傳送門：戳這裏
將邊按從小到大排序，模擬Kruskal，設當前要合併的2個集合爲x和y
設要佔領當前這條邊，需要花費w
①如果要佔領這條邊，則花費爲min(b[v]) * max(w, max(a[u]))，其中u和v是集合x和y中的點
②如果不佔領這條邊，則花費爲f[x] + f[y]
#include<bits/stdc++.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define x first
#define y second
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define per(i,a,b) for(int i=(a)-1;i>=(b);--i)
#define fuck(x) cout<<'['<<#x<<' '<<(x)<<']'<<endl
#define clr(a,b) memset(a,b,sizeof(a))
#define cp(a,b) memcpy(a,b,sizeof(b))
#define eps 1e-10
#define pb push_back
#define speed_read ios::sync_with_stdio(false), cin.tie(0);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> pii;
typedef unsigned int ui;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int MX = 3e5 + 5;
struct Edge {
int u, v, w;
bool operator<(const Edge& _A) const {
return w < _A.w;
}
} E[MX];
ll f[MX];
int p[MX], a[MX], b[MX];
int find(int x) {return p[x] == x ? x : (p[x] = find(p[x]));}
ll solve(int n, int m) {
sort(E + 1, E + m + 1);
rep(i, 1, m + 1) {
int u = E[i].u, v = E[i].v, w = E[i].w;
int x = find(u), y = find(v);
if(x == y) continue;
int ta = max(w, max(a[x], a[y]));
int tb = min(b[x], b[y]);
f[x] = min(f[x] + f[y], (ll) ta * tb);
p[y] = x;
a[x] = max(a[x], a[y]);
b[x] = min(b[x], b[y]);
}
ll ans = 0;
rep(i, 1, n + 1) if(find(i) == i) ans += f[i];
return ans;
}
int main() {
#ifdef local
freopen("in.txt", "r", stdin);
#endif // local
int n, m; cin >> n >> m;
rep(i, 1, n + 1) scanf("%d%d", &a[i], &b[i]);
rep(i, 1, m + 1) scanf("%d%d%d", &E[i].u, &E[i].v, &E[i].w);
rep(i, 1, n + 1) p[i] = i, f[i] = (ll) a[i] * b[i];
printf("%lld\n", solve(n, m));
return 0;
}