Codeforces Round #485 (Div. 2)

Codeforces Round #485 (Div. 2)

https://codeforces.com/contest/987

A

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define IT set<ll>::iterator
 #define sqr(x) ((x)*(x))
 #define pb push_back
 #define eb emplace_back
 #define maxn 1000006
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef pair<int,int> pii;
 typedef pair<ll,ll>pll;
 typedef pair<ll,int> pli;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long MOD=;
 const double oula=0.57721566490153286060651209;
 using namespace std;

 map<string,string>mp;
 int main(){
     std::ios::sync_with_stdio(false);
     int n;
     cin>>n;
     string s[];
     mp["purple"]="Power";
     mp["green"]="Time";
     mp["blue"]="Space";
     mp["orange"]="Soul";
     mp["red"]="Reality";
     mp["yellow"]="Mind";
     for(int i=;i<n;i++){
         cin>>s[i];
         mp[s[i]]="";
     }
     cout<<-n<<endl;
     for(auto it:mp){
         if(it.second!="") cout<<it.second<<endl;
     }
 }

B

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define IT set<ll>::iterator
 #define sqr(x) ((x)*(x))
 #define pb push_back
 #define eb emplace_back
 #define maxn 1000006
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef pair<int,int> pii;
 typedef pair<ll,ll>pll;
 typedef pair<ll,int> pli;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long MOD=;
 const double oula=0.57721566490153286060651209;
 using namespace std;

 map<string,string>mp;
 int main(){
     std::ios::sync_with_stdio(false);
     ll x,y;
     cin>>x>>y;
     if(x==y||(x==&&y==)||(x==&&y==)){cout<<"="<<endl;return ;}
     if(x==&&y==||x==){cout<<"<"<<endl;return ;}
     if(x==&&y==||y==){cout<<">"<<endl;return ;}
     else if(x<y){cout<<">"<<endl;return ;}
     else if(x>y){cout<<"<"<<endl;return ;}
     return ;
 }

C

题意：给n个数，每个位置有两个属性s,c，要求选择3个位置i,j,k，i<j<k且si<sj<sk 且ci+cj+ck的值最小

思路：原本想了个三重for循环暴力，但是看了数据觉得不可行，然后发现，如果枚举中间那个数j,那么i往前枚举，k往后枚举，这样只要O（n^2）的时间复杂度，可以通过该题

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define IT set<ll>::iterator
 #define sqr(x) ((x)*(x))
 #define pb push_back
 #define eb emplace_back
 #define maxn 1000006
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef pair<int,int> pii;
 typedef pair<ll,ll>pll;
 typedef pair<ll,int> pli;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long MOD=;
 const double oula=0.57721566490153286060651209;
 using namespace std;

 int n;
 struct sair{
     int pos,v;
     bool operator<(const sair&b)const{
         return pos<b.pos;
     }
 }a[];
 vector<sair>ve;

 int main(){
     std::ios::sync_with_stdio(false);
     cin>>n;
     for(int i=;i<=n;i++){
         cin>>a[i].pos;
     }
     for(int i=;i<=n;i++){
         cin>>a[i].v;
     }
     ll ans=0x3f3f3f3f3f3f3f3f;
     for(int i=;i<n;i++){
         ll Min1=0x3f3f3f3f3f3f3f3f,Min2=0x3f3f3f3f3f3f3f3f;
         int posl=i-,posr=i+;
         while(posl>=){
             if(a[posl].pos<a[i].pos){
                 if(Min1>a[posl].v){
                     Min1=a[posl].v;
                 }
             }
             posl--;
         }
         while(posr<=n){
             if(a[posr].pos>a[i].pos){
                 if(Min2>a[posr].v){
                     Min2=a[posr].v;
                 }
             }
             posr++;
         }
         if(Min1!=0x3f3f3f3f3f3f3f3f&&Min2!=-0x3f3f3f3f3f3f3f3f){
             ans=min(ans,Min1+Min2+a[i].v);
         }
     }
     if(ans==0x3f3f3f3f3f3f3f3f) cout<<-<<endl;
     else cout<<ans<<endl;
 }

D

题意：一些公司将在某地举办展览会，该地有n个城市，有m条双向道路。有k种类型的物品，每个城市可以生产出一个类型的物品。举办展览会需要有s种物品。每种物品运输需要一定的费用，费用等于路径的长度，问在n个城市举办展览会的最少费用

思路：把每种物品跑最短路即可

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define IT set<ll>::iterator
 #define sqr(x) ((x)*(x))
 #define pb push_back
 #define eb emplace_back
 #define maxn 5000005
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef pair<int,int> pii;
 typedef pair<ll,ll>pll;
 typedef pair<ll,int> pli;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long MOD=;
 const double oula=0.57721566490153286060651209;
 using namespace std;

 int n,m,s,k;
 int a[];
 vector<int>ve[];
 int book[];
 int dis[][];

 void bfs(){
     queue<int>Q;
     memset(dis,-,sizeof(dis));
     for(int i=;i<=s;i++){
         while(!Q.empty()) Q.pop();
         for(int j=;j<=n;j++){
             if(a[j]==i){
                 Q.push(j);
                 dis[j][i]=;
             }
         }
         while(!Q.empty()){
             int ss=Q.front();
             Q.pop();
             for(auto au:ve[ss]){
                 if(dis[au][i]==-){
                     dis[au][i]=dis[ss][i]+;
                     Q.push(au);
                 }
             }
         }
     }
     ll ans;
     for(int i=;i<=n;i++){
         ans=;
         sort(dis[i]+,dis[i]+s+);
         for(int j=;j<=k;j++){
             ans+=dis[i][j];
         }
         cout<<ans<<" ";
     }
 }

 int main(){
     std::ios::sync_with_stdio(false);
     cin>>n>>m>>s>>k;
     for(int i=;i<=n;i++){
         cin>>a[i];
     }
     int x,y;
     for(int i=;i<=m;i++){
         cin>>x>>y;
         ve[x].pb(y);
         ve[y].pb(x);
     }
     bfs();
 }

E

题意：有1-n按顺序排列的数，A进行3*n操作，每次交换两个数，B进行7*n+1操作，给个1-n的排列，问是谁打乱的

思路：每次交换逆序对都会加一或减一，且从一个序列变回该序列需要至少两次操作，所以判断逆序对和n的奇偶性即可

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define IT set<ll>::iterator
 #define sqr(x) ((x)*(x))
 #define pb push_back
 #define eb emplace_back
 #define maxn 5000005
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef pair<int,int> pii;
 typedef pair<ll,ll>pll;
 typedef pair<ll,int> pli;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long MOD=;
 const double oula=0.57721566490153286060651209;
 using namespace std;

 int n;
 int a[];
 int tree[];

 int lowbit(int x){return x&(-x);}
 int ask(int x){int ans=;while(x){ans+=tree[x];x-=lowbit(x);}return ans;}
 void add(int x){while(x<=n){tree[x]+=;x+=lowbit(x);}}

 int main(){
     std::ios::sync_with_stdio(false);
     cin>>n;
     for(int i=;i<=n;i++){
         cin>>a[i];
     }
     int sum=;
     for(int i=n;i;i--){
         sum+=ask(a[i]);
         add(a[i]);
     }
     sum&=,n&=;
     if(sum==n) cout<<"Petr"<<endl;
     else cout<<"Um_nik"<<endl;
 }

F

题意：有m个整数，每个整数都在0~2^n-1之间，以每个整数为顶点建立一个无向图，当x&y==0时，则认为x,y之间存在一条边。计算图中联通块的数量。

思路：1010和0101符合条件，那1010和0001，0100，0000也符合条件，所以当一个数为x时，直接找～x（就是x转二进制后每位数与1异或后的值）的子集，然后搜索找联通块即可

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define IT set<ll>::iterator
 #define sqr(x) ((x)*(x))
 #define pb push_back
 #define eb emplace_back
 #define maxn 5000005
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef pair<int,int> pii;
 typedef pair<ll,ll>pll;
 typedef pair<ll,int> pli;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long MOD=;
 const double oula=0.57721566490153286060651209;
 using namespace std;

 int n,m;
 int book[maxn],a[maxn],num[maxn];

 void bfs(int x){
     queue<int>Q;
     Q.push(x);
     book[x]=;
     while(!Q.empty()){
         int s=Q.front();
       //  cout<<s<<endl;
         Q.pop();
         for(int i=;i<n;i++){
             if(s&(<<i)){
                 int tmp=s-(<<i);
                 if(!book[tmp]){
                     book[tmp]=;
                     Q.push(tmp);
                     if(num[tmp]){
                         tmp=(<<n)--tmp;
                         if(!book[tmp]){
                             book[tmp]=;
                             Q.push(tmp);
                         }
                     }
                 }
             }
         }
     }
 }

 int main(){
     std::ios::sync_with_stdio(false);
     cin>>n>>m;
     for(int i=;i<=m;i++){
         cin>>a[i];
         num[a[i]]=;
     }
     int ans=;
     for(int i=;i<=m;i++){
         if(!book[a[i]]){
             ans++;
             book[a[i]]=;
             int tmp=(<<n)--a[i];
             bfs(tmp);
         }
     }
     cout<<ans<<endl;
 }