链剖-进阶ing-填坑-NOIP2013-货车运输

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#include <ctime>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define sizen 10
#define sizem 10
#define sizew 32767
#define sizeq 20
int main(){
    freopen("car.in","w",stdout);
    srand(time(NULL));
    int n=rand()%sizen+5,m=rand()%sizem+3;
    cout<<n<<" "<<m<<endl;
    while(m--){
        int u=rand()%n+1,v=rand()%n+1;
        if(u==v)u=(u-2+n)%n+1;
        cout<<u<<" "<<v<<" "<<rand()%sizew+10<<endl;
    }
    int q=rand()%sizeq+10;
    cout<<q<<endl;
    while(q--){
        int u=rand()%n+1,v=rand()%n+1;
        if(u==v)u=(u-1+n)%n;
        cout<<u<<" "<<v<<endl;
    }
  return 0;
}

 

 

 

似乎官方给的是倍增lca

不管了，最近练习链剖，以后有时间在补倍增的写法

就是边权下放成点权，然后树链剖分套一颗线段树就可以了

开始sb似的建成一颗大树，其实直接利用Kruskal的并查集查询是否在同一子树就好了[森林x1森林x2森林x3重要的事说三遍]

ac code 莫名压行，不压不爽

 

 1 //边权的下放，可能是这题唯一的细节吧
 2 #include <cstdio>
 3 #include <algorithm>
 4 #define E(u,v,w) e[++cnt]=(edge){v,w,head[u]},head[u]=cnt
 5 using namespace std;
 6 int GI(){
 7   int x=0,c=getchar(),f=0;
 8   while(c<'0'||c>'9'){if(c=='-')f=1;c=getchar();}
 9   while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
10   return f?-x:x;
11 }
12 const int maxn=20000,maxm=80000;
13 struct edge{int v,w,next;}e[maxm*2];
14 struct data{int u,v,w;}a[maxm];
15 int n,m,q,head[maxn],cnt,f[maxn],vis[maxn];
16 int find(int x){return f[x]==x?x:f[x]=find(f[x]);}  
17 bool operator<(data x,data y){return x.w>y.w;}
18 void add(int u,int v,int w){E(u,v,w),E(v,u,w);}
19 void kruskal(){
20     sort(a+1,a+m+1);
21     for(int i=1;i<=n;i++)f[i]=i;
22     for(int i=1,tot=0;i<=m;i++){
23         int u=a[i].u,v=a[i].v;
24         int x=find(u),y=find(v);
25         if(x!=y)f[x]=y,add(a[i].u,a[i].v,a[i].w),++tot;
26         if(tot==n-1)return;
27     }
28 }
29 int dfn[maxn],dep[maxn],fa[maxn],top[maxn],son[maxn],rak[maxn],siz[maxn],idx,W[maxn];
30 void dfs1(int u,int _fa){
31     siz[u]=1,fa[u]=_fa,dep[u]=dep[_fa]+1;
32     for(int i=head[u];i;i=e[i].next)
33         if(e[i].v!=_fa){
34             dfs1(e[i].v,u),siz[u]+=siz[e[i].v],W[e[i].v]=e[i].w;//细节1
35             if(!son[u]||siz[e[i].v]>siz[son[u]])son[u]=e[i].v;
36         }
37 }
38 void dfs2(int u,int _top){
39     dfn[u]=++idx,rak[dfn[u]]=u,top[u]=_top;
40     if(son[u])dfs2(son[u],_top);
41     for(int i=head[u];i;i=e[i].next)
42         if(e[i].v!=fa[u]&&e[i].v!=son[u])dfs2(e[i].v,e[i].v);
43 }
44 #define ls (x<<1)
45 #define rs (x<<1|1)
46 #define mid ((l+r)>>1)
47 int Min[maxn<<2];
48 void up(int x){Min[x]=min(Min[ls],Min[rs]);}
49 void build(int x,int l,int r){
50     if(l==r)Min[x]=W[rak[l]];
51     else build(ls,l,mid),build(rs,mid+1,r),up(x);
52 }
53 int MIN(int x,int l,int r,int L,int R){
54     if(L>R)return 1<<30;
55     if(L<=l&&r<=R)return Min[x];
56     if(R<=mid)return MIN(ls,l,mid,L,R);
57     if(L>mid)return MIN(rs,mid+1,r,L,R); 
58     return min(MIN(ls,l,mid,L,R),MIN(rs,mid+1,r,L,R));
59 }
60 int jump(int x,int y){
61     if(find(x)!=find(y))return -1;
62     int ans=1<<30;
63     while(top[x]!=top[y]){
64         if(dep[top[x]]<dep[top[y]])swap(x,y);
65         ans=min(ans,MIN(1,1,n,dfn[top[x]],dfn[x]));
66         x=fa[top[x]];
67     }
68     if(dep[x]>dep[y])swap(x,y);
69     ans=min(ans,MIN(1,1,n,dfn[x]+1,dfn[y]));//细节2 
70     return ans;
71 }
72 int main(){
73     freopen("car.in","r",stdin);
74     freopen("car.out","w",stdout);
75     int i;n=GI(),m=GI();
76     for(i=1;i<=m;i++)a[i].u=GI(),a[i].v=GI(),a[i].w=GI();
77     for(kruskal(),i=1;i<=n;i++)if(!dfn[i])dfs1(i,0),dfs2(i,i);
78     for(build(1,1,n),q=GI(),i=1;i<=q;i++)printf("%d\n",jump(GI(),GI()));
79     return 0;
80 }

 

posted @ 2017-04-13 13:28  墨鳌  阅读(282)  评论(0)  编辑  收藏  举报