Codeforces 362D Fools and Foolproof Roads 构造题

题目链接：点击打开链接

题意：

给定n个点 m条边的无向图 需要在图里增加p条边 使得图最后连通分量数为q

问是否可行，不可行输出NO

可行输出YES，并输出添加的p条边。

set走起。。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;
#define N 123456

#define ll __int64
ll n,m,p,q;
struct Edge{
	ll from, to, dis;
}edge[N*2];
ll edgenum;
void add(ll u,ll v,ll dis){
	Edge E={u,v,dis};
	edge[edgenum++] = E;
}
ll f[N];
ll find(ll x){return x==f[x]?x:f[x]=find(f[x]);}
void Union(ll x,ll y){
	ll fx = find(x), fy = find(y);
	if(fx==fy)return ;
	if(fx<fy)swap(fx,fy);
	f[fx]=fy;
}
set<ll>myset;
set<ll>::iterator pp;
ll siz[N];
vector<int>L,R;
struct node{
	ll fa, val;
	bool operator<(const node&x)const{
		if(x.val==val)return x.fa<fa;
		return x.val>val;
	}
	node(ll x=0,ll y = 0):fa(x),val(y){}
};
set<node>hehe;
set<node>::iterator dd;
void init(){
	hehe.clear();
	L.clear(); R.clear();
	memset(siz, 0, sizeof siz);
	myset.clear();
	for(ll i = 1; i <= n; i++)f[i]=i;
	edgenum = 0;
}

void go(){
	dd = hehe.begin();
	node x = *dd;
	hehe.erase(dd);
	dd = hehe.begin();
	node y = *dd;
	hehe.erase(dd);
	Union(x.fa,y.fa);
	ll now = min((ll)1000000000, x.val+y.val+1);
	node z = node(find(x.fa),x.val+y.val+now);
	hehe.insert(z);
	L.push_back(x.fa); R.push_back(y.fa);
	add(x.fa,y.fa,1);
}
int main(){
	ll i, j, u, v, d;
	while(~scanf("%I64d %I64d %I64d %I64d",&n,&m,&p,&q)){
		init();
		while(m--){
			scanf("%I64d %I64d %I64d",&u,&v,&d);
			add(u,v,d);
			Union(u,v);
		}
		for(i = 1; i <= n; i++)find(i);
		for(i = 1; i <= n; i++)myset.insert(f[i]);
		for(i = 0; i < edgenum; i++){
			siz[f[edge[i].from]]+=edge[i].dis;
		}
		if(myset.size()<q){puts("NO");continue;}
		if(myset.size()==q)
		{
			if(p && edgenum==0)puts("NO");
			else 
				{
					puts("YES");
					while(p--){
						cout<<edge[0].from<<" "<<edge[0].to<<endl;
					}
			}
			continue;
		}
		ll ned = myset.size()-q;
		if(ned>p){puts("NO");continue;}
		p-=ned;

		for(pp=myset.begin(); pp!=myset.end(); pp++)hehe.insert(node(*pp,siz[*pp]));
		while(ned--)go();

		puts("YES");
		for(i = 0; i < L.size(); i++)cout<<L[i]<<" "<<R[i]<<endl;
		while(p--)
			cout<<edge[0].from<<" "<<edge[0].to<<endl;
	}
	return 0;
}

Codeforces 362D Fools and Foolproof Roads 构造题,布布扣,bubuko.com