【NOIP2012】疫情控制

题面

https://www.luogu.org/problem/P1084

题解

也算二分的经典题吧。

关于这道题有一个尴尬的故事，我口胡出了二分答案，然后以为自己做出来了，$gyf$说还要贪心，然后我把他$D$了，其实是我错了。。。。。

先二分答案，尽可能的往上走，然后让最上的支援最下的，若此时支援点只有一支队伍，必须保证其他的人到这里更近才能更新。

#include<cstdio>
#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<queue>
#define ri register int
#define ll long long
#define N 50050
using namespace std;
int fa[N],n,m,f[N][17];
int que[N],nque[N],tong[N],top[N];
long long dis[N];
bool imp[N],reach[N];
vector<int> to[N],len[N];

struct node1 {
  int x;
  ll walked;
  bool operator < (const node1 &rhs) const {
    return walked>rhs.walked;
  }
};

struct node2 {
  int x;
  bool operator < (const node2 &rhs) const {
    return dis[x]<dis[rhs.x];
  }
};

void maketree(int x,int ff,int l) {
  dis[x]=dis[ff]+l;
  f[x][0]=fa[x]=ff;
  for (ri i=1;i<=16;i++) f[x][i]=f[f[x][i-1]][i-1];
  for (ri i=to[x].size()-1;i>=0;i--) if (to[x][i]!=ff) maketree(to[x][i],x,len[x][i]);
}

void dfs(int x,int tp) {
  imp[x]=1;  top[x]=tp;
  if (to[x].size()>2) return;
  for (ri i=to[x].size()-1;i>=0;i--) if (to[x][i]!=fa[x]) dfs(to[x][i],tp);
}

bool can(long long cc) {
  priority_queue<node1> yinyu; while (!yinyu.empty()) yinyu.pop();
  priority_queue<node2> ineed; while (!ineed.empty()) ineed.pop();
  memset(reach,0,sizeof(reach));
  memset(tong,0,sizeof(tong));
  for (ri i=1;i<=m;i++) {
    int x=que[i],nx=x;
    for (ri j=16;j>=0;j--) if (f[nx][j]!=1 && dis[x]-dis[f[nx][j]]<=cc) nx=f[nx][j];
    nque[i]=nx;
    yinyu.push((node1){nx,dis[x]});
  }
  for (ri i=1;i<=m;i++) if (imp[nque[i]]) {
    int nx=nque[i];
    reach[top[nx]]=1;
    tong[top[nx]]++;
  }

  for (ri i=to[1].size()-1;i>=0;i--) if (!reach[to[1][i]]) {
    ineed.push((node2){to[1][i]});
  }
  
  if (ineed.empty()) return 1;

  while (!yinyu.empty()) {
    node2 need=ineed.top();
    node1 cur=yinyu.top();
    if (cur.walked*1LL+dis[need.x]>cc) return 0;
    if (tong[top[cur.x]]==1 && dis[top[cur.x]]>dis[need.x]) {
      yinyu.pop();
      continue;
    }
    ineed.pop(); yinyu.pop();
    if (--tong[top[cur.x]]==0) ineed.push((node2){top[cur.x]});
    if (ineed.empty()) return 1;
  }

  return 0;
}

int main(){
  scanf("%d",&n);
  for (ri i=1;i<n;i++) {
    int u,v,w;
    scanf("%d %d %d",&u,&v,&w);
    to[u].push_back(v); len[u].push_back(w);
    to[v].push_back(u); len[v].push_back(w);
  }
  maketree(1,1,0);
  scanf("%d",&m);
  for (ri i=1;i<=m;i++) scanf("%d",&que[i]);
  long long l=0,r=N*1e9;
  long long ans=-1;
  for (ri i=to[1].size()-1;i>=0;i--) dfs(to[1][i],to[1][i]);
  while (l<=r) {
    long long mid=(l+r)/2;
    if (can(mid)) ans=mid,r=mid-1;else l=mid+1;
  }
  cout<<ans<<endl;
}