A. Andryusha and Socks 思路：模拟即可
#include<bits/stdc++.h>using namespace std;const int maxn=1e5+10;bool vis[maxn];int main(){ int n; scanf("%d",&n); memset(vis,false,sizeof(vis)); int now=0,maxx=0; for(int i=0;i<2*n;i++){ int temp; scanf("%d",&temp); if(!vis[temp]){ now+=1; vis[temp]=true; }else{ now-=1; } if(now>maxx) maxx=now; } PRintf("%d",maxx);}B. The Meeting Place Cannot Be Changed 思路：二分搜索，如果时间t可以满足所有人到达同一个点，则检查t/2是否也可以，然后根据题目要求精度达到1e-7退出
#include<bits/stdc++.h>using namespace std;const int maxn=60000+10;struct p{ int dis,v;};p per[maxn];int n;int md,sd;double s;double eps=1e-7;bool check(double t){ double a,b; for(int i=1;i<=n;i++){ double x=(double)per[i].dis-(double)per[i].v*t; double y=(double)per[i].dis+(double)per[i].v*t; if(i==1){ a=x,b=y; }else{ if(a>y || b<x) return false; if(a<=x) a=x; if(b>=y) b=y; } } return true;}int main(){ scanf("%d",&n); md=-1,sd=0x3f3f3f3f; for(int i=1;i<=n;i++){ scanf("%d",&per[i].dis); if(per[i].dis>md) md=per[i].dis; if(per[i].dis<sd) sd=per[i].dis; } for(int i=1;i<=n;i++){ scanf("%d",&per[i].v); } double l,r,ans; l=0,r=1e9; while(r-l>=eps){ double mid=(l+r)/2; if(check(mid)){ ans=mid; r=mid; }else l=mid; } printf("%.7f",ans);}C. Andryusha and Colored Balloons 思路：树上dfs，每个节点的颜色是不和父亲节点和祖父亲节点相同的最小的未使用颜色。
#include<bits/stdc++.h>using namespace std;const int maxn=2*1e5+10;vector<int>E[maxn];int n;int ans[maxn];int fa[maxn];set<int>sum;void dfs(int now){ int cnt=1; for(int i=0;i<E[now].size();i++){ int p=E[now][i]; if(p!=fa[now]){ while(cnt==ans[now] || cnt==ans[fa[now]]) cnt++; ans[p]=cnt; sum.insert(cnt); fa[p]=now; dfs(p); cnt++; } }}int main(){ scanf("%d",&n); for(int i=1;i<n;i++){ int a,b; scanf("%d%d",&a,&b); E[a].push_back(b); E[b].push_back(a); } ans[1]=1; fa[1]=1; dfs(1); sum.insert(1); printf("%d/n",sum.size()); for(int i=1;i<=n;i++) printf("%d ",ans[i]);}新闻热点
疑难解答
