牛客小白月赛96 D 最小连通代价

题目在这里

题意: 加边是所有点连通,没有重边和自环,问最小代价
加边规则:两点权值奇偶性相同代价为a,否则为b
− 100 ≤ a , b ≤ 100 -100\leq a,b \leq100 100a,b100

分析:
这题就是一个分类讨论,先读进来统计奇数点和偶数点
n a na na为奇偶性相同的点的连边, n b nb nb为奇偶性不同的点的连边, j i ji ji为奇数点, o u ou ou为偶数点
1. a ≤ 0 1.a\leq 0 1.a0 and b ≤ 0 : b\leq0: b0:
所有可以连的边都连上
n a = C ( 2 j i ) + C ( 2 o u ) na=C\binom{2}{ji}+C\binom{2}{ou} na=C(ji2)+C(ou2)
n b = j i ∗ o u nb=ji*ou nb=jiou

2. a ≤ 0 2.a\leq 0 2.a0 and b > 0 : b>0: b>0:
n a = C ( 2 j i ) + C ( 2 o u ) na=C\binom{2}{ji}+C\binom{2}{ou} na=C(ji2)+C(ou2)
如果全是技术点或者偶数点,则不需要奇数点和偶数点连一条边
i f ( m i n ( j i , o u ) ) n b = 1 ; if(min(ji,ou)) nb = 1; if(min(ji,ou))nb=1;

3. a > 0 3.a> 0 3.a>0 and b ≤ 0 : b\leq0: b0:
如果既有奇数点又有偶数点,则它们之间的所有边都要,否则我只要n-1条边因为代价大于0
i f ( m i n ( j i , o u ) ) if(min(ji,ou)) if(min(ji,ou)) n b = j i ∗ o u ; nb = ji*ou; nb=jiou;
e l s e else else n a = n − 1 ; na = n-1; na=n1;

4. a > 0 4.a> 0 4.a>0 and b > 0 : b>0: b>0:
如果全是奇数或者偶数点,只取n-1条边即可
否则:
如果a<=b: n a = n − 2 ; na = n-2; na=n2; n b = 1 nb = 1 nb=1
如果a>b: n b = n − 1 nb = n-1 nb=n1即可

#include<bits/stdc++.h>
using namespace std;
using i64 = long long;
using i128 = __int128;
#define ios ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);

int n,a,b;
void solve(){
    cin>>n>>a>>b;
    long long ji = 0,ou = 0;
    for(int i = 1;i<=n;++i){
        int x;cin>>x;
        ji+=(x%2==1);
        ou+=(x%2==0);
    }
    long long na = 0,nb =0;
    if(a<=0&&b<=0){
        na = ji*(ji-1)/2+ou*(ou-1)/2;
        nb = ji*ou;
    }else if(a<=0){
        na = (ji-1)*ji/2+ou*(ou-1)/2;
        if(min(ji,ou)) nb = 1;
    }else if(b<=0){
        if(min(ji,ou)) nb = ji*ou;
        else na = n-1;
    }else{
        if(min(ji,ou)==0){
            na = n-1;
        }else{
            if(a<=b){
                na = n-2;
                nb = 1;
            }else nb = n-1;
        }
    }
    cout<<na*a+nb*b<<"\n";
}

int main(){
    ios;
    int t;
    cin>>t;
    while(t--){
        solve();
    }
    return 0;
}

没开long long WA了好几发

### 关于小白109的信息 目前并未找到关于小白109的具体比信息或题解内容[^5]。然而,可以推测该事可能属于网举办的系列算法竞之一,通常这类比会涉及数据结构、动态规划、图论等经典算法问题。 如果要准备类似的事,可以通过分析其他场次的比题目来提升自己的能力。例如,在小白13中,有一道与二叉树相关的题目,其核心在于处理树的操作以及统计最终的结果[^3]。通过研究此类问题的解决方法,能够帮助理解如何高效地设计算法并优化时间复杂度。 以下是基于已有经验的一个通用解决方案框架用于应对类似场景下的批量更新操作: ```python class TreeNode: def __init__(self, id): self.id = id self.weight = 0 self.children = [] def build_tree(n): nodes = [TreeNode(i) for i in range(1, n + 1)] for node in nodes: if 2 * node.id <= n: node.children.append(nodes[2 * node.id - 1]) if 2 * node.id + 1 <= n: node.children.append(nodes[2 * node.id]) return nodes[0] def apply_operations(root, operations, m): from collections import defaultdict counts = defaultdict(int) def update_subtree(node, delta): stack = [node] while stack: current = stack.pop() current.weight += delta counts[current.weight] += 1 for child in current.children: stack.append(child) def exclude_subtree(node, total_nodes, delta): nonlocal root stack = [(root, False)] # (current_node, visited) subtree_size = set() while stack: current, visited = stack.pop() if not visited and current != node: stack.append((current, True)) for child in current.children: stack.append((child, False)) elif visited or current == node: if current != node: subtree_size.add(current.id) all_ids = {i for i in range(1, total_nodes + 1)} outside_ids = all_ids.difference(subtree_size.union({node.id})) for idx in outside_ids: nodes[idx].weight += delta counts[nodes[idx].weight] += 1 global nodes nodes = {} queue = [root] while queue: curr = queue.pop(0) nodes[curr.id] = curr for c in curr.children: queue.append(c) for operation in operations: op_type, x = operation.split(' ') x = int(x) target_node = nodes.get(x, None) if not target_node: continue if op_type == '1': update_subtree(target_node, 1) elif op_type == '2' and target_node is not None: exclude_subtree(target_node, n, 1) elif op_type == '3': path_to_root = [] temp = target_node while temp: path_to_root.append(temp) if temp.id % 2 == 0: parent_id = temp.id // 2 else: parent_id = (temp.id - 1) // 2 if parent_id >= 1: temp = nodes[parent_id] else: break for p in path_to_root: p.weight += 1 counts[p.weight] += 1 elif op_type == '4': pass # Implement similarly to other cases. result = [counts[i] for i in range(m + 1)] return result ``` 上述代码片段展示了针对特定类型的树形结构及其操作的一种实现方式。尽管它并非直接对应小白109中的具体题目,但它提供了一个可借鉴的设计思路。 ####
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