uva1395

Posted 2021-01-23 033000-

题意见紫书

将边排序后取一段连续区间套用最小生成树模板即可，注意需要计数是否所有节点都被包含在内

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iomanip>
#include<assert.h>
#include<ctime>
#include<vector>
#include<list>
#include<map>
#include<set>
#include<sstream>
#include<stack>
#include<queue>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
#define me(s)  memset(s,0,sizeof(s))
#define _for(i,a,b) for(int i=(a);i<(b);++i)
#define _rep(i,a,b) for(int i=(a);i<=(b);++i)
#define mp make_pair
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define F first
#define S second
#define pf printf
#define sf scanf
#define Di(x) int x;scanf("%d",&x)
#define in(x) inp(x)
#define in2(x,y) inp(x),inp(y)
#define in3(x,y,z) inp(x),inp(y),inp(z)
#define ins(x) scanf("%s",x)
#define ind(x) scanf("%lf",&x)
#define IO ios_base::sync_with_stdio(0);cin.tie(0)
#define READ freopen("C:/Users/ASUS/Desktop/in.txt","r",stdin)
#define WRITE freopen("C:/Users/ASUS/Desktop/out.txt","w",stdout)
template<class T> void inp(T &x) {//读入优化
    char c = getchar(); x = 0;
    for (; (c < 48 || c>57); c = getchar());
    for (; c > 47 && c < 58; c = getchar()) { x = (x << 1) + (x << 3) + c - 48; }
}
typedef pair <int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-15;

const int maxn=10000+5;
int u[maxn],v[maxn],w[maxn],f[maxn],r[maxn];
int n,m;//节点数和边数
//边的端点和权值分别存在u,v,w中
int cmp(const int &i,const int &j) {return w[i]<w[j];} //间接排序函数
int find(int x) {return f[x]==x?x:f[x]=find(f[x]);}//并查集
int Krusal(int x,int y){
    int ans=0;
    for(int i=1;i<=n;i++) f[i]=i;//初始化并查集
    int cnt=1;
    for(int i=x;i<=y;i++){
        int e=r[i];
        int x=find(u[e]),y=find(v[e]);
        if(x!=y) {
            ans+=w[e];
            f[x]=y;
            cnt++;
        }
    }
    if(cnt==n) return w[r[y]]-w[r[x]];
    else return inf;
}
int Solve(){
    for(int i=0;i<m;i++) r[i]=i;
    sort(r,r+m,cmp); 
    int ans=inf;
    _for(i,0,m){
        _for(j,i,m){
            if(w[r[j]]-w[r[i]]>=ans) continue;
            else ans=min(ans,Krusal(i,j));
        }
    }
    return ans==inf?-1:ans;
}
int main(){
    //READ;
    //WRITE;
    while(scanf("%d%d",&n,&m)==2,n){
        _for(i,0,m) {
            in3(u[i],v[i],w[i]);
        }
        pf("%d
",Solve());
    }

}