luogu题解P2502[HAOI2006]旅行--最小生成树变式

题目链接

https://www.luogu.org/problemnew/show/P2502

分析

一个很\(naive\)的做法是从\(s\)到\(t\)双向BFS这当然会TLE

这时我就有个想法就是二分套二分边下标来求得一个比值,同时排序后从小到大枚举每一条边作为最小值,同时再枚举每一条边,如果边权之比小于比值就连起来用并查集维护连通性,可是这个时间复杂度\(O(m^2 log^2m \ \alpha(n))\)过不去QAQ

然后想为什么不直接枚举每条边作为最小值,同时搞一颗以这条边为最小值且联通s,t的最小生成树呢,因为边是排序好的,这样答案是单调的,且正确性是显然的时间复杂度\(O(m^2)\).

代码

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cctype>
#include <cmath>
#define ll long long
#define ri register int
using std::sort;
using std::min;
using std::max;
using std::swap;
template <class T>inline void read(T &x){
	x=0;int ne=0;char c;
	while(!isdigit(c=getchar()))ne=c=='-';
	x=c-48;
	while(isdigit(c=getchar()))x=(x<<3)+(x<<1)+c-48;
	x=ne?-x:x;return ;
}
const int maxm=5005;
const int maxn=505;
const int inf=0x7fffffff;
struct Edge{
	int x,y,dis;
	bool operator <(const Edge &b)const{
		return dis<b.dis;
	}
}edge[maxm];
int num_edge=0;
int n,m,s,t;
int fa[maxn];
int get(int x){return fa[x]==x?fa[x]:(fa[x]=get(fa[x]));}
int gcd(int a,int b){
	return b?gcd(b,a%b):a;
}
int main(){
	int x,y,v,xx,yy;
	bool flag=0;
	read(n),read(m);
	for(ri i=1;i<=m;i++){
		read(x),read(y),read(v);
		edge[i].x=x,edge[i].y=y,edge[i].dis=v;
	}
	read(s),read(t);
	sort(edge+1,edge+1+m);
	int mx,cnt=0;
	double mi=inf;
	int fz,fm;
	for(ri i=1;i<=m;i++){
		mx=-inf,flag=0;
		for(ri j=1;j<=n;j++)fa[j]=j;
		for(ri j=i;j<=m;j++){
			x=edge[j].x,y=edge[j].y,v=edge[j].dis;
			xx=get(x),yy=get(y);
			if(xx==yy)continue;
			fa[xx]=yy;
			mx=max(mx,v);
			if(get(s)==get(t)){
				flag=1;break;
			}//if(cnt==n-1)break;
		}
		if(i==1&&get(s)!=get(t)){
			puts("IMPOSSIBLE");
			return 0;
		}
		else if(flag){
			double tmp=(double)mx/edge[i].dis;
			//printf("%d %d %lf\n",mx,edge[i].dis,tmp);
			if(tmp<mi){
				flag=1;
				mi=tmp;
				fm=edge[i].dis,fz=mx;
			}
		}
	}
	int GCD=gcd(fz,fm);
	fm=fm/GCD,fz=fz/GCD;
	if(fm==1)printf("%d\n",fz);
	else printf("%d/%d\n",fz,fm);
	return 0;
}