最小生成树。每次加一条边之后判断一下集合是否符合条件即可。
#include<iostream>
#include<string>
#include<cstdio>
#include<set>
#include<stack>
#include<list>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<fstream>
using namespace std;
typedef long long ll;
//ifstream cin;
const int maxn = 5010,INF = 0x3f3f3f3f;
int n,m;
struct node{
int u,v,w;
bool operator < (const node &a) const{
return w > a.w;
}
};
vector<node> e;
vector<pair<int,int>> r[maxn];
int fa[maxn];
int son[maxn];
int find(int x){
if (x == fa[x]) return x;
else return fa[x] = find(fa[x]);
}
bool check(int x){
int mi = INF,mx = 0;
for(int u = 1;u <= n;++u){
if (find(u) == x){
for(int i = 0;i < r[u].size();++i){
int p = r[u][i].first,w = r[u][i].second;
if (find(p) == x){
mi = min(mi,w);
}
else{
mx = max(mx,w);
}
if (mi <= mx) return false;
}
}
}
return true;
}
void kruskal(){
int ans = 0;
for(int i = 1;i <= n;++i){
fa[i] = i;
son[i] = 1;
}
for(int k = 0;k < m;++k){
int u = e[k].u,v = e[k].v;
u = find(u),v = find(v);
if (u != v){
fa[u] = v;
son[v] += son[u];
if (check(v)){
ans += son[v];
}
}
}
printf("%d\n",ans);
}
void init(){
cin >> n >> m;
int a,b,w;
e.clear();
for(int i = 1;i <= n;++i) r[i].clear();
for(int i = 0;i < m;++i){
cin >> a >> b >> w;
e.push_back({a,b,w});
r[a].push_back({b,w});
r[b].push_back({a,w});
}
sort(e.begin(),e.end());
kruskal();
}
int main(){
//cin.open("in.txt");
int t;
cin >> t;
while(t--){
init();
}
}