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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#451375#5148. Tree DistanceWilliamxzhTL 0ms92052kbC++234.1kb2024-06-23 10:18:482024-06-23 10:18:48

Judging History

你现在查看的是最新测评结果

  • [2024-06-23 10:18:48]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:92052kb
  • [2024-06-23 10:18:48]
  • 提交

answer

#include <bits/stdc++.h>
#define il inline
#define pii pair<int,int>
#define fi first
#define se second
using namespace std;
typedef long long ll;
il int read(){
    int x=0,c=getchar(),f=0;
    while(!isdigit(c)) f|=c=='-',c=getchar();
    while(isdigit(c)) x=x*10+c-48,c=getchar();
    return f?-x:x;
}
const int N=2e5+5,M=1e6+5,K=5e6+5;const ll inf=1e18;
int n,q,fa[N],de[N],size[N],dfn[N],top[N],son[N];ll s[N];
int vis[N],siz[N],mx[N],rt,tot;ll d[N],ans[M];
vector<pii> e[N],g[N];
il void add(int x,int y,int z){e[x].push_back({y,z});}
struct node{
    int u,v;ll w;
    il node(){u=v=0,w=0ll;}
    il node(int u,int v,ll w) : u(u),v(v),w(w) {}
    bool operator<(const node &x) const {return v<x.v;}
}f[K];int k;
void DFS(int x,int fath){
    int y,z;fa[x]=fath,de[x]=de[fath]+1,siz[x]=1;
    for(auto it:e[x]){
        y=it.fi,z=it.se;if(y==fath) continue;
        s[y]=s[x]+1ll*z,DFS(y,x),siz[x]+=siz[y];
        if(siz[y]>siz[son[x]]) son[x]=y;
    }
}
void DFS1(int x,int t){
    top[x]=t;
    if(son[x]) DFS1(son[x],t);
    for(auto it:e[x]) if(it.fi!=fa[x] && it.fi!=son[x]) DFS1(it.fi,it.fi);
}
il int lca(int x,int y){
    while(top[x]!=top[y]){
        if(de[top[x]]<de[top[y]]) swap(x,y);
        x=fa[top[x]];
    }return de[x]<de[y]?x:y;
}
il ll dis(int x,int y){return s[x]+s[y]-2ll*s[lca(x,y)];}
void dfs0(int x,int fath){
    ++tot;
    for(auto it:e[x])if(it.fi!=fath && !vis[it.fi]) dfs0(it.fi,x);
}
void dfs(int x,int fath){
    int y;siz[x]=1,mx[x]=0;
    for(auto it:e[x]){
        y=it.fi;if(y==fath) continue;
        dfs(y,x),siz[x]+=siz[y],mx[x]=max(mx[x],siz[y]);
    }mx[x]=max(mx[x],tot-siz[x]);
    if(!rt || mx[x]<mx[rt]) rt=x;
}
int m,a[N],st[N],TOP,bel[N],cur;
void dfs1(int x,int fath){
    int y,z;a[++m]=x,bel[x]=cur;
    for(auto it:e[x]){
        y=it.fi,z=it.se;if(y==fath || vis[y]) continue;
        d[y]=d[x]+1ll*z,dfs1(y,x);
    }
}
il void calc(int x){
    int y,z,u,v,w,p;m=0,d[x]=0ll;
    for(auto it:e[x]){
        y=it.fi,z=it.se;if(vis[y]) continue;
        d[y]=1ll*z,cur=y,dfs1(y,x);
    }
    sort(a+1,a+1+m);
    TOP=0;
    for(int i=1;i<=m;++i){
        while(TOP && d[st[TOP]]<d[a[i]]) --TOP;
        if(TOP && bel[st[TOP]]!=bel[a[i]]) f[++k]={st[TOP],a[i],d[st[TOP]]+d[a[i]]};
        st[++TOP]=a[i];
    }TOP=0;
    for(int i=1;i<=m;++i){
        while(TOP && d[st[TOP]]>d[a[i]]) --TOP;
        if(TOP && d[st[TOP]]!=d[a[i]] && bel[st[TOP]]!=bel[a[i]]) f[++k]={st[TOP],a[i],d[st[TOP]]+d[a[i]]};
        st[++TOP]=a[i];
    }TOP=0;
    for(int i=1;i<=m;++i){
        u=x,v=a[i];if(u>v) swap(u,v);
        f[++k]={u,v,d[a[i]]};
    }
}
void solve(int x){
    int y;vis[x]=1,calc(x);
    for(auto it:e[x]){
        y=it.fi;if(vis[y]) continue;
        rt=tot=0,dfs0(y,x),dfs(y,x),solve(rt);
    }
}
ll t[N<<2];
il void pushup(ll x){t[x]=min(t[x<<1],t[x<<1|1]);}
void build(int x,int l,int r){
    t[x]=inf;if(l==r) return ;
    int mid=(l+r)>>1;
    build(x<<1,l,mid),build(x<<1|1,mid+1,r);
    pushup(x);
}
void update(int x,int l,int r,int p,ll v){
    if(l==r){t[x]=min(t[x],v);return ;}
    int mid=(l+r)>>1;
    if(p<=mid) update(x<<1,l,mid,p,v);
    else update(x<<1|1,mid+1,r,p,v);
    pushup(x);
}
ll query(int x,int l,int r,int L,int R){
    if(l>=L && r<=R) return t[x];
    int mid=(l+r)>>1;
    if(L>mid) return query(x<<1|1,mid+1,r,L,R);
    else if(R<=mid) return query(x<<1,l,mid,L,R);
    else return min(query(x<<1,l,mid,L,R),query(x<<1|1,mid+1,r,L,R));
}
int x,y,z,p;
int main(){
    scanf("%d",&n);
    for(int i=1;i<n;++i) x=read(),y=read(),z=read(),add(x,y,z),add(y,x,z);
    scanf("%d",&q);
    for(int i=1;i<=q;++i) x=read(),y=read(),g[y].push_back({x,i});
    DFS(1,0),DFS1(1,1),build(1,1,n);
    rt=0,tot=n,dfs(1,0),solve(rt);
    sort(f+1,f+1+k);
    for(int i=1;i<=n;++i){
        //for(auto it:f[i]) update(1,1,n,it.fi,it.se);
        while(p<k && f[p+1].v<=i) ++p,update(1,1,n,f[p+1].u,f[p+1].w);
        for(auto it:g[i]) {
            if(it.fi==i) ans[it.se]=-1;
            else ans[it.se]=query(1,1,n,it.fi,i-1);
        }
    }
    for(int i=1;i<=q;++i) printf("%lld\n",ans[i]);
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 92052kb

input:

5
1 2 5
1 3 3
1 4 4
3 5 2
5
1 1
1 4
2 4
3 4
2 5

output:

-1
3
7
7
2

result:

ok 5 number(s): "-1 3 7 7 2"

Test #2:

score: -100
Time Limit Exceeded

input:

199999
31581 23211 322548833
176307 196803 690953895
34430 82902 340232856
36716 77480 466375266
7512 88480 197594480
95680 61864 679567992
19572 14126 599247796
188006 110716 817477802
160165 184035 722372640
23173 188594 490365246
54801 56250 304741654
10103 45884 643490340
127469 154479 214399361...

output:


result: