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Codeforces Round 411 Div.2 题解

A    Fake NPstandard input/output1 s, 256 MB     Submit  Add to favourites     x3673B    3-palindromestandard input/output1 s, 256 MB     Submit  Add to favourites     x3760C    Find Amirstandard input/output1 s, 256 MB     Submit  Add to favourites     x3503D    Minimum number of stepsstandard input/output1 s, 256 MB     Submit  Add to favourites     x2242E    Ice cream coloringstandard input/output2 s, 256 MB     Submit  Add to favourites     x142F    Expected diameter of a treestandard input/output3 s, 256 MB     Submit  Add to favourites     x28

考的时候写出来了前4道...    ORZ yzy Rank7

A题

考的时候SB 了   r-l<=100的时候搞了个暴力判了判..

应该是这样的..

//By SiriusRen#include <map>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;map<int,int>mp,remp;int l,r;int main(){	scanf("%d%d",&l,&r);	if(l==r){		printf("%d\n",l);	}	else{		puts("2");	}}

  

B  构造   aabbaabb...

//By SiriusRen#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n,a[200050];int main(){	scanf("%d",&n);	a[1]=0,a[2]=1;	for(int i=3;i<=n;i++){		if(a[i-2]==0)a[i]=1;		else a[i]=0;	}	for(int i=1;i<=n;i++)printf("%c",a[i]+‘a‘);}

C  贪心 1->n->2->n-1.....

判一下奇偶

//By SiriusRen#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n;int main(){	scanf("%d",&n);	printf("%d\n",n/2-1+(n%2)); }

  

D  乱搞

//By SiriusRen#include <cstdio>#include <cstring>using namespace std;const int N=1000050,p=1000000007;char s[N];int n,ans,res;int main(){	scanf("%s",s+1),n=strlen(s+1);	for(int i=n;i;i--)		if(s[i]==‘b‘)res=(res+1)%p;		else (ans+=res)%=p,(res+=res)%=p;	printf("%d\n",ans);}

  

E  max(s[i]) 贪心染色  别问我为什么

//By SiriusRen#include<bits/stdc++.h>#define N 300005using namespace std;vector<int> G[N],E[N];int n,m,ans,col[N],cnt,vis[N];void dfs(int t,int fa){	int i,cnt=1;	for(i=0;i<E[t].size();i++)		if(col[E[t][i]]) vis[col[E[t][i]]]=t;	for(i=0;i<E[t].size();i++)		if(!col[E[t][i]]){			while(vis[cnt]==t) cnt++;			vis[cnt]=t,col[E[t][i]]=cnt;		}	for(i=0;i<G[t].size();i++)		if(G[t][i]!=fa)dfs(G[t][i],t);}int main(){	int i,x,y;	scanf("%d %d",&n,&m);	for(i=1;i<=n;i++){		scanf("%d",&x),ans=max(ans,x);		while(x--)scanf("%d",&y),E[i].push_back(y);	}	for(i=1;i<n;i++)scanf("%d%d",&x,&y),G[x].push_back(y),G[y].push_back(x);	dfs(1,0);	printf("%d\n",max(ans,1));	for(i=1;i<=m;i++)printf("%d ",max(col[i],1));	return 0;}

  

保存每个点能走的最远距离 (可以O(1)求)

搞个前缀和之类的东西维护一下

每回 O(小块)的复杂度  再开个set记录一下答案 就(卡)过去了..

//By SiriusRen#include <set>#include <vector>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N=200050;vector<int>vec[N],dis[N],sum[N];struct Node{int x,y;double ans;}JY;bool operator<(Node a,Node b){if(a.x!=b.x)return a.x<b.x;return a.y<b.y;}set<Node>s;set<Node>::iterator it;int n,m,q,first[N],nxt[N],v[N],w[N],tot,root,xx,yy,vis[N];int maxx[N][2],rec[N][2],f[N],Root[N],vv[N],fi,flag,size[N];void add(int x,int y){w[tot]=1,v[tot]=y,nxt[tot]=first[x],first[x]=tot++;}int find(int x){return  x==f[x]?x:f[x]=find(f[x]);}void get_dia(int x,int fa){	rec[x][0]=rec[x][1]=x;	for(int i=first[x];~i;i=nxt[i])if(v[i]!=fa){		get_dia(v[i],x);		if(maxx[v[i]][0]+w[i]>maxx[x][0])			maxx[x][1]=maxx[x][0],rec[x][1]=rec[x][0],			maxx[x][0]=maxx[v[i]][0]+w[i],rec[x][0]=rec[v[i]][0];		else if(maxx[v[i]][0]+w[i]>maxx[x][1])			maxx[x][1]=maxx[v[i]][0]+w[i],rec[x][1]=rec[v[i]][0];	}if(maxx[x][0]+maxx[x][1]>maxx[root][0]+maxx[root][1])root=x;}void DFS(int x,int fa,int deep){	vv[x]=max(vv[x],deep);	if(flag)vec[fi].push_back(vv[x]),size[fi]++;	for(int i=first[x];~i;i=nxt[i])if(v[i]!=fa){		DFS(v[i],x,deep+w[i]);	}}int main(){	memset(first,-1,sizeof(first));	scanf("%d%d%d",&n,&m,&q);	for(int i=1;i<=n;i++)f[i]=i;	for(int i=1;i<=m;i++)		scanf("%d%d",&xx,&yy),add(xx,yy),add(yy,xx),f[find(xx)]=find(yy);	for(int i=1;i<=n;i++){		fi=find(i);		if(!rec[fi][0]){			root=flag=0,get_dia(i,0),Root[fi]=root;			DFS(rec[root][0],0,0),flag=1,DFS(rec[root][1],0,0);		}	}	for(int i=1;i<=n;i++){		fi=find(i);		if(!vis[fi]){			vis[fi]=1;			dis[fi].resize(size[fi]+1),sum[fi].resize(size[fi]+1);			for(int j=0;j<vec[fi].size();j++)				dis[fi][vec[fi][j]]++,sum[fi][vec[fi][j]]+=vec[fi][j];			for(int j=size[fi]-1;j>=0;j--)				dis[fi][j]+=dis[fi][j+1],sum[fi][j]+=sum[fi][j+1];		}	}	while(q--){		scanf("%d%d",&xx,&yy);		int fx=find(xx),fy=find(yy);		if(fx==fy){puts("-1");continue;}		if(vec[fx].size()>vec[fy].size())swap(xx,yy),swap(fx,fy);		JY.x=fx,JY.y=fy;		if((it=s.find(JY))!=s.end()){printf("%.10lf\n",it->ans);continue;}		int dis1=maxx[Root[fx]][0]+maxx[Root[fx]][1];		int dis2=maxx[Root[fy]][0]+maxx[Root[fy]][1];		int Dis=max(dis1,dis2);		long long ans=0;		for(int i=0;i<vec[fx].size();i++){			ans+=sum[fy][Dis-vec[fx][i]]+1ll*dis[fy][Dis-vec[fx][i]]*(vec[fx][i]+1);			ans+=1ll*Dis*(size[fy]-dis[fy][Dis-vec[fx][i]]);		}		JY.ans=(double)ans/size[fx]/size[fy],s.insert(JY);		printf("%.10lf\n",(double)ans/size[fx]/size[fy]);	}//	for(int i=1;i<=n;i++)printf("%d ",vv[i]);}

  

 

Codeforces Round 411 Div.2 题解