Codeforces Round #670 (Div. 2) ABC 题解
A. Subset Mex
题意：把一个集合（可能有重复元素）分成两部分，使得每部分的缺少的最小非负整数的和最大。
其中在原序列里面就缺少的那个最小非负整数肯定是躲不掉的。就先把这个数之前的所有数放在一个集合，另外的放另一个集合。这样贪心是最优的。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll a[maxn];
ll Map[150];
int main()
{
int kase;
cin>>kase;
while(kase--)
{
ll n = read();
mem(Map,0);
rep(i,1,n) a[i] = read(), Map[a[i]]++;
ll obj = 0;
rep(i,0,n) if(!Map[i]) {obj = i; break;}
rep(i,0,obj) Map[i]--;
rep(i,1,n) if(Map[a[i]]==0) a[i] = -1;
mem(Map,0);
rep(i,1,n) Map[a[i]]++;
ll obj1 = 0;
rep(i,0,n) if(!Map[i]) {obj1 = i; break; }
cout<<obj+obj1<<'\n';
}
return 0;
}
B. Maximum Product
题意：给一个序列，找出一个五元组，使得其积最大。
思路：贪心。分类讨论一下
全负
四负一正
三负二正
二负三正
一负四正
全正
的情况。如果结果是负数的就挑各自绝对值最小的，反之就挑绝对值最大的几个。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 2e5+200;
const ll inf=0x3f3f3f3f3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll a[maxn];
ll dp[maxn];
vector<ll> neg;
vector<ll> pos;
int main()
{
int kase;
cin>>kase;
while(kase--)
{
neg.clear(), pos.clear();
ll n = read();
rep(i,1,n) a[i] = read();
rep(i,1,n)
{
if(a[i]<0) neg.pb(a[i]);
else pos.pb(a[i]);
}
ll ans = -inf;
if(neg.size()+pos.size()>=5)
{
sort(neg.begin(), neg.end());
sort(pos.begin(),pos.end());
if(neg.size()>=4&&pos.size()>=1)
ans = max((ll)neg[0]*neg[1]*neg[2]*neg[3]*pos[pos.size()-1], ans);
if(neg.size()>=3&&pos.size()>=2)
ans = max((ll)neg[neg.size()-1]*neg[neg.size()-2]*neg[neg.size()-3]*pos[0]*pos[1],ans);
if(neg.size()>=2&&pos.size()>=3)
ans = max((ll)neg[0]*neg[1]*pos[pos.size()-1]*pos[pos.size()-2]*pos[pos.size()-3],ans);
if(neg.size()>=1&&pos.size()>=4)
ans = max((ll)neg[neg.size()-1]*pos[0]*pos[1]*pos[2]*pos[3],ans);
if(pos.size()>=5)
ans = max((ll)pos[pos.size()-1]*pos[pos.size()-2]*pos[pos.size()-3]*pos[pos.size()-4]*pos[pos.size()-5],ans);
if(neg.size()>=5)
ans = max((ll)neg[neg.size()-1]*neg[neg.size()-2]*neg[neg.size()-3]*neg[neg.size()-4]*neg[neg.size()-5], ans);
}
cout<<ans<<'\n';
}
return 0;
}
C. Link Cut Centroids
题意：给一颗树，定义一个特殊点为删除它和连接的边后剩下子树最大节点数最小。现在让你在保持树结构的前提下，改一条边，使得这个特殊点唯一。
思路：其实这个特殊点就是重心。
重心怎么求呢？
随便找个结点当根，先统计所有结点的子树包含结点，然后用一个变量存每个结点子树数量的最大值。其中最大值最小的那个即是重心。
重心有个性质就是如果不唯一，只能最多存在两个，且相邻。
那么这个题求完重心后就看看唯一与否，是的话直接随便找条和它相连的边删掉再连回去。不是的话就把另一个重心的另外一个相连点连到当前这个上。
view code
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };
ll Map[maxn];
vector<vector<ll> > D(maxn);
ll siz[maxn];
ll son[maxn];
bool vis[maxn];
ll n;
ll dfs(ll x)
{
ll cur = 0;
ll ma = 0;
vis[x] = 1;
for(int i=0 ;i<D[x].size(); i++)
{
ll y = D[x][i];
if(vis[y]) continue;
cur += dfs(y);
ma = max(ma,siz[y]);
}
son[x] = ma = max(ma, n-cur-1);
siz[x] = cur+1;
return cur+1;
}
int main()
{
int kase;
cin>>kase;
while(kase--)
{
n = read();
rep(i,0,n+1) Map[i] = 0, D[i].clear(), vis[i] = 0, son[i] = 0, siz[i] = 0;
rep(i,1,n-1)
{
ll x = read(), y = read();
D[x].pb(y);
D[y].pb(x);
}
ll m = dfs(1);
ll obj = 1;
ll mi = inf;
rep(i,1,n)
{
Map[son[i]]++;
if(son[i]<mi) mi = son[i], obj = i;
}
if(Map[mi]==1)
{
ll v = D[obj][0];
cout<<obj<<' '<<v<<'\n';
cout<<obj<<' '<<v<<'\n';
}
else
{
ll v = 1;
for(int i=0; i<D[obj].size(); i++)
if(son[D[obj][i]]==mi) {v=D[obj][i]; break;}
ll vv = 1;
for(int i=0; i<D[v].size(); i++) if(D[v][i]!=obj) {vv=D[v][i]; break;}
cout<<vv<<' '<<v<<'\n';
cout<<vv<<' '<<obj<<'\n';
}
}
return 0;
}
