有一個 n 個點， m 條邊的圖，現在多加入一條邊 <u, v>，求最少要刪去多少條邊，才能使邊 <u, v> 既可能出現在最小生成樹上，也可能出現在最大生成樹上。

設後加入的邊的長度為 L，那麼只有原圖中小於 L 的邊不能再用的時候，<u, v> 才可能出現在最小生成樹上，用最小割可求。最大生成樹時同理。兩次答案相加。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX_N = 20005, MAX_M = 200005;
const int inf = 1000000007;

struct node {
	int v, w, next;
}E[MAX_M << 1];
struct arr {
	int x, y, k;
}R[MAX_M];
int head[MAX_N], cur[MAX_N], top = 0;
int n, m, s, t, l, ans = 0;
int dep[MAX_N], q[MAX_M];

inline void add(int u, int v, int x) {
	int t1 = x << 1, t2 = t1 | 1;
	E[t1].v = v; E[t1].w = 1; E[t1].next = head[u]; head[u] = t1;
	E[t2].v = u; E[t2].w = 1; E[t2].next = head[v]; head[v] = t2;
}
inline bool cmp(arr a, arr b) {
	return a.k < b.k;
}
inline bool bfs() {
	memset(dep, -1, sizeof(dep));
	int bg = 0, ed = 1;
	q[ed] = s; dep[s] = 0;
	while (bg < ed) {
		int x = q[++ bg];
		for (int i = head[x]; i != -1; i = E[i].next) 
			if (dep[E[i].v] == -1 && E[i].w) {
				dep[E[i].v] = dep[x] + 1; q[++ ed] = E[i].v;
				if (E[i].v == t) return 1;
			}
	}
	return 0;
}
int dfs(int x, int mx) {
	if (x == t) return mx;
	int ret;
	for (int &i = cur[x]; i != -1; i = E[i].next) 
		if (E[i].w && dep[E[i].v] == dep[x] + 1 && (ret = dfs(E[i].v, min(mx, E[i].w)))) {
			E[i].w -= ret; E[i ^ 1].w += ret;
			return ret;
		}
	return 0;
}
void dinic() {
	int ret; 
	while (bfs()) {
		for (int i = 0; i <= n; i ++) cur[i] = head[i];
		while (ret = dfs(s, inf)) ans += ret;
	}
}
void init() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i ++) 
		scanf("%d%d%d", &R[i].x, &R[i].y, &R[i].k);
	scanf("%d%d%d", &s, &t, &l);
}
void doit() {
	sort(R + 1, R + 1 + m, cmp);
	memset(head, -1, sizeof(head)); top = 0;
	for (int i = 1; i <= m; i ++) {
		if (R[i].k < l) add(R[i].x, R[i].y, top ++);
		else break;
	}
	dinic();
	memset(head, -1, sizeof(head)); top = 0;
	for (int i = m; i >= 1; i --) {
		if (R[i].k > l) add(R[i].x, R[i].y, top ++);
		else break;
	}
	dinic();
	printf("%d\n", ans);
}
int main() {
	init();
	doit();
	return 0;
}