Codeforces Round #411 (Div. 2)-优快云博客

本文提供了CodeForces 805比赛的A-F题详细解答过程及代码实现，涵盖贪心算法、字符串处理、图论、树结构等核心算法知识点。

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拿游子小号和大家打了一场codeforces，感觉自己被肝爆了qwqqq，wxl求轻虐...qwq

貌似是让求区间[l,r]中出现次数最多的约数...那就是2呗...当然l==r时是l

//codeforces 805A
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
int n,l,r;
int main()
{
 int i,j;
 scanf("%d%d",&l,&r);
 if(l==r) printf("%d\n",l);
 else printf("2\n");
 return 0;
}

构造一个ab序列，不能出现长度为3的回文

那就aabbaabb无限循环...

//codeforces 805B
#include<cstdio>
#include<cstring>//codeforces 805D
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mod=1e9+7;
const int mxn=1000005;
ll n,num,cnt,res,ans;
ll pw[mxn];
char s[mxn];
int main()
{
 int i,j;
 scanf("%s",s+1);
 n=strlen(s+1);
 pw[0]=1;
 fo(i,1,n) pw[i]=(pw[i-1]*2)%mod;
 fo(i,1,n)
 {
  if(s[i]=='a') cnt++;
  else if(s[i]=='b' && (s[i+1]=='a' || i==n))
  {
   res++;
   ans=(ans+res*(pw[cnt]-1)%mod)%mod;
   res=0;
  }
  else res++;
 }
 printf("%lld\n",ans);
 return 0;
}

贪心的来吧...1->n->2->n-1->3->n-2...

大力观察得出这样是最优解→_→



//codeforces 805C
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
int n;
int main()
{
 int i,j;
 scanf("%d",&n);
 if(n&1) printf("%d\n",n/2);
 else printf("%d\n",n/2-1);
 return 0;
}

给一个ab序列，如果出现ab就变成bba，问变几次就不需要再变了...

大力观察么...

//codeforces 805D
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mod=1e9+7;
const int mxn=1000005;
ll n,num,cnt,res,ans;
ll pw[mxn];
char s[mxn];
int main()
{
 int i,j;
 scanf("%s",s+1);
 n=strlen(s+1);
 pw[0]=1;
 fo(i,1,n) pw[i]=(pw[i-1]*2)%mod;
 fo(i,1,n)
 {
  if(s[i]=='a') cnt++;
  else if(s[i]=='b' && (s[i+1]=='a' || i==n))
  {
   res++;
   ans=(ans+res*(pw[cnt]-1)%mod)%mod;
   res=0;
  }
  else res++;
 }
 printf("%lld\n",ans);
 return 0;
}

这个题...观察到联通块的性质，所以暴力dfs+贪心染色就好啦

//codeforces 805E. Ice cream coloring
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=300005;
bool bo[mxn],wxl[mxn];
int n,m,cnt,tot,ans=1;
vector <int> e[mxn];
int head[mxn],color[mxn];
struct node {int to,next;} f[mxn<<1];

inline void add(int u,int v)
{
 f[++cnt].to=v,f[cnt].next=head[u],head[u]=cnt;
}

inline void dfs(int u,int fa)
{
 int tmp=1,v;
 for(int i=0;i<e[u].size();i++)
 {
  v=e[u][i];
  bo[color[v]]=1;
 }
 for(int i=0;i<e[u].size();i++)
 {
  v=e[u][i];
  while(bo[tmp]) tmp++;
  if(!color[v]) color[v]=tmp++;
 }
 for(int i=0;i<e[u].size();i++)
 {
  v=e[u][i];
  bo[color[v]]=0;
 }
 for(int i=head[u];i;i=f[i].next)
 {
  v=f[i].to;
  if(v==fa) continue;
  dfs(v,u);
 }
}

int main()
{
 int i,j,x,u,v;
 scanf("%d%d",&n,&m);
 fo(i,1,n)
 {
  scanf("%d",&x);
  while(x--)
  {
   scanf("%d",&v);
   e[i].push_back(v);
  }
 }
 fo(i,2,n)
 {
  scanf("%d%d",&u,&v);
  add(u,v),add(v,u);
 }
 dfs(1,0);
 fo(i,1,m) ans=max(ans,color[i]);
 printf("%d\n",ans);
 fo(i,1,m) printf("%d ",max(color[i],1));
 printf("\n");
 return 0;
}

调了一上午嗯嗯嗯

这题超讨厌...树的题完全不会...真的好难啊qwqqq

做出来还是不会证明复杂度qwqqq

总的来说：map大法好，vector把你保

//codeforces 805F
#include<map>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=100005;
double x,ans;
vector <int> e[mxn],res[mxn];
map < pair<int,int>,double > mp;
bool vis[mxn],dia[mxn];
int n,m,q,fff,cnt,tot,mx;
int head[mxn],p[mxn][3],len[mxn],last[mxn];
int father[mxn],pre[mxn],dist[mxn],size[mxn];
struct node {int to,next,from;} f[mxn<<1];
inline void add(int u,int v)
{
 f[++cnt].to=v,f[cnt].next=head[u],head[u]=cnt;
}
inline int find(int x)
{
 if(x==father[x]) return father[x];
 return father[x]=find(father[x]);
}
inline void dfs1(int u,int lenth)
{
 vis[u]=1,size[fff]++;
 if(lenth>=mx) mx=lenth,p[fff][1]=u;
 for(int i=head[u];i;i=f[i].next)
 {
  int v=f[i].to;
  if(!vis[v]) dfs1(v,lenth+1);
 }
}
inline void dfs2(int u,int lenth)
{
 vis[u]=1;
 if(lenth>=len[fff]) p[fff][2]=u;
 len[fff]=max(len[fff],lenth);
 for(int i=head[u];i;i=f[i].next)
 {
  int v=f[i].to;
  if(!vis[v]) dfs2(v,lenth+1);
 }
}
inline void dfs3(int u)  //标记最长链
{
 vis[u]=1;
 if(u==p[fff][2])
 {
  dia[u]=1,dist[u]=len[fff];
  int num=0;
  while(pre[u])
    num++,u=pre[u],dia[u]=1,dist[u]=max(num,len[fff]-num);
  return;
 }
 for(int i=head[u];i;i=f[i].next)
 {
  int v=f[i].to;
  if(v==pre[u]) continue;
  pre[v]=u;
  dfs3(v);
 }
}
inline void dfs4(int u,int lenth)
{
 vis[u]=1,dist[u]=lenth;
 e[fff].push_back(lenth);
 for(int i=head[u];i;i=f[i].next)
 {
  int v=f[i].to;
  if(dia[v] || vis[v]) continue;
  dfs4(v,lenth+1);
 }
}
inline double solve(int t1,int t2)
{
 int i,j,l,r;
 ll sum=0,tmp=max(len[t1],len[t2]);
 for(l=0;l<(int)e[t1].size();l++)
 {
  r=upper_bound(e[t2].begin(),e[t2].end(),tmp-e[t1][l]-1)-e[t2].begin();
  sum+=r*tmp+res[t2][size[t2]-1]+(size[t2]-r)*(e[t1][l]+1);
  if(r) sum-=res[t2][r-1];
 }
 return (double)sum/((double)size[t1]*size[t2]);
}
int main()
{
 int i,j,u,v,t1,t2,z;
 scanf("%d%d%d",&n,&m,&q);
 fo(i,1,n) father[i]=i;
 fo(i,1,m)
 {
  scanf("%d%d",&u,&v);
  add(u,v),add(v,u);
  int t1=find(u),t2=find(v);
  if(t1!=t2) father[t1]=t2;
 }
 fo(i,1,n) if(!vis[i])
   fff=find(i),mx=0,dfs1(i,0);
 memset(vis,0,sizeof vis);
 fo(i,1,n) if(!vis[i]) fff=find(i),dfs2(p[fff][1],0);
 memset(vis,0,sizeof vis);
 fo(i,1,n) if(!vis[i]) fff=find(i),dfs3(p[fff][1]);
 memset(vis,0,sizeof vis);
 fo(i,1,n) if(!vis[i] && dia[i]) fff=find(i),dfs4(i,dist[i]);
 fo(fff,1,n) if(father[fff]==fff)
 {
  sort(e[fff].begin(),e[fff].end());
  res[fff].push_back(e[fff][0]);
  for(j=1;j<e[fff].size();j++)
    res[fff].push_back(res[fff][j-1]+e[fff][j]);
 }
 while(q--)
 {
  scanf("%d%d",&u,&v);
  int t1=find(u),t2=find(v);
  if(size[t2]<size[t1]) swap(t1,t2);
  if(t1==t2)
  {
   printf("-1\n");
   continue;
  }
  if(mp.count(make_pair(t1,t2)))
  {
   printf("%.10lf\n",mp[make_pair(t1,t2)]);
   continue;
  }
  ans=mp[make_pair(t1,t2)]=solve(t1,t2);
  printf("%.10lf\n",ans);
 }
 return 0;
}
/*
5 2 1
2 4
4 3
2 5
*/