Codeforces Round #544 (Div. 3) D F1 F2

题目链接：D. Zero Quantity Maximization

#include <bits/stdc++.h>
using namespace std;
#define maxn 200005
#define LL long long
#define pii pair<LL,LL>
map<pair<LL,LL>,LL>mp,mpp;
LL a[maxn],b[maxn];
LL gcd(LL a,LL b){
    return b?gcd(b,a%b):a;
}
int main(){
   LL n;
   scanf("%lld",&n);
   ;j<n;j++){
      scanf("%lld",&a[j]);
   }
   ;k<n;k++){
      scanf("%lld",&b[k]);
   }
   LL mx = ,ans = ;
   ;j<n;j++){
      LL z = gcd(abs(a[j]),abs(b[j]));
      ){
         a[j] = -a[j];
         b[j] = -b[j];
      }
      ){
         ) ans++;
         continue;
      }
      ){
           mp[make_pair(1LL*,1LL*)]++;
           mx = max(mx, mp[make_pair(1LL*,1LL*)]);
           continue;
      }
      mp[make_pair(a[j]/z,b[j]/z)]++;
      mx = max(mx, mp[make_pair(a[j]/z,b[j]/z)]);
   }
   printf("%lld\n",mx+ans);
}

F1： Spanning Tree with Maximum Degree

判断一下 bfs 就ok

#include<bits/stdc++.h>
using namespace std;
#define maxn 200005
#define LL long long
vector<int>q[maxn];
int fa[maxn];
bool flag[maxn];
void bfs(int u){
    queue<int>Q;
    Q.push(u);
    while(!Q.empty()){
        u = Q.front();
        Q.pop();
        ;j<q[u].size();j++){
            int v = q[u][j];
            if(!flag[v]){
                flag[v] = ;
                Q.push(v);
                printf("%d %d\n",u,v);
            }
        }
    }
}
int main(){
    memset(fa,,sizeof(fa));
    memset(flag,,sizeof(flag));
    ,;
    scanf("%d%d",&n,&m);
    ;j<m;j++){
       int u,v;
       scanf("%d%d",&u,&v);
       q[u].push_back(v);
       q[v].push_back(u);
       fa[u]++;
       fa[v]++;
       if(mx<fa[u]){
          mx = fa[u];
          in = u;
       }
       if(mx<fa[v]){
          mx = fa[v];
          in = v;
       }
    }
//    cout<<in<<endl;
    flag[;
    bfs(in);
}

F2 ：F2. Spanning Tree with One Fixed Degree

先判断 节点的度够不够  不够直接 输出 NO

dfs 判断1有多少个连通块 如果连通块数量 大于d 直接输出NO

最后 添加 在 1 节点添加 d-连通块个数个 点   bfs

#include <bits/stdc++.h>
using namespace std;
#define maxn 200005
#define LL long long
#define pii pair<int,int>
vector<int>q[maxn],z;
vector<pii>lis;
bool fa[maxn];
int n,m,d;
void dfs(int u){
    ;j<q[u].size();j++){
        int v = q[u][j];
        if(!fa[v]){
            fa[v] = ;
            dfs(v);
        }
    }
}
bool bfs(int u){
    queue<int>Q;
    ;
    Q.push(u);
    while(!Q.empty()){
        u = Q.front();
        Q.pop();
        ;j<q[u].size();j++){
            int v = q[u][j];
            if(!fa[v]){
                fa[v] = ,sum++;
                Q.push(v);
                //cout<<u<<" "<<v<<endl;
                lis.push_back(make_pair(u,v));
                ){
                    d--;
                    ) break;
                }
            }
        }
    }
    //cout<<sum<<" "<<n-1<<endl;
    )) ;
    ;
}
int main(){
    scanf("%d%d%d",&n,&m,&d);
    memset(fa,,sizeof(fa));
    ;
    ;j<m;j++){
        int u,v;
        scanf("%d%d",&u,&v);
        q[u].push_back(v);
        q[v].push_back(u);
        ||v==) sum++;
    }
    if(sum<d){
        printf("NO\n");
        ;
    }
    fa[]=;
    set<int>se;
    ;j<q[].size();j++){
        ][j];
        if(!fa[v]){
            fa[v]=;
            se.insert(v);
            dfs(v);
        }
    }
    //cout<<se.size()<<endl;
    if(se.size()>d){
        printf("NO\n");
        ;
    }
    //cout<<*se.begin()<<endl;
    ;j<q[].size()&&se.size()<d;j++){
        se.insert(q[][j]);
    }

    q[].clear();
    for(auto &i:se){
        //cout<<i<<endl;
        q[].push_back((int)i);
    }
    memset(fa,,sizeof(fa));
    fa[] = ;
    bfs();
    printf("YES\n");
    ;j<lis.size();j++){
        pii pa = lis[j];
        printf("%d %d\n",pa.first,pa.second);
    }
}