牛客小白月赛22

牛客小白月赛22

水水水----放点稍微不水的

A.操作序列
在这里插入图片描述
模拟+stl
看着题意打就行,主要输入恶心,所以上榜了

#pragma GCC optimize(3,"Ofast","inline")  	//G++
#include<bits/stdc++.h>
#define TEST freopen("C:\\Users\\hp\\Desktop\\ACM\\in.txt","r",stdin);
#define mem(a,x) memset(a,x,sizeof(a))
#define debug(x) cout << #x << ": " << x << endl;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fcout cout<<setprecision(4)<<fixed
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;


const int inf=0x3f3f3f3f;
const ll INF=0x7fffffffffffffff;
const int mod=1e9+7;
const int maxn = 1e6+5;
const double eps=1e-8;

template<typename T> void read(T &x) {
    x = 0;
    char ch = getchar();
    ll f = 1;
    while(!isdigit(ch)) {
        if(ch == '-')f*=-1;
        ch=getchar();
    }
    while(isdigit(ch)) {
        x = x*10+ch-48;
        ch=getchar();
    }
    x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) {
    read(first);
    read(args...);
}
int sgn(double a) {
    return a<-eps?-1:a<eps?0:1;
}
int num[105];
int get_cin(){
    string str;
    getline(cin,str);
    if(str[0]=='-')
        return -1;
    int n=str.size();
    int op=1,pos;
    for(int i=0;i<n;i++){
        if(str[i]==' ')
            op=2,pos=i;
    }
    if(op==1){
        num[1]=0;
        for(int i=0;i<n;i++){
            num[1]=num[1]*10+str[i]-'0';
        }
    }
    else{
        num[1]=num[2]=0;
        for(int i=0;i<pos;i++){
            num[1]=num[1]*10+str[i]-'0';
        }
        for(int i=pos+1;i<n;i++){
            num[2]=num[2]*10+str[i]-'0';
        }
    }
    return op;
}
set<int>s;
map<int,int>mp;
void solve(){
    int op=get_cin();
    if(op==2){
        int t=num[1],c=num[2];
//        auto it=s.lower_bound(t-30);
//        if(it==s.end()||*it>t+30){
//            s.insert(t);
//            mp[t]=c;
//        }
        if(c&&s.lower_bound(t-30)==s.upper_bound(t+30)){
            s.insert(t);
            mp[t]=c;
        }
    }
    else if(op==1){
        int t=num[1];
        cout<<mp[t]<<"\n";
    }
    else
    {
        if(s.size()==0){
            cout<<"skipped\n";
            return ;
        }
        cout<<mp[*s.begin()]<<"\n";
        mp[*s.begin()]=0;
        s.erase(s.begin());
    }
}
int main(){
    int T;
    scanf("%d",&T);
    getchar();
    while(T--)
        solve();

}

B
在这里插入图片描述
树的直径—树形DP

#pragma GCC optimize(3,"Ofast","inline")  	//G++
#include<bits/stdc++.h>
#define mem(a,x) memset(a,x,sizeof(a))
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pi;

const ll inf=0x3f3f3f3f;
const ll maxn = 1e6+5;

template<typename T> void read(T &x){
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) {
    read(first);
    read(args...);
}

ll n, m;
ll ans=-1e15;
vector<ll> edge[maxn];
void add(ll x,ll y){
    edge[x].push_back(y);
    edge[y].push_back(x);
}
ll dp[maxn][2];// 0最长 1次长
ll a[maxn];
void dfs(ll u,ll fa){

    dp[u][0]  =a[u];
    dp[u][1] = -inf;
    for(auto v:edge[u]){
        if(v==fa) continue;
        dfs(v,u);
        if(dp[v][0]+a[u]>dp[u][0]) dp[u][1]=dp[u][0],dp[u][0]=dp[v][0]+a[u];
        else if(dp[v][0]+a[u]>dp[u][1])
            dp[u][1] = dp[v][0] + a[u];
    }
    ans = max(dp[u][0] + dp[u][1]-a[u], ans);
}
main()
{
    read(n);
    for (ll i = 1; i <= n;i++){
        read(a[i]);
        ans = max(a[i], ans);
        //   dp[i][0] = dp[i][1] = -inf;
    }
    for (ll i = 1; i <= n - 1; i++)
    {
        ll x, y;
        read(x, y);
        add(x, y);
    }
    dfs(1, 0);
    cout << ans << "\n";
}

C
在这里插入图片描述
所有状态预处理一遍即可

//#pragma GCC optimize(3,"Ofast","inline")  	//G++
//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include <cstdlib>
#include <iomanip>
#define TEST freopen("C:\\Users\\hp\\Desktop\\ACM\\in.txt","r",stdin);
#define mem(a,x) memset(a,x,sizeof(a))
#define debug(x) cout << #x << ": " << x << endl;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fcout cout << setprecision(4) << fixed;
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;


const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn = 1e6+5;
const double eps=1e-8;

template<typename T> void read(T &x){
    x = 0;char ch = getchar();int f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) {
    read(first);
    read(args...);
}
int sgn(double a){
    return a<-eps?-1:a<eps?0:1;
}
int a[maxn],dp[maxn];
void dfs(int now,int pos){

    if(now==11){
        int sum = 0;
        for (int i = 1;i<=12;i++)
            sum = sum + (a[i]);
        dp[pos] = min(dp[pos], sum);
        return ;
    }
    dfs(now + 1,pos);
    if(a[now]==1&&a[now+1]==1&&a[now+2]==0){

        a[now] = 0, a[now + 1] = 0, a[now + 2] = 1;
        dfs(max(now-2,1) ,pos);
        a[now] = 1, a[now + 1] = 1, a[now + 2] = 0;
    }
    if(a[now]==0&&a[now+1]==1&&a[now+2]==1){
        a[now] = 1, a[now + 1] = 0, a[now + 2] = 0;
        dfs(max(now-2,1) ,pos);
        a[now] = 0, a[now + 1] = 1, a[now + 2] = 1;
    }
}
void init(){
    
    for (int i = 0;i<(1<<12);i++){
        for(int j=0,k=12;j<12;j++,k--){
            a[k] =((i >> j) & 1);
        }
        dp[i] = inf;
        dfs(1, i);
    }
}
char s[maxn];
void solve(){
    int sum = 0;
    scanf("%s", s + 1);
    for(int i=1;i<=12;i++){
        sum = sum * 2 + (s[i]=='o');
    }
    cout << dp[sum] << "\n";
}
int main(){
    init();
    int T;
    read(T);
    while(T--)
        solve();
}

D.dfs暴力 水题
F签到题 水题
G暴力 水题

H
在这里插入图片描述
区间合并+差分数组 好像还是水题

//#pragma GCC optimize(3,"Ofast","inline")      //G++
//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include <cstdlib>
#include <iomanip>
#define TEST freopen("C:\\Users\\hp\\Desktop\\ACM\\in.txt","r",stdin);
#define mem(a,x) memset(a,x,sizeof(a))
#define debug(x) cout << #x << ": " << x << endl;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fcout cout << setprecision(4) << fixed;
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;
 
 
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn = 1e6+5;
const double eps=1e-8;
 
template<typename T> void read(T &x){
    x = 0;char ch = getchar();int f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) {
    read(first);
    read(args...);
}
int sgn(double a){
    return a<-eps?-1:a<eps?0:1;
}
int n, m;
int d[maxn];
map<int,vector<pair<int, int> > > mp;
void add(int l,int r){
    d[l]++, d[r + 1]--;
}
void work(vector<pair<int,int> >v){
    sort(v.begin(), v.end());
    int l=maxn,r=-maxn;
    for(auto it:v){
        int ll = it.first;
        int rr = it.second;
        if(ll>r){
            if(l!=maxn){
                add(l, r);
            }
            l = ll, r = rr;
        }
        else{
            l=min(l,ll);
            r = max(r, rr);
        }
    }
    add(l, r);
}
void solve(){
    read(n, m);
    for (int i = 1; i <= n;i++){
        int l, r, dd;
        read(l, r, dd);
        mp[dd].push_back({l, r});
    }
    for(auto it:mp){
        work(mp[it.first]);
    }
    int ans, sum = 0;
    for (int i = 1;i<=n;i++){
        d[i] += d[i - 1];
        if(d[i]>sum){
            sum=d[i];
            ans = i;
        }
    }
    cout << ans << "\n";
}
int main(){
    solve();
}

I不会
J模拟—水题

### 关于牛客小白109的信息 目前并未找到关于牛客小白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牛客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,在牛客小白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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