$ \color{red} {solution:} $
对于每条树边\(i\),其边权只可能变小,对于非树边\(j\),其边权只可能变大,所以对于任意非树边覆盖的树边有 \(wi - di <= wj + dj\),
变形一下 \(wi - wj <= di +dj\), 而这一部分正是带权二分图匹配,可以使用\(KM\)算法
不会\(KM\)算法的同学这里有篇算法介绍;
#include <bits/stdc++.h>
using namespace std;
const int maxn = 10010, inf = 0x3f3f3f3f;
template <typename T> inline bool check_Max(T &x, const T &y) {
return x < y ? x = y, false : true;
}
template <typename T> inline bool check_Min(T &x, const T &y) {
return x > y ? x = y, false : true;
}
struct edge {int u, v, w; } E[maxn];
int v[maxn], to[maxn], head[maxn], pos[maxn], p;
inline void link(int a, int b, int c) {
v[++ p] = b; to[p] = head[a]; head[a] = p; pos[p] = c;
}
inline void init() {
p = 1; memset(head, 0, sizeof(head));
}
int dep[maxn], fa[maxn], rfe[maxn];
inline void dfs(int u, int f) {
dep[u] = dep[f] + 1; fa[u] = f;
for ( int i = head[u]; i; i = to[i]) if( v[i] ^ f)
rfe[v[i]] = i >> 1, dfs(v[i], u);
}
int a[1010][1010], pre[maxn], slack[maxn], vis[maxn], mat[maxn], tot;
int lv[maxn], rv[maxn];
inline void aug(int s) {
for ( int i = 0; i <= tot; ++ i) slack[i] = inf, vis[i] = pre[i] = 0;
int u = 0; mat[u] = s;
do {
int now = mat[u], d = inf, nxt; vis[u] = 1;
for ( int v = 1; v <= tot; ++ v) if( !vis[v]){
if( !check_Min(slack[v], lv[now] + rv[v] - a[now][v]))
pre[v] = u;
if( !check_Min(d, slack[v])) nxt = v;
}
for ( int i = 0; i <= tot; ++ i)
if( vis[i]) lv[mat[i]] -= d, rv[i] += d;
else slack[i] -= d;
u = nxt;
} while ( mat[u]);
while ( u) mat[u] = mat[pre[u]], u = pre[u];
}
int n, m;
int main() {
#ifndef ONLINE_JUDGE
freopen("1.in","r",stdin);
#endif
init();
scanf("%d%d", &n, &m); tot = max(n-1, m-n+1);
for ( int i = 1; i <= m; ++ i) {
scanf("%d%d%d", &E[i].u, &E[i].v, &E[i].w);
if( E[i].u > E[i].v) swap(E[i].u, E[i].v);
}
for ( int i = 1, u, v; i < n; ++ i) {
scanf("%d%d", &u, &v); if( u > v) swap(u, v);
for ( int k = 1; k <= m; ++ k) if( E[k].u == u && E[k].v == v) {
vis[k] = 1, link(u, v, E[k].w), link(v, u, E[k].w); break;
}
}
dfs(1, 0);
int num = 0;
for ( int i = 1; i <= m; ++ i) if( !vis[i]) {
int u = E[i].u, v = E[i].v, c; ++ num;
while ( u ^ v) {
if( dep[u] < dep[v]) swap(u, v);
c = rfe[u];
a[c][num] = pos[c << 1] - E[i].w; u = fa[u];
}
}
for ( int i = 1; i <= tot; ++ i)
for ( int k = 1; k <= tot; ++ k) check_Max(lv[i], a[i][k]);
for ( int i = 1; i < n; ++ i) aug(i);
int ans = 0;
for ( int i = 1; i <= tot; ++ i) ans += lv[i] + rv[i];
printf("%d\n", ans);
return 0;
}