CF125E MST Company

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 ;
}