T1.http://ezoj.org.cn/problem/382
At first glance is a single-source shortest, but built directly FIG complexity is m ^ 2, obviously does not meet the requirements.
Consider a new set for each virtual point within the set of all points connected to a bidirectional edge (the right but only one value), the complexity O (nlgm), the score 100 is desirable.
//两岸 #include<cstdio> #include<algorithm> #include<queue> #include<cstring> #define ll long long #define FN "way" #define inf 824602269000 using namespace std; inline void Open() { freopen(FN".in","r",stdin); freopen(FN".out","w",stdout); } inline ll read() { bool f=1;ll s=0;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=0;ch=getchar();} while(ch>='0'&&ch<='9')s=(s<<3)+(s<<1)+ch-'0',ch=getchar(); return f?s:-s; } priority_queue<pair <ll,ll> >q; ll b[100010],n,m,t,s,cnt,ct,head[100010],dis[100010]; bool vis[100010]; struct Node { ll to,nxt,val; }a[200010]; inline void add(ll from,ll to,ll val) { a[++cnt].to=to; a[cnt].val=val; a[cnt].nxt=head[from]; head[from]=cnt; } int main() { //Open(); n=read(),m=read();ct=n; for(int i=1;i<=m;i++) { int t=read();ct++; for(int j=1;j<=t;j++)b[j]=read(); ll x=read(); for(int j=1;j<=t;j++)add(b[j],ct,0),add(ct,b[j],x); } s=read(),t=read(); for(int i=1;i<=ct;i++){dis[i]=inf;vis[i]=0;} q.push(make_pair(0,s));dis[s]=0; while(!q.empty()) { ll x=q.top().second; q.pop(); if(vis[x])continue;vis[x]++; for(int i=head[x];i;i=a[i].nxt) { ll y=a[i].to; if(dis[y]>dis[x]+a[i].val)dis[y]=dis[x]+a[i].val,q.push(make_pair(-dis[y],y)); } } if(dis[t]==inf)dis[t]=-1; printf("%lld",dis[t]); return 0; }
zyb tell us optimize the construction drawing can be done nlgq + m, but this and we have to do it?
T2.http://ezoj.org.cn/problem/383
Not difficult to find (in fact very difficult) for all the points can constitute diameter can be divided into two sets, taking the diameter of the path between two points is a tree from either each set, so every time we maintain that this two sets can be completed within the time limit maintenance operations.
//两岸 #include<cstdio> #include<algorithm> #include<cstring> #include<vector> #define ll long long #define FN "lct" using namespace std; inline void Open() { freopen(FN".in","r",stdin); freopen(FN".out","w",stdout); } ll lg[500010],n,cnt=0,mx=0; vector<ll>s1,s2; struct Poi { ll dep,fa[100]; }d[500010]; inline ll read() { bool f=1;ll s=0;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=0;ch=getchar();} while(ch>='0'&&ch<='9')s=(s<<3)+(s<<1)+ch-'0',ch=getchar(); return f?s:-s; } void build(ll x,ll fa) { d[x].fa[0]=fa,d[x].dep=d[fa].dep+1; for(ll i=1;(1<<i)<=d[x].dep;i++)d[x].fa[i]=d[d[x].fa[i-1]].fa[i-1]; } inline ll lca(ll x,ll y) { ll ans=0; if(d[x].dep<d[y].dep)swap(x,y); while(d[x].dep>d[y].dep)ans+=1<<(lg[d[x].dep-d[y].dep]-1),x=d[x].fa[lg[d[x].dep-d[y].dep]-1]; if(x==y)return ans; for(ll i=lg[d[x].dep]-1;i>=0;i--) if(d[x].fa[i]!=d[y].fa[i])ans+=1<<(i+1),x=d[x].fa[i],y=d[y].fa[i]; return ans+2; } int main() { //Open(); n=read()+1;lg[1]=1; s1.push_back(1);d[1].fa[0]=0;d[1].dep=1; for(ll i=2;i<=n;i++) { lg[i]=lg[i-1]+(1<<lg[i-1]==i); ll x=read();build(i,x); ll lc1=s1.empty()?0:lca(s1[0],i),lc2=s2.empty()?0:lca(s2[0],i); if(max(lc1,lc2)>mx) { mx=max(lc1,lc2); if(lc1==mx){for(ll j=0;j<s2.size();j++)if(lca(s2[j],i)==mx)s1.push_back(s2[j]);s2.clear();} else {for(ll j=0;j<s1.size();j++)if(lca(s1[j],i)==mx)s2.push_back(s1[j]);s1.clear();} } if(max(lc1,lc2)==mx)(lc1==mx?s2:s1).push_back(i); printf("%d\n",s1.size()+s2.size()); } return 0; }
zyb tell us Mang lct can live on this question, but we have it and what does it matter?
T3.http://ezoj.org.cn/problem/384
No make cooing ...... ......