BZOJ1050 HAOI2006旅行（最小生成树+LCT）

　　暴力枚举路径中权值最小边是哪个，然后求出边权不小于它的边的最小生成树，即可保证该路径上最大值最小。暴力当然可以过，这里使用LCT维护。注意数据中有自环。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 510
#define M 5010
#define lson tree[k].ch[0]
#define rson tree[k].ch[1]
#define lself tree[tree[k].fa].ch[0]
#define rself tree[tree[k].fa].ch[1]
int n,m,s,t,ansx,ansy,v[N+M];
struct edge
{
    int x,y,z;
    bool operator <(const edge&a) const
    {
        return z<a.z;
    }
}e[M];
struct data{int ch[2],fa,rev,x;
}tree[N+M];
void up(int k)
{
    tree[k].x=k;
    if (v[tree[lson].x]>v[tree[k].x]) tree[k].x=tree[lson].x;
    if (v[tree[rson].x]>v[tree[k].x]) tree[k].x=tree[rson].x;
}
void rev(int k){if (k) swap(lson,rson),tree[k].rev^=1;}
void down(int k){if (tree[k].rev) rev(lson),rev(rson),tree[k].rev=0;}
bool isroot(int k){return lself!=k&&rself!=k;}
int whichson(int k){return rself==k;}
void push(int k){if (!isroot(k)) push(tree[k].fa);down(k);}
void move(int k)
{
    int fa=tree[k].fa,gf=tree[fa].fa,p=whichson(k);
    if (!isroot(fa)) tree[gf].ch[whichson(fa)]=k;tree[k].fa=gf;
    tree[fa].ch[p]=tree[k].ch[!p],tree[tree[k].ch[!p]].fa=fa;
    tree[fa].fa=k,tree[k].ch[!p]=fa;
    up(fa),up(k);
}
void splay(int k)
{
    push(k);
    while (!isroot(k))
    {
        int fa=tree[k].fa;
        if (!isroot(fa))
            if (whichson(k)^whichson(fa)) move(k);
            else move(fa);
        move(k);
    }
}
void access(int k){for (int t=0;k;t=k,k=tree[k].fa) splay(k),tree[k].ch[1]=t,up(k);}
void makeroot(int k){access(k),splay(k),rev(k);}
int findroot(int k){access(k),splay(k);for (;lson;k=lson) down(k);splay(k);return k;}
void link(int x,int y){makeroot(x);tree[x].fa=y;}
void cut(int x,int y){makeroot(x);access(y);splay(y);tree[y].ch[0]=tree[x].fa=0;up(y);}
int query(int x,int y){makeroot(x);access(y);splay(y);return tree[y].x;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj1050.in","r",stdin);
    freopen("bzoj1050.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    n=read(),m=read();
    for (int i=1;i<=m;i++)
    e[i].x=read(),e[i].y=read(),e[i].z=read();
    s=read(),t=read();
    sort(e+1,e+m+1);
    ansx=1000000000,ansy=1;
    for (int i=1;i<=n;i++) v[i]=-1;
    for (int i=1;i<=m;i++) v[i+n]=e[i].z;
    for (int i=m;i;i--)
    if (e[i].x!=e[i].y)
    {
        if (findroot(e[i].x)==findroot(e[i].y))
        {
            int x=query(e[i].x,e[i].y);
            cut(x,e[x-n].x),cut(x,e[x-n].y);
        }
        link(i+n,e[i].x),link(i+n,e[i].y);
        if (findroot(s)==findroot(t))
        {
            int x=v[query(s,t)];
            if (1ll*x*ansy<1ll*ansx*e[i].z) ansx=x,ansy=e[i].z;
        }
    }
    if (ansx==1000000000) cout<<"IMPOSSIBLE";
    else if (ansx%ansy==0) cout<<ansx/ansy;
    else cout<<ansx/gcd(ansx,ansy)<<'/'<<ansy/gcd(ansx,ansy);
    return 0;
}