Codeforces Beta Round #14 (Div. 2)

Codeforces Beta Round #14 (Div. 2)

http://codeforces.com/contest/14

A

找最大最小的行列值即可

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 500005
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 string str[];

 int main(){
     #ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     int n,m;
     cin>>n>>m;
     int hangmin=,hangmax=-,liemin=,liemax=-;
     for(int i=;i<n;i++) cin>>str[i];
     for(int i=;i<n;i++){
         for(int j=;j<m;j++){
             if(str[i][j]=='*'){
                 hangmin=min(hangmin,i);
             }
         }
     }
     for(int i=;i<n;i++){
         for(int j=;j<m;j++){
             if(str[i][j]=='*'){
                 hangmax=max(hangmax,i);
             }
         }
     }
     for(int i=;i<n;i++){
         for(int j=;j<m;j++){
             if(str[i][j]=='*'){
                 liemin=min(liemin,j);
             }
         }
     }
     for(int i=;i<n;i++){
         for(int j=;j<m;j++){
             if(str[i][j]=='*'){
                 liemax=max(liemax,j);
             }
         }
     }
    // cout<<hangmin<<" "<<hangmax<<" "<<liemin<<" "<<liemax<<endl;
     for(int i=hangmin;i<=hangmax;i++){
         for(int j=liemin;j<=liemax;j++){
             cout<<str[i][j];
         }
         cout<<endl;
     }
 }

B

水题

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 500005
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 int n,x;
 int a[];

 int main(){
     #ifndef ONLINE_JUDGE
       //  freopen("1.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     cin>>n>>x;
     int u,v;
     for(int i=;i<=n;i++){
         cin>>u>>v;
         if(u>v) swap(u,v);
         a[u]++,a[v+]--;
     }
     int ans=0x3f3f3f3f;
     int num=;
     for(int i=;i<=;i++){
         num+=a[i];
         if(num==n){
             ans=min(ans,abs(i-x));
         }
     }
     if(ans==0x3f3f3f3f) cout<<-<<endl;
     else cout<<ans<<endl;
 }

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

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 500005
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 map<pair<int,int>,int>mp;
 map<pair<int,int>,int>::iterator it;
 map< pair< pair<int,int>,pair<int,int> >,int >book;

 int main(){
     #ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     int a,b,c,d;
     int flag=;
     for(int i=;i<;i++){
         cin>>a>>b>>c>>d;
         mp[make_pair(a,b)]++;
         mp[make_pair(c,d)]++;
         if(a==c&&b==d) flag=;
         vector<pair<int,int> >v;
         v.push_back(make_pair(a,b));
         v.push_back(make_pair(c,d));
         sort(v.begin(),v.end());
         if(book[make_pair(v[],v[])]==){
             book[make_pair(v[],v[])]=;
         }
         else{
             flag=;
         }
     //    cout<<v[0].first<<" "<<v[0].second<<" "<<v[1].first<<" "<<v[1].second<<endl;
     }
     for(it=mp.begin();it!=mp.end();it++){
         if(it->second!=){
             flag=;
         }
     }
     if(flag){
         cout<<"NO"<<endl;
     }
     else{
         vector<pair<int,int> >ve;
         for(it=mp.begin();it!=mp.end();it++){
             ve.push_back(it->first);
         }
         sort(ve.begin(),ve.end());
         if(ve[].first==ve[].first&&ve[].first==ve[].first&&
            ve[].second==ve[].second&&ve[].second==ve[].second){
             cout<<"YES"<<endl;
            }
         else cout<<"NO"<<endl;
     }
 }

 /*
 0 0 0 2
 2 0 2 2
 0 2 0 0
 2 2 2 0
 */

D

枚举删边，然后跑dfs

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 500005
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 int n;
 vector<int>ve[];
 vector<pair<int,int> >d;

 int r;

 int dfs(int pos,int pre){
     int len1=,len2=;
     int tmp=;
     for(int i=;i<ve[pos].size();i++){
         if(ve[pos][i]!=pre){
             tmp=max(tmp,dfs(ve[pos][i],pos));
             if(r>len1){len2=len1,len1=r;}
             else{len2=max(len2,r);}
         }
     }
     tmp=max(tmp,len1+len2);
     r=len1+;
     return tmp;
 }

 int main(){
     #ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
     #endif
     cin>>n;
     int a,b;
     for(int i=;i<n;i++){
         cin>>a>>b;
         ve[a].push_back(b);
         ve[b].push_back(a);
         d.push_back(make_pair(a,b));
     }
     int ans=;
     for(int i=;i<d.size();i++){
         int t1=dfs(d[i].first,d[i].second);
         int t2=dfs(d[i].second,d[i].first);
       //  cout<<t1<<" "<<t2<<" "<<d[i].first<<" "<<d[i].second<<endl;
         ans=max(ans,t1*t2);
     }
     cout<<ans<<endl;

 }

E

DP,细节在代码里

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 500005
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 ll dp[][][][];///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）

 int main(){
     #ifndef ONLINE_JUDGE
        // freopen("1.txt","r",stdin);
     #endif
     int n,m;
     for(int i=;i<;i++) dp[][][i][]=;
     for(int i=;i<=;i++){
         for(int j=;j<=;j++){
             for(int k=;k<=;k++){
                 for(int l=;l<k;l++){///更新上升的尖峰: 1：下降峰加个上升   2：上升再加一个上升
                     dp[i][j][k][]+=dp[i-][j-][l][]+dp[i-][j][l][];
                 }
                 if(i==){///2步不可能的情况
                     dp[i][j][k][]=;
                 }
                 else{
                     for(int l=k+;l<=;l++){
                         ///更新下降的尖峰(1:下降峰再加一个下降,2:上升峰加一个下降)
                         dp[i][j][k][]+=dp[i-][j][l][]+dp[i-][j][l][];
                     }
                 }
             }
         }
     }
     cin>>n>>m;
     ll ans=;
     for(int i=;i<=;i++){
         ans+=dp[n][m][i][];
     }
     cout<<ans<<endl;
 }