国王的烦恼

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题意

n 个点 m 条边，每条边有一个权值表示该边消失的时间，求有新的点不联通的时刻的数量。

题解

以时间从大到小排序，用并查集反向建图，注意同时刻只能计一次。

代码

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#include <bits/stdc++.h>
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define inlld(lld) scanf("%lld", &lld)
#define inlf(f) scanf("%lf", &f)
#define ind(d) scanf("%d", &d)
#define ins(s) scanf("%s", s)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef std::pair<long long, long long> pll;
typedef std::vector<long long> vll;
typedef std::pair<int, int> pii;
typedef unsigned long long ull;
typedef std::vector<int> vii;
typedef long double db;
typedef long long ll;
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const ll mod = 1e9+7;
const int N = 1.1e4;
const db eps = 1e-8;
using namespace std;
int sign(db a) { return a < -eps ? -1 : a > eps; }
int db_cmp(db a, db b){ return sign(a-b); }

int uni[N];
int find_r(int x){
    if(uni[x]==x)return x;
    else return uni[x]=find_r(uni[x]);
}
bool merge(int a,int b){
    int fa=find_r(a),fb=find_r(b);
    if(fa==fb)return 0;
    uni[fa]=fb;
    return 1;
}
struct edg{
    int u,v,w;
    bool operator < (const edg y)const{
        return w>y.w;
    }
}edge[10*N];
int main() {
#ifdef PerpEternal
    // freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n,m;
    scanf("%d%d",&n,&m);
    for1(i,n)uni[i]=i;
    for0(i,m)scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
    sort(edge,edge+m);
    int ans=0,time=0;
    for0(i,m){
        // cout<<edge[i].u<<' '<<edge[i].v<<' '<<edge[i].w<<endl; 
        if(merge(edge[i].u,edge[i].v)){
            if(edge[i].w!=time){
                time=edge[i].w;
                ans++;
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}