AT2172 Shik and Travel

题目描述：

luogu

题解：

二分+暴力$vector$+$dfs$。

记录下所有可能的子树内合法方案，双指针+归并合并。

代码：

#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 200050;
template<typename T>
inline void read(T&x)
{
    T f = 1,c = 0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){c=c*10+ch-'0';ch=getchar();}
    x = f*c;
}
int n,fa[N],ch[N][2];
ll v[N],dep[N];
void DFS(int u)
{
    if(!ch[u][0])return ;
    dep[ch[u][0]]=dep[u]+v[ch[u][0]],DFS(ch[u][0]);
    dep[ch[u][1]]=dep[u]+v[ch[u][1]],DFS(ch[u][1]);
}
struct Pair
{
    ll x,y;
    Pair(){}
    Pair(ll x,ll y):x(x),y(y){}
}g1[N],g2[N],g[N];
vector<Pair>ve[N];
ll mid;
int Merge(int len)
{
    int i = 1,j = len,k = 0;
    while(i<=len&&j>=1)
    {
        if(g1[i].x<=g2[j].x)g[++k]=g1[i++];
        else g[++k]=g2[j--];
    }
    while(i<=len)g[++k]=g1[i++];
    while(j>=1)g[++k]=g2[j--];
    return k;
}
void dfs(int u)
{
    if(!ch[u][0]){ve[u].push_back(Pair(dep[u],dep[u]));return ;}
    int ls = ch[u][0],rs = ch[u][1];
    dfs(ls),dfs(rs);int k=0;
    if(ve[ls].size()>ve[rs].size())swap(ls,rs);
    for(int i=0,j=-1,l1=ve[ls].size(),l2=ve[rs].size();i<l1;i++)
    {
        while(j<l2-1&&ve[ls][i].y+ve[rs][j+1].x-2ll*dep[u]<=mid)j++;
        if(~j)g1[++k]=Pair(ve[ls][i].x,ve[rs][j].y),g2[k]=Pair(ve[rs][j].y,ve[ls][i].x);
    }
    int len = Merge(k);
    for(int i=1;i<=len;i++)
        if(i==1||g[i].y<g[i-1].y)
            ve[u].push_back(g[i]);
}
bool check()
{
    for(int i=1;i<=n;i++)
        ve[i].clear();
    dfs(1);
    return (int)ve[1].size()>0;
}
int main()
{
    read(n);
    for(int i=2;i<=n;i++)
    {
        read(fa[i]),read(v[i]);
        if(!ch[fa[i]][0])ch[fa[i]][0]=i;
        else ch[fa[i]][1]=i;
    }
    DFS(1);
    ll l = 0,r = 200000ll*n,ans = r;
    while(l<=r)
    {
        mid = (l+r)>>1;
        if(check())ans=mid,r=mid-1;
        else l=mid+1;
    }
    printf("%lld
",ans);
    return 0;
}