i = 1; i <= N; ++i) {
assert(vis[i]);
cout << S[i] << "\n";
}
}