Codeforces Beta Round #29 (Div. 2, Codeforces format)

Posted 2021-02-09 fighting-sh

Codeforces Beta Round #29 (Div. 2, Codeforces format)

http://codeforces.com/contest/29

A

 1 #include<bits/stdc++.h>
 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 pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 const ull MOD=257;
11 /*#ifndef ONLINE_JUDGE
12         freopen("1.txt","r",stdin);
13 #endif */
14 
15 
16 int main(){
17     #ifndef ONLINE_JUDGE
18       //  freopen("1.txt","r",stdin);
19     #endif
20     std::ios::sync_with_stdio(false);
21     int n;
22     cin>>n;
23     int a[105],b[105];
24     for(int i=1;i<=n;i++){
25         cin>>a[i]>>b[i];
26     }
27     int flag=0;
28     for(int i=1;i<=n;i++){
29         for(int j=1;j<=n;j++){
30             if(i!=j){
31                 if(a[i]+b[i]==a[j]&&a[j]+b[j]==a[i]){
32                     flag=1;
33                 }
34             }
35         }
36     }
37     if(flag) cout<<"YES"<<endl;
38     else cout<<"NO"<<endl;
39 }

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B

模拟

 1 #include<bits/stdc++.h>
 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 pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 const ull MOD=257;
11 /*#ifndef ONLINE_JUDGE
12         freopen("1.txt","r",stdin);
13 #endif */
14 
15 
16 int main(){
17     #ifndef ONLINE_JUDGE
18       //  freopen("1.txt","r",stdin);
19     #endif
20     std::ios::sync_with_stdio(false);
21     double l,d,v,g,r;
22     cin>>l>>d>>v>>g>>r;
23     double ans=0;
24     if(v*g>d) ans=l/v;
25     else{
26         l-=d;
27         double t=d/v;
28         ans=t;
29         double tt=g+r;
30         int flag=0;
31         while(t>=0){
32             t-=g;
33             if(t>=0){
34                 t-=r;
35             }
36             if(t<0) flag=1;
37         }
38       //  cout<<t<<endl;
39         if(flag==1){
40             ans-=t;
41         }
42         ans+=l/v;
43     }
44     printf("%.7f
",ans);
45 }

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C

找出一个度为1的点，然后跑dfs。因为值为1e9，所以需要离散化

 1 #include<bits/stdc++.h>
 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 pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 const ull MOD=257;
11 /*#ifndef ONLINE_JUDGE
12         freopen("1.txt","r",stdin);
13 #endif */
14 
15 int n;
16 vector<int>ve;
17 int a[100005];
18 int b[100005];
19 int d[100005];
20 vector<int>V[100005];
21 
22 int getid(int x){
23     return lower_bound(ve.begin(),ve.end(),x)-ve.begin()+1;
24 }
25 
26 void dfs(int pos,int pre){
27     cout<<ve[pos-1]<<" ";
28     for(int i=0;i<V[pos].size();i++){
29         if(V[pos][i]!=pre){
30             dfs(V[pos][i],pos);
31         }
32     }
33 }
34 
35 int main(){
36     #ifndef ONLINE_JUDGE
37       //  freopen("1.txt","r",stdin);
38     #endif
39     std::ios::sync_with_stdio(false);
40     cin>>n;
41     for(int i=1;i<=n;i++){
42         cin>>a[i]>>b[i];
43         ve.pb(a[i]);
44         ve.pb(b[i]);
45     }
46     sort(ve.begin(),ve.end());
47     ve.erase(unique(ve.begin(),ve.end()),ve.end());
48     int pos,pos1,pos2;
49     for(int i=1;i<=n;i++){
50         pos1=getid(a[i]);
51         pos2=getid(b[i]);
52         d[pos1]++,d[pos2]++;
53         V[pos1].pb(pos2);
54         V[pos2].pb(pos1);
55     }
56     for(int i=1;i<=n;i++){
57         pos1=getid(a[i]);
58         pos2=getid(b[i]);
59         if(d[pos1]==1){
60             pos=pos1;
61             break;
62         }
63         if(d[pos2]==1){
64             pos=pos2;
65             break;
66         }
67     }
68     dfs(pos,0);
69 }

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D

跑两个叶子结点的最短路，看看最后的个数是不是2*n-1

 1 #include<bits/stdc++.h>
 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 pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 const ull MOD=257;
11 /*#ifndef ONLINE_JUDGE
12         freopen("1.txt","r",stdin);
13 #endif */
14 
15 int n;
16 vector<int>ve[305];
17 int vis[305];
18 int d[305];
19 vector<int>ans;
20 
21 
22 void dfs(int pos,int pre,int last,vector<int>tmp){
23     vis[pos]=1;
24     if(pre!=0) {
25         tmp.pb(pos);
26     }
27     if(pos==last){
28         for(int i=0;i<tmp.size();i++){
29             ans.pb(tmp[i]);
30         }
31         return;
32     }
33     for(int i=0;i<ve[pos].size();i++){
34         if(!vis[ve[pos][i]]&&ve[pos][i]!=pre){
35             dfs(ve[pos][i],pos,last,tmp);
36         }
37     }
38 }
39 
40 int main(){
41     #ifndef ONLINE_JUDGE
42         freopen("1.txt","r",stdin);
43     #endif
44     std::ios::sync_with_stdio(false);
45     cin>>n;
46     int u,v;
47     for(int i=1;i<n;i++){
48         cin>>u>>v;
49         ve[u].pb(v);
50         ve[v].pb(u);
51         d[u]++,d[v]++;
52     }
53     int co=0;
54     for(int i=2;i<=n;i++){
55         if(d[i]==1){
56             co++;
57         }
58     }
59     vector<int>son;
60     son.pb(1);
61     for(int i=1;i<=co;i++){
62         cin>>u;
63         son.pb(u);
64     }
65     son.pb(1);
66     ans.pb(1);
67     for(int i=0;i<son.size()-1;i++){
68         memset(vis,0,sizeof(vis));
69         vector<int>tmp;
70         tmp.clear();
71         dfs(son[i],0,son[i+1],tmp);
72     }
73     if(ans.size()==2*n-1){
74         for(int i=0;i<ans.size();i++){
75             cout<<ans[i]<<" ";
76         }
77         cout<<endl;
78     }
79     else{
80         cout<<-1<<endl;
81     }
82 }

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E

bfs搜索，思路很有趣

标记是用A和B的相对位置进行标记的，懂了这一点这题就解决了

 1 #include<bits/stdc++.h>
 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 pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 /*#ifndef ONLINE_JUDGE
11         freopen("1.txt","r",stdin);
12 #endif */
13 int n,m;
14 vector<int>ve[505];
15 int pre[505][505][2];
16 bool book[505][505];
17 vector<int>ans[2];
18 
19 bool bfs(){
20     pre[1][n][0]=-1;
21     pair<int,int>p,pp;
22     p=make_pair(1,n);
23     queue<pair<int,int> >Q;
24     Q.push(p);
25     while(!Q.empty()&&pre[n][1][0]==0){
26         p=Q.front();
27         Q.pop();
28         int p1=p.first,p2=p.second;
29         for(int i=0;i<ve[p1].size();i++){
30             if(!book[ve[p1][i]][p2]){
31                 book[ve[p1][i]][p2]=1;
32                 for(int j=0;j<ve[p2].size();j++){
33                     if(ve[p1][i]!=ve[p2][j]){
34                         if(pre[ve[p1][i]][ve[p2][j]][0]==0){
35                             pre[ve[p1][i]][ve[p2][j]][0]=p1;
36                             pre[ve[p1][i]][ve[p2][j]][1]=p2;
37                             pp=make_pair(ve[p1][i],ve[p2][j]);
38                             Q.push(pp);
39                         }
40                     }
41                 }
42             }
43         }
44     }
45     if(pre[n][1][0]==0) return false;
46     int x=n,y=1,z;
47     while(x>0){
48         ans[0].pb(x);
49         ans[1].pb(y);
50         z=x;
51         x=pre[x][y][0];
52         y=pre[z][y][1];
53     }
54     return true;
55 }
56 
57 int main(){
58     #ifndef ONLINE_JUDGE
59         freopen("1.txt","r",stdin);
60     #endif
61     std::ios::sync_with_stdio(false);
62     cin>>n>>m;
63     int u,v;
64     for(int i=1;i<=m;i++){
65         cin>>u>>v;
66         ve[u].pb(v);
67         ve[v].pb(u);
68     }
69     if(bfs()){
70         cout<<ans[0].size()-1<<endl;
71         for(int i=0;i<2;i++){
72             for(int j=ans[i].size()-1;j>=0;j--){
73                 cout<<ans[i][j]<<" ";
74             }
75             cout<<endl;
76         }
77     }
78     else{
79         cout<<-1<<endl;
80     }
81 
82 }

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