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;
}