Codeforces Beta Round #14 (Div. 2)

A

 找最大最小的行列值即可

1 #include
      
        2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9         freopen("1.txt","r",stdin);10 #endif */11 12 string str[55];13 14 int main(){15     #ifndef ONLINE_JUDGE16         freopen("1.txt","r",stdin);17     #endif18     std::ios::sync_with_stdio(false);19     int n,m;20     cin>>n>>m;21     int hangmin=105,hangmax=-1,liemin=105,liemax=-1;22     for(int i=0;i
       
        >str[i];23     for(int i=0;i
       
      

 

B

水题

1 #include
      
        2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9         freopen("1.txt","r",stdin);10 #endif */11 12 int n,x;13 int a[1005];14 15 int main(){16     #ifndef ONLINE_JUDGE17       //  freopen("1.txt","r",stdin);18     #endif19     std::ios::sync_with_stdio(false);20     cin>>n>>x;21     int u,v;22     for(int i=1;i<=n;i++){23         cin>>u>>v;24         if(u>v) swap(u,v);25         a[u]++,a[v+1]--;26     }27     int ans=0x3f3f3f3f;28     int num=0;29     for(int i=0;i<=1001;i++){30         num+=a[i];31         if(num==n){32             ans=min(ans,abs(i-x));33         }34     }35     if(ans==0x3f3f3f3f) cout<<-1<
      

 

C

判断四条线段是否可以构成与坐标轴平行或垂直的正方形，感觉数据有点坑

1 #include
      
        2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9         freopen("1.txt","r",stdin);10 #endif */11 12 map
       
        
         ,int>mp;13 map
         
          
           ,int>::iterator it;14 map< pair< pair
           
            ,pair
            
              >,int >book;15 16 int main(){17 #ifndef ONLINE_JUDGE18 freopen("1.txt","r",stdin);19 #endif20 std::ios::sync_with_stdio(false);21 int a,b,c,d;22 int flag=0;23 for(int i=0;i<4;i++){24 cin>>a>>b>>c>>d;25 mp[make_pair(a,b)]++;26 mp[make_pair(c,d)]++;27 if(a==c&&b==d) flag=1;28 vector
             
              
                >v;29 v.push_back(make_pair(a,b));30 v.push_back(make_pair(c,d));31 sort(v.begin(),v.end());32 if(book[make_pair(v[0],v[1])]==0){33 book[make_pair(v[0],v[1])]=1;34 }35 else{36 flag=1;37 }38 // cout<
               
                <<" "<
                
                 <<" "<
                 
                  <<" "<
                  
                   <
                   
                    second!=2){42 flag=1;43 }44 }45 if(flag){46 cout<<"NO"<
                    
                      >ve;50 for(it=mp.begin();it!=mp.end();it++){51 ve.push_back(it->first);52 }53 sort(ve.begin(),ve.end());54 if(ve[0].first==ve[1].first&&ve[2].first==ve[3].first&&55 ve[0].second==ve[2].second&&ve[1].second==ve[3].second){56 cout<<"YES"<
                    
                   
                  
                 
                
               
              
             
            
           
          
         
        
       
      

 

D

枚举删边，然后跑dfs

1 #include
      
        2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9         freopen("1.txt","r",stdin);10 #endif */11 12 int n;13 vector
       
        ve[205];14 vector
        
         
           >d;15 16 int r;17 18 int dfs(int pos,int pre){19     int len1=0,len2=0;20     int tmp=0;21     for(int i=0;i
          
           len1){len2=len1,len1=r;}25             else{len2=max(len2,r);}26         }27     }28     tmp=max(tmp,len1+len2);29     r=len1+1;30     return tmp;31 }32 33 int main(){34     #ifndef ONLINE_JUDGE35         freopen("1.txt","r",stdin);36     #endif37     cin>>n;38     int a,b;39     for(int i=1;i
           
            >a>>b;41 ve[a].push_back(b);42 ve[b].push_back(a);43 d.push_back(make_pair(a,b));44 }45 int ans=0;46 for(int i=0;i
           
          
         
        
       
      

 

E

DP,细节在代码里

1 #include
      
        2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define maxn 500005 7 typedef long long ll; 8 /*#ifndef ONLINE_JUDGE 9         freopen("1.txt","r",stdin);10 #endif */11 12 ll dp[25][15][5][2];///分别表示第i个数，第j个尖峰数，尖峰高度和上升（1）或下降（0）13 14 int main(){15     #ifndef ONLINE_JUDGE16        // freopen("1.txt","r",stdin);17     #endif18     int n,m;19     for(int i=1;i<5;i++) dp[1][1][i][1]=1;20     for(int i=2;i<=20;i++){21         for(int j=1;j<=10;j++){22             for(int k=1;k<=4;k++){23                 for(int l=1;l
       
        >n>>m;39     ll ans=0;40     for(int i=1;i<=4;i++){41         ans+=dp[n][m][i][0];42     }43     cout<
        
         <