Codeforces Round 798 (Div. 2) (C 树形dp D 曼哈顿距离转换 E 位运算构造)

C：dp以f[i]为根的时候能获得多少个节点，那么dp就是全部儿子里面找一个切掉，其他就是

f[v]的总和了

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+10,M=2*N,mod=1e9+7;
#define int long long
typedef long long LL;
typedef pair<int, int> PII;
typedef unsigned long long ULL;
using Node=tuple<int,int,int>;
const long long inf=1e9;

int n,m,k;
vector<int> g[N];
int sz[N];
int son[N];
int f[N];
void dfs(int u,int fa)
{
    sz[u]=1;
    int s=0;
    for(auto v:g[u])
    {
        if(v==fa) continue;
        dfs(v,u);
        sz[u]+=sz[v];
        s+=f[v];
        //f[v] sz[v]
    }
    for(auto v:g[u]){
        if(v==fa) continue;
        f[u]=max(f[u],s-f[v]+sz[v]-1);
    }
}
void solve()
{
    cin>>n;
    for(int i=1;i<=n;i++) g[i].clear(),f[i]=0;
    for(int i=1;i<n;i++){
        int a,b;cin>>a>>b;
        g[a].push_back(b);g[b].push_back(a);
    }
    dfs(1,0);
    cout<<f[1]<<"\n";
}
//1 2 3 4
signed main()
{
    cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
    int t=1;

    cin>>t;

    while(t--) solve();
}

D:曼哈顿距离两点最大值转换成max(max(xi+yi)-min(xi+yi), max(xi-yi)-min(xi-yi))

直接拿个multiset维护前面几个最小值和最大即可

#include<bits/stdc++.h>
using namespace std;
const int N = 2010+10,M=2*N,mod=1e9+7;

typedef long long LL;
typedef pair<int, int> PII;
typedef unsigned long long ULL;
using Node=tuple<int,int,int>;
const long long inf=1e9;

int n,m,k;
char a[N][N];
void solve()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++)
            cin>>a[i][j];
    }
    //|xi-xj|+|yi-yj|
    //(xi+yi)-(xj+yj)
    //(xj+yj)-(xi+yi)
    //(xi-yi)-(xj-yj)
    //(xj-yj)-(xi-yi)
    //max({x+y}-({x+y}))
    //max({x-y}-(x-y))
    //任意两个点距离是
    //max{ max(xi+yi)-min(xj+yj)}
    // max(xi-yi)-min(xj-yj)
    multiset<int> x,y;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(a[i][j]=='B')
            {
                x.insert(i+j);
                auto it=x.begin();
                it++;
                it++;it++;
                while(x.size()>=10)
                {
                    x.extract(*it);
                }
                y.insert(i-j);
                it=y.begin();
                it++;
                it++;it++;
                while(y.size()>=10)
                {
                    y.extract(*it);
                }
            }
        }
    }
    if(x.size()==0)
    {
        cout<<0<<"\n";return ;
    }
    if(x.size()==1){
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(a[i][j]=='B'){
                    cout<<i<<" "<<j<<"\n";return ;
                }
            }
        }
    }
    int mx=inf;
    int idx,idy;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++)
        {
            if(a[i][j]=='B'){
                x.extract(i+j);
                y.extract(i-j);
            }
            int xx=i+j;
            int yy=i-j;
            int cnt=max(
                {
                xx-(*x.begin()),
                (*x.rbegin())-xx,
                yy-(*y.begin()),
                (*y.rbegin())-yy
            });
           // cout<<cnt<<" "<<mx<<" "<<i<<" "<<j<<"\n";
            mx=min(cnt,mx);
            if(cnt==mx){
                idx=i;idy=j;
            }
            if(a[i][j]=='B'){
                x.insert(i+j);
                y.insert(i-j);
            }
        }
    }
    cout<<idx<<" "<<idy<<"\n";

}
//1 2 3 4
signed main()
{
    cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
    int t=1;

    cin>>t;

    while(t--) solve();
}

E：先全部把0变成1，然后就找最大lowbit，至少可以通过两次构造

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+10,M=2*N,mod=1e9+7;
#define int long long
typedef long long LL;
typedef pair<int, int> PII;
typedef unsigned long long ULL;
using Node=tuple<int,int,int>;
const long long inf=1e9;

int n,m,k;
int a[N],res[N];
class dsu {
    public:
    vector<int> p;
    int n;
 
    dsu(int _n) : n(_n) {
        p.resize(n);
        iota(p.begin(), p.end(), 0);
    }
 
    inline int get(int x) {
        return (x == p[x] ? x : (p[x] = get(p[x])));
    }
 
    inline bool unite(int x, int y) {
        x = get(x);
        y = get(y);
        if (x != y) {
            p[x] = y;
            return true;
        }
        return false;
    }
};
int lowbit(int x)  // 返回末尾的1
{
    return x & -x;
}
int pos[N],pre[N],suf[N];
int ans;
void solve()
{
    cin>>n;
    ans=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        if(a[i]==0) a[i]=1,ans++;
    }
    int cnt=n;
    dsu d(n+10);
    for(int j=0;j<=29;j++)
    {
        int x=0;
        for(int i=1;i<=n;i++){
            if(a[i]>>j&1){
                if(x==0) x=i;
                else d.unite(x,i);
            }
        }
    }
    cnt=0;
    for(int i=1;i<=n;i++)
    if(d.get(i)==i) cnt++;
    if(cnt==1)
    {
        return ;
    }
    ans++;
    int mx=0,tot=0,qwq=0;
    for(int i=1;i<=n;i++)
    mx=max(mx,lowbit(a[i]));
    
    for(int i=1;i<=n;i++){
        if(lowbit(a[i])==mx){
            pos[++tot]=i;
        }else qwq|=a[i];
    }
    if(cnt==2){
        int x=d.get(pos[1]);
        for(int i=1;i<=n;i++)
        {
            if(d.get(i)!=x&&(a[i]&1)){
                a[pos[1]]++;
                return ;
            }
        }
    }
    if(tot==1){
        a[pos[1]]--;
        return ;
    }
    pre[0]=suf[tot+1]=0;
    for(int i=1;i<=tot;i++){
        pre[i]=pre[i-1]|a[pos[i]];
    }
    for(int i=tot;i>=1;i--){
        suf[i]=suf[i+1]|a[pos[i]];
    }
    for(int i=1;i<=tot;i++){
        if((pre[i-1]|suf[i+1])&((a[pos[i]]^mx)|qwq)){
          a[pos[i]]--;return ;  
        }
    }
    ans++;
    a[pos[2]]++;a[pos[1]]--;
}
//1 2 3 4
signed main()
{
    cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
    int t=1;

    cin>>t;

    while(t--){
        solve();
        cout<<ans<<"\n";
        for(int i=1;i<=n;i++){
            cout<<a[i]<<" \n"[i==n];
        }
    }
}