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换句话说就是要求一号点的出度为k的最短路

容易发现:当连在1号节点的几个边同时加一个数的时候,1号店的度会减少,反之,当同时增大一个k的时候,1号点的度会增大

因此,我们考虑二分一个数作为连在1号节点的那几条边修改的数值,进行二分查找出答案即可

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#define maxn 5500
#define maxm 1100000
#define rep(x,y,z) for(int x = y ; x <= z ; x ++)
using namespace std ;
int n , m , kk ;
struct dy{
    int u , v , w , id ;
    int operator < (const dy &x) const {
        return w == x.w ? u < x.u : w < x.w ;
    }
}a[maxm] ;
vector<int>ans ;
int fa[maxn] ;
int find(int x) {
    return x == fa[x] ? x : fa[x] = find(fa[x]) ;
}
void unionn(int x ,int y){
    fa[find(x)] = find(y) ;
}
int read() ;
int ksm(int k) {
    ans.clear() ;
    rep(i,1,n) fa[i] = i ;
    int cnt = 0 , in = 0 ;
    rep(i,1,m) {
        if(a[i].u == 1 || a[i].v == 1) {
            a[i].w += k ;
        }
    }sort(a+1,a+1+m) ;
    rep(i,1,m) {
        int u = a[i].u , v = a[i].v , w = a[i].w ;
        int fu = find(u) , fv = find(v) ;
        if(fu == fv) continue ; 
        fa[fv] = fu ;
        if(u == 1 || v == 1) in ++  ;
        ans.push_back(a[i].id) ;
        cnt ++ ;
        if(cnt == n-1) break ;
    }
    rep(i,1,m) {
        if(a[i].u == 1 || a[i].v == 1) a[i].w -= k ;
    }return in ;
    
}
int deal(int k) {
    ans.clear() ;
    rep(i,1,n) fa[i] = i ;
    int cnt = 0 , in = 0 ;
    rep(i,1,m) {
        if(a[i].u == 1 || a[i].v == 1) {
            a[i].w += k ;
        }
    }sort(a+1,a+1+m) ;
    rep(i,1,m) {
        int u = a[i].u , v = a[i].v , w = a[i].w ;
        int fu = find(u) , fv = find(v) ;
        if(fu == fv) continue ;
        if(a[i].u == 1 && in == kk) continue ;
        fa[fv] = fu ;
        if(u == 1 || v == 1) in ++ ;
        cnt ++ ;
        ans.push_back(a[i].id) ;
    }
    rep(i,1,m) {
        if(a[i].u == 1 || a[i].v == 1) a[i].w -= k ;
    }
    if(cnt != n-1 || in != kk) {
        puts("-1") ;
        exit(0) ;
    } 
    cout << n-1 << endl ;
    for(int i = 0 ; i < ans.size() ; i ++) cout << ans[i] << " " ;
    puts("") ;exit(0) ;
    return in ; 
}
int main () {
    n = read() , m = read() , kk = read() ;
    rep(i,1,m) {
        a[i].u = read() , a[i].v = read() , a[i].w = read() , a[i].id = i ;
        if(a[i].u > a[i].v) swap(a[i].u,a[i].v) ;
    }int l = -1e5 , r = 1e5 , Ans = -1e9 ;
    while(l <= r) {
        int mid = (l+r) / 2 ;
        if(ksm(mid) < kk) r = mid - 1 ;
        else {
            l = mid + 1 , Ans = mid ; 
        }
    }
    deal(Ans) ;
    return 0 ;
}
int read() {
    int x = 0 , f = 1 ; char s = getchar() ;
    while(s > '9' || s < '0') {if(s == '-') f = -1 ; s = getchar() ;}
    while(s <='9' && s >='0') {x = x * 10 + (s-'0'); s = getchar() ;}
    return x*f ;
}