Codeforces Round 411 Div.2 题解

A
Fake NP
standard input/output
1 s, 256 MB     Submit  Add to favourites     x3673
B
3-palindrome
standard input/output
1 s, 256 MB     Submit  Add to favourites     x3760
C
Find Amir
standard input/output
1 s, 256 MB     Submit  Add to favourites     x3503
D
Minimum number of steps
standard input/output
1 s, 256 MB     Submit  Add to favourites     x2242
E
Ice cream coloring
standard input/output
2 s, 256 MB     Submit  Add to favourites     x142
F
Expected diameter of a tree
standard input/output
3 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 300005
using 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;
}

F

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]);
}