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 }
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 }
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 }
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 }
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 }