ACM-ICPC2018 青岛赛区网络预赛-B- Red Black Tree

Posted 2021-01-07 tetew

题目描述

BaoBao has just found a rooted tree with n vertices and (n-1) weighted edges in his backyard. Among the vertices, m of them are red, while the others are black. The root of the tree is vertex 1 and it’s a red vertex.
Let’s define the cost of a red vertex to be 0, and the cost of a black vertex to be the distance between this vertex and its nearest red ancestor.
Recall that 

• The length of a path on the tree is the sum of the weights of the edges in this path.
• The distance between two vertices is the length of the shortest path on the tree to go from one vertex to the other.
• Vertex u is the ancestor of vertex v if it lies on the shortest path between vertex v and the root of the tree (which is vertex 1 in this problem).

As BaoBao is bored, he decides to play q games with the tree. For the i-th game, BaoBao will select ki vertices vi,1, vi,2, . . . , vi,ki on the tree and try to minimize the maximum cost of these ki vertices by changing at most one vertex on the tree to a red vertex.
Note that

• BaoBao is free to change any vertex among all the n vertices to a red vertex, NOT necessary among the ki vertiecs whose maximum cost he tries to minimize.
• All the q games are independent. That is to say, the tree BaoBao plays with in each game is always the initial given tree, NOT the tree modi?ed during the last game by changing at most one vertex.

Please help BaoBao calculate the smallest possible maximum cost of the given ki vertices in each game after changing at most one vertex to a red vertex.

输入

There are multiple test cases. The first line of the input is an integer T, indicating the number of test cases. For each test case:
The first line contains three integers n, m and q (2≤m≤n≤105, 1≤q≤2×105), indicating the size of the tree, the number of red vertices and the number of games.
The second line contains m integers r1, r2, . . . , rm (1 = r1 < r2 <...< rm≤n), indicating the red vertices.
The following (n-1) lines each contains three integers ui, vi and wi (1≤ui, vi≤n, 1≤wi≤109),indicating an edge with weight wi connecting vertex ui and vi in the tree.
For the following q lines, the i-th line will first contain an integer ki (1≤ki≤n). Then ki integers vi,1, vi,2, . . . , vi,ki follow (1≤vi,1 < vi,2 < ... < vi,ki≤n), indicating the vertices whose maximum cost BaoBao has to minimize.
It’s guaranteed that the sum of n in all test cases will not exceed 106, and the sum of ki in all test cases will not exceed 2×106.

输出

For each test case output q lines each containing one integer, indicating the smallest possible maximum cost of the ki vertices given in each game after changing at most one vertex in the tree to a red vertex.

样例输入

2
12 2 4
1 9
1 2 1
2 3 4
3 4 3
3 5 2
2 6 2
6 7 1
6 8 2
2 9 5
9 10 2
9 11 3
1 12 10
3 3 7 8
4 4 5 7 8
4 7 8 10 11
3 4 5 12
3 2 3
1 2
1 2 1
1 3 1
1 1
2 1 2
3 1 2 3

样例输出

4
5
3
8
0
0
0

提示

The first sample test case is shown above. Let’s denote C(v) as the cost of vertex v.
For the 1st game, the best choice is to make vertex 2 red, so that C(3) = 4, C(7) = 3 and C(8) = 4. So the answer is 4.
For the 2nd game, the best choice is to make vertex 3 red, so that C(4) = 3, C(5) = 2, C(7) = 4 and C(8) = 5. So the answer is 5.
For the 3rd game, the best choice is to make vertex 6 red, so that C(7) = 1, C(8) = 2, C(10) = 2 and C(11) = 3. So the answer is 3.
For the 4th game, the best choice is to make vertex 12 red, so that C(4) = 8, C(5) = 7 and C(12) = 0.
So the answer is 8.

 

某个题解qwq http://www.cnblogs.com/taozi1115402474/p/9678091.html

二分总是过不了呀呀呀呀

自闭了啊啊啊啊 

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=200000+10;
int T,n,m,q,cnt,tot;
bool red[N];
int last[N],pos[N],f[N],rmq[N],mm[N],dp[N][20],a[N];
ll cost[N],cost_t[N];
struct tree{
    int v,w,nex;
}t[N];
bool cmp(int a,int b)
{
    return cost_t[a]>cost_t[b];
}
void add(int x,int y,int z)
{
    cnt++;
    t[cnt].v=y;
    t[cnt].nex=last[x];
    last[x]=cnt;
    t[cnt].w=z;
}
void dfs(int x,int fa,int deep,ll dis,ll dis1)
{
    if (red[x]) dis1=0;
    cost[x]=dis; cost_t[x]=dis1;
    pos[x]=tot; f[tot]=x; rmq[tot++]=deep;
    for (int i=last[x];i;i=t[i].nex)
    {
        if (t[i].v==fa) continue;
        dfs(t[i].v,x,deep+1,dis+t[i].w,dis1+t[i].w);
        f[tot]=x;
        rmq[tot++]=deep;
    }
}
void ST(int n)
{
    mm[0]=-1;
    for (int i=1;i<=n;i++)
    {
        mm[i]=((i&(i-1))==0) ? mm[i-1]+1:mm[i-1];
        dp[i][0]=i;
    }
    for (int j=1;j<=mm[n];j++)
        for (int i=1;i+(1<<j)-1<=n;i++)
        dp[i][j]=rmq[dp[i][j-1]]<rmq[dp[i+(1<<(j-1))][j-1]] ? dp[i][j-1] : dp[i+(1<<(j-1))][j-1];
}
int query(int a,int b)
{
    a=pos[a]; b=pos[b];
    if (a>b) swap(a,b);
    int k=mm[b-a+1];
    int ret=rmq[dp[a][k]]<=rmq[dp[b-(1<<k)+1][k]] ? dp[a][k] : dp[b-(1<<k)+1][k];
    return f[ret];
}
int main()
{
    scanf("%d",&T);
    while (T--)
    {
        int x,y,z,k;
        cnt=0; tot=1;
        memset(red,0,sizeof(red));
        memset(last,0, sizeof(last));
  
        scanf("%d%d%d",&n,&m,&q);
        for (int i=1;i<=m;i++)
        {
            scanf("%d",&x);
            red[x]=true;
        }
        for (int i=1;i<n;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z); add(y,x,z);
        }
         
        cost[1]=cost_t[1]=cost_t[n+1]=0;
        dfs(1,-1,1,0,0);
        ST(tot-1);
  
        while (q--)
        {
            scanf("%d",&k);
            for (int i=1;i<=k;i++) scanf("%d",&a[i]);
             
            sort(a+1,a+1+k,cmp);  a[k+1]=n+1;
             
            ll ans=cost_t[a[2]],lon,maxx=0;
            int fa=a[1];
            for (int i=2;i<=k;i++)
            {
                int new_fa=query(fa,a[i]);
                int dep1=rmq[pos[fa]],dep2=rmq[pos[new_fa]];
                if (dep2<dep1) maxx+=cost[fa]-cost[new_fa];
  
                lon=min(cost_t[a[i]],cost[a[i]]-cost[new_fa]);
                maxx=max(maxx,lon);
  
                fa=new_fa;
                ans=min(ans,max(maxx,cost_t[a[i+1]]));
            }
            printf("%lld
",ans);
        }
    }
    return 0;
}

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