ABC267 - C,D,E,F Solutions

目录

• ABC267 - C,D,E,F Solutions
  • C - Index × A(Continuous ver.)
    • Problem Statement
    • Solution
    • Implementation
  • D - Index × A(Not Continuous ver.)
    • Problem Statement
    • Solution
    • Implementation
  • E - Erasing Vertices 2
    • Problem Statement
    • Solution
    • Implementation
  • F - Exactly K Steps
    • Problem Statement
    • Solution
    • Implementation

ABC267 - C,D,E,F Solutions

C - Index × A(Continuous ver.)

Problem Statement

You are given an integer sequence \(A=(A_1,A_2,\dots,A_N)\) of length \(N\).

Find the maximum value of \(\sum_{i=1}^{M} i \times B_i\) for a contiguous subarray \(B=(B_1,B_2,\dots,B_M)\) of \(A\) of length \(M\).

Constraints

• \(1 \le M \le N \le 2 \times 10^5\)
• \(-2 \times 10^5 \le A_i \le 2 \times 10^5\)

Solution

不妨令 \(R_k=\sum_{i=1}^{M} i \times B_i\)，其中 \(B=(A_k,A_{k+1},\dots,A_{k+M-1})\)。直接计算是 \(O(N^2)\) 的，需要进行优化。

首先可以考虑 \(R_k\) 与 \(R_{k+1}\) 的关系，会发现 \(R_{k+1}-R_k=M\times A_{k+M}-\sum_{i=k}^{k+M-1}A_i\)，于是简单前缀和就解决了。

另外也可以直接求，对 \(A_i\) 与 \(i\times A_i\) 都求前缀和，\(R_k\) 即为两者做差的形式，此处不再赘述。

Implementation

signed main(){
    n=rd();m=rd();jk(i,1,n)a[i]=rd();
    jk(i,1,n)s[i]=s[i-1]+a[i];
    jk(i,1,m)R+=a[i]*i;r=R;
    jk(i,m+1,n)r=std::max(r,R=R-(s[i-1]-s[i-m-1])+m*a[i]);
    P(r);
}

D - Index × A(Not Continuous ver.)

Problem Statement

You are given an integer sequence \(A=(A_1,A_2,\dots,A_N)\) of length \(N\).

Find the maximum value of \(\sum_{i=1}^{M} i \times B_i\) for a (not necessarily contiguous) subsequence \(B=(B_1,B_2,\dots,B_M)\) of length \(M\) of \(A\).

Constraints

• \(1 \le M \le N \le 2000\)
• \(-2 \times 10^5 \le A_i \le 2 \times 10^5\)

Solution

和上题不一样的地方在于 \(B\) 的选择可以不连续，不难想到可以通过 dp 进行求解，于是令 \(f(n,m)\) 表示 \((A_1,A_2,\dots,A_n)\) 中，选择了 \(m\) 个数时 \(\sum_{i=1}^{m} i \times B_i\) 最大的值。故转移方程为 \(f(n,m)=\max(f(n-1,m),f(n-1,m-1)+m*A[n])\)。

Implementation

\(A_i\) 可能为负数，需要注意边界条件。

#define IN -1000000000000000000ll
signed main(){
    n=rd();m=rd();jk(i,1,n)a[i]=rd();
    jk(i,1,n)jk(j,1,n)f[i][j]=IN;
    f[0][1]=IN; //Important!
    jk(i,1,n)jk(j,1,i)f[i][j]=std::max(f[i-1][j],f[i-1][j-1]+j*a[i]);
    P(f[n][m]);
}

E - Erasing Vertices 2

Problem Statement

You are given a simple undirected graph with \(N\) vertices and \(M\) edges. The \(i\)-th edge connects Vertices \(U_i\) and \(V_i\). Vertex \(i\) has a positive integer \(A_i\) written on it.

You will repeat the following operation \(N\) times.

• Choose a Vertex \(x\) that is not removed yet, and remove Vertex \(x\) and all edges incident to Vertex \(x\). The cost of this operation is the sum of the integers written on the vertices directly connected by an edge with Vertex \(x\) that are not removed yet.

We define the cost of the entire \(N\) operations as the maximum of the costs of the individual operations. Find the minimum possible cost of the entire operations.

Constraints

• \(1≤N≤2×10^5\)
• \(0 \le M \le 2 \times 10^5\)
• \(1 \le A_i \le 10^9\)
• The given graph is simple.

Solution

Minimize the maximum. 经典二分答案，需要注意删掉一个点时会导致相邻点的代价变小，所以开一个队列，在删掉后将相邻满足条件的点加到队列里。

Implementation

bool ck(int x){
    memcpy(s,S,(n+1)*sizeof(int));memset(de,0,sizeof de);
    std::queue<int>q;
    jk(i,1,n)if(s[i]<=x)q.push(i);
    while(!q.empty()){
        int u=q.front();q.pop();if(de[u])continue;de[u]=1;
        for(int i=h[u];i;i=e[i].n){
            int v=e[i].v;if(de[v])continue;
            if((s[v]-=V[u])<=x)q.push(v);
        }
    }
    jk(i,1,n)if(!de[i])return 0;
    return 1;
}

signed main(){
    n=rd();m=rd();jk(i,1,n)V[i]=rd();
    int u,v;jk(i,1,m){u=rd();v=rd();ad(u,v);ad(v,u);}
    jk(u,1,n)for(int i=h[u];i;i=e[i].n)S[u]+=V[e[i].v];
    int l=0,r=MX*2,mi;while(l<r){mi=(l+r)>>1;if(ck(mi))r=mi;else l=mi+1;}
    P(l);
}

F - Exactly K Steps

Problem Statement

You are given a tree with \(N\) vertices. The vertices are numbered \(1, \dots, N\), and the \(i\)-th (\(1 \leq i \leq N - 1\)) edge connects Vertices \(A_i\) and \(B_i\).

We define the distance between Vertices \(u\) and \(v\) on this tree by the number of edges in the shortest path from Vertex \(u\) to Vertex \(v\).

You are given \(Q\) queries. In the \(i\)-th (\(1 \leq i \leq Q\)) query, given integers \(U_i\) and \(K_i\), print the index of any vertex whose distance from Vertex \(U_i\) is exactly \(K_i\). If there is no such vertex, print -1.

Constraints

• \(2≤N≤2×10^5\)
• The given graph is a tree.
• \(1 \leq Q \leq 2 \times 10^5\)

Solution

需要知道一个性质：令直径两端点为 \(X\) 与 \(Y\)，则对于树中的任意节点 \(u\)，\(u\) 到 \(X\) 的路径与 \(u\) 到 \(Y\) 的路径中必有一条为 \(u\) 点出发可达的最长路径，此处性质在两边dfs求直径的方法中被证明过。

于是先求出直径，然后分别以 \(X\) 与 \(Y\) 为根建两棵树，使用树上倍增找点。

另：官方题解中把询问离线下来，\(O(N+Q)\) 地解决了，回来补下这个做法。

Implementation

void df(int u,int f,int o){
    D[u][o]=D[f][o]+1;F[u][0][o]=f;
    if(o)jk(i,1,L)F[u][i][o]=F[F[u][i-1][o]][i-1][o];
    for(int i=h[u];i;i=e[i].n)if(e[i].v!=f)df(e[i].v,u,o);
}

void pr(){
    int mx,mi;df(1,0,0);
    mx=D[1][0],mi=1;jk(i,2,n)if(D[i][0]>mx)mx=D[i][0],mi=i;
    memset(D,0,sizeof D);X=mi;df(mi,0,0);
    mx=D[1][0],mi=1;jk(i,2,n)if(D[i][0]>mx)mx=D[i][0],mi=i;
    Y=mi;df(X,0,1);df(Y,0,2);
}

int fi(int x,int d,int o){jk(i,0,L)if((d>>i)&1)x=F[x][i][o];return x;}
int wk(int x,int d){jk(o,1,2)if(D[x][o]>d)return fi(x,d,o);return -1;}