luogu 1967 货车运输(最大瓶颈生成树+LCA)

Posted 2020-09-17 freeloop

题意:给出一颗n个点的图，q个询问，每次询问u到v的路径中最小的边最大是多少。

图的最大瓶颈生成树有一个性质，对于该图的任意两个点，在树中他们之间路径的最小边最大。

由于这个图不一定联通，于是我们对它的联通块都求一次最大瓶颈生成树。

每次询问就变成了在最大瓶颈生成树上找出u到v路径的最小边。

这个显然可以用LCA维护点到它的2^x祖先之间的边的最小值来解决。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-7
# define MOD 1024523
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar(‘-‘); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+‘0‘);
}
const int N=10005;
//Code begin...

const int DEG=20;
struct Edge{int p, next, w;}edge[N<<1];
struct Node{int u, v, w;}node[N*10];
int head[N], cnt=1, F[N], from[N];
int fa[N][DEG], mi[N][DEG], deg[N];
queue<int>que;

void add_edge(int u, int v, int w){edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++;}
bool comp(Node a, Node b){return a.w>b.w;}
int find(int x){return F[x]==-1?x:F[x]=find(F[x]);}
void Kruscal(int n){
    mem(F,-1); sort(node+1,node+2*n+1,comp);
    FOR(i,1,2*n) {
        int u=node[i].u, v=node[i].v, t1=find(u), t2=find(v);
        if (t1!=t2) add_edge(u,v,node[i].w), add_edge(v,u,node[i].w), F[t1]=t2;
    }
}
void BFS(int root){
    deg[root]=0; fa[root][0]=root; mi[root][0]=INF; que.push(root);
    while (!que.empty()) {
        int tmp=que.front(); que.pop();
        FO(i,1,DEG) fa[tmp][i]=fa[fa[tmp][i-1]][i-1], mi[tmp][i]=min(mi[tmp][i-1],mi[fa[tmp][i-1]][i-1]);
        for (int i=head[tmp]; i; i=edge[i].next) {
            int v=edge[i].p;
            if (v==fa[tmp][0]) continue;
            deg[v]=deg[tmp]+1; fa[v][0]=tmp; mi[v][0]=edge[i].w; que.push(v);
        }
    }
}
int LCA(int u, int v){
    int ans=INF;
    if (deg[u]>deg[v]) swap(u,v);
    int hu=deg[u], hv=deg[v], tu=u, tv=v;
    for (int det=hv-hu, i=0; det; det>>=1, ++i) if (det&1) ans=min(ans,mi[tv][i]), tv=fa[tv][i];
    if (tu==tv) return ans;
    for (int i=DEG-1; i>=0; --i) {
        if (fa[tu][i]==fa[tv][i]) continue;
        ans=min(ans,mi[tu][i]); ans=min(ans,mi[tv][i]);
        tu=fa[tu][i]; tv=fa[tv][i];
    }
    return min(ans,min(mi[tu][0],mi[tv][0]));
}
int main ()
{
    int n, m, q, u, v;
    scanf("%d%d",&n,&m);
    FOR(i,1,m) scanf("%d%d%d",&node[i].u,&node[i].v,&node[i].w);
    Kruscal(m);
    int pos=0;
    FOR(i,1,n) {
        int v=find(i);
        if (from[v]==0) from[v]=1, BFS(v);
    }
    scanf("%d",&q);
    while (q--) {
        scanf("%d%d",&u,&v);
        if (find(u)!=find(v)) {puts("-1"); continue;}
        printf("%d\n",LCA(u,v));
    }
    return 0;
}

View Code