Codeforces Round 1023 (Div. 2)

Maximum Subarray Sum

将所有的0位置填-1e18,就不用分段求最大值m
k<m，不行，怎么加都可以取当前的一段达到m
k==m，就这样就可以
k>m,可以这样构造
找到最左边的0,前缀和后缀最大加上他是k即可，别的再怎么加也达不到k

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define pii pair<int,int>
#define ll long long
#define pb push_back
#define ft first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define int long longint n,k;
const int N=200010;
int a[N];int f[N];int g[N];
void solve(){
cin>>n>>k;
string b;cin>>b;b="a"+b;for(int i=1;i<=n;i++)cin>>a[i];for(int i=1;i<=n;i++)if(b[i]=='0')a[i]=-1e17;int m=0;
for(int i=1;i<=n;i++){f[i]=max(f[i-1]+a[i],a[i]);m=max(m,f[i]);  //cout<<f[i]<<" ";
}
if(m>k){no;
}else if(m==k){yes;for(int i=1;i<=n;i++)cout<<a[i]<<" ";cout<<'\n';
}else {int st=1;while(st<=n&&b[st]=='1')st++;if(st>n){no;return ;}g[n+1]=0;for(int i=n;i>=1;i--){g[i]=max(g[i+1]+a[i],a[i]);}a[st]=k-max(0ll,f[st-1])-max(0ll,g[st+1]);
yes;
for(int i=1;i<=n;i++)cout<<a[i]<<" ";cout<<'\n';
}}
signed main(){std::ios::sync_with_stdio(false);int T=1;cin>>T;while(T--){solve();}
}

D. Apple Tree Traversing

答案是每次找端点最大的直径，
每次找未访问过的连通块直径,标记这个直径上的点为访问过
O(n根号n)类似长链剖分

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define pii pair<int,int>
#define ll long long
#define pb push_back
#define ft first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3fconst int N=150010;
vector<int> G[N];int n;
int mx=0;
int dep[N];int pre[N];bool vis[N];
void dfs1(int u,int fa){// cout<<"dfs1:"<<u<<'\n';dep[u]=dep[fa]+1;if(dep[u]>dep[mx]){mx=u;}else if(dep[u]==dep[mx]){if(u>mx)mx=u;}for(auto v:G[u]){if(vis[v])continue;if(v==fa)continue;dep[v]=dep[u]+1;dfs1(v,u);}
}
void dfs2(int u,int fa){pre[u]=fa;dep[u]=dep[fa]+1;if(dep[u]>dep[mx]){mx=u;
}else if(dep[u]==dep[mx]){if(u>mx)mx=u;
}for(auto v:G[u]){if(vis[v])continue;if(v==fa)continue;dfs2(v,u);}
}void solve(){
cin>>n;for(int i=1;i<=n;i++){vis[i]=0;G[i].clear();
}
for(int i=1;i<=n-1;i++){int u,v;cin>>u>>v;G[u].pb(v);G[v].pb(u);
}vector<pair<int,pii>> ans;
while(1){int cur=-1;for(int i=1;i<=n;i++){if(!vis[i]){cur=i;break;}}if(cur==-1)break;mx=0;dfs1(cur,0);//cout<<"fin1"<<'\n';int vl=mx;mx=0;dfs2(vl,0);//cout<<"fin2"<<'\n';int vr=mx;int t=vr;int c=0;while(t){vis[t]=1;c++;t=pre[t];}if(vl<vr)swap(vl,vr);ans.pb({c,{vl,vr}});
}
sort(ans.begin(),ans.end());reverse(ans.begin(),ans.end());
for(auto tem:ans){
cout<<tem.ft<<" "<<tem.se.ft<<" "<<tem.se.se<<' ';
}
}
signed main(){std::ios::sync_with_stdio(false);int T=1;cin>>T;while(T--){solve();}
}