2018 ACM 网络选拔赛 青岛赛区

一些题目的代码被网站吞了……

Problem B. Red Black Tree

http://acm.zju.edu.cn/onlinejudge/searchProblem.do?contestId=1&titlefrom=0&authorfrom=0&sourcefrom=0&query=The+2018+ACM-ICPC+Asia+Qingdao+Regional+Contest%2C+Online

http://acm.zju.edu.cn/pcpst/index.html

http://acm.zju.edu.cn/contest-materials/qdol2018/qdol2018_problems.pdf

注意是找祖先(红色孩子节点无效)！

先按照花费对节点从大到小进行排序，依次增加一个点，这个点的花费采用从该点到所有新加入点的最近公共祖先的距离，直到遇到花费不可减少，则停止。

注意maxlen！

 #include <bits/stdc++.h>
 #define ll long long
 using namespace std;
 const int maxn=1e5+;
 const double minv=1e-;
 const ll inf=1e18;

 vector<pair<int,ll> >e[maxn];
 int r[maxn],dep[maxn<<],p[maxn<<],lef[maxn],er[],point[maxn],ind,f[maxn<<][];
 ll cost[maxn],len[maxn];
 bool vis[maxn];

 void dfs(int d,ll c,ll l,int deep)
 {
     vis[d]=;
     if (r[d]==)
         c=;
     cost[d]=c;
     len[d]=l;
     dep[++ind]=deep;
     p[ind]=d;
     lef[d]=ind;
     vector<pair<int,ll> >::iterator j;
     for (j=e[d].begin();j!=e[d].end();j++)
         if (!vis[j->first])
         {
             dfs(j->first,c+j->second,l+j->second,deep+);
             dep[++ind]=deep;
             p[ind]=d;
         }
 }

 int cmp(int a,int b)
 {
     return cost[a]>cost[b];
 }

 int main()
 {
     int t,n,m,q,u,v,d,deep,i,j,k;
     ll w,result,maxlen;
     for (i=;i<=;i++)
         er[i]=<<i;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%d%d%d",&n,&m,&q);
         for (i=;i<=n;i++)
         {
             r[i]=;
             vis[i]=;
             e[i].clear();
         }
         for (i=;i<=m;i++)
             scanf("%d",&d),r[d]=;
         for (i=;i<n;i++)
         {
             scanf("%d%d%lld",&u,&v,&w);
             e[u].push_back({v,w});
             e[v].push_back({u,w});
         }
         ind=;
         dfs(,,,);
         for (j=;j<=ind;j++)
             f[j][]=j;
         deep=(int)(log(ind)/log()+minv);
         for (i=;i<=deep;i++)
             for (j=,k=er[i-]+;j<=ind-er[i]+;j++,k++)
                 f[j][i]=(dep[f[j][i-]]<dep[f[k][i-]])?f[j][i-]:f[k][i-];
         while (q--)
         {
             scanf("%d",&m);
             for (i=;i<=m;i++)
                 scanf("%d",&point[i]);
             sort(point+,point+m+,cmp);
             if (m==)
                 result=;
             else
             {
                 result=cost[point[]];
                 d=point[];
             }
             maxlen=len[point[]];
             for (i=;i<=m;i++)
             {
                 u=lef[d];
                 v=lef[point[i]];
                 if (u>v)
                     swap(u,v);
                 deep=(int)(log(v-u+)/log()+minv);
                 d=(dep[f[u][deep]]<dep[f[v-er[deep]+][deep]])?f[u][deep]:f[v-er[deep]+][deep];
                 d=p[d];
                 maxlen=max(maxlen,len[point[i]]);
                 w=maxlen-len[d];
                 if (i==m || w>=cost[point[i+]])
                 {
                     result=min(result,w);
                     break;
                 }
                 result=min(result,cost[point[i+]]);
             }
             printf("%lld\n",result);
         }
     }
     return ;
 }
 /*
 10
 10 2 100
 7 10
 1 2 3
 1 3 4
 1 4 1
 2 5 1
 3 6 2
 5 7 3
 3 8 2
 4 9 3
 9 10 10
 5 1 2 3 4 5

 */

另外：

对花费采用二分

参见 https://blog.csdn.net/qq_40993793/article/details/82762766，

每次询问O( klog(w*n) )，(其中排序那块我没看懂，这里的时间复杂度没有算排序)

sum(k)<=2e6,n<=1e5,w<=1e9,

因此log(w*n)=47,maxtime=9.4*10^7，存在超时的可能性(虽然假设的有点。。。)。不太推荐。。。

在zoj测试时，超时了

 #include <bits/stdc++.h>
 #define ll long long
 using namespace std;
 const int maxn=1e5+;
 const double minv=1e-;
 const ll inf=1e18;

 vector<pair<int,ll> >e[maxn];
 int red[maxn],dep[maxn<<],p[maxn<<],lef[maxn],er[],point[maxn],ind,f[maxn<<][];
 ll cost[maxn],len[maxn];
 bool vis[maxn];

 void dfs(int d,ll c,ll l,int deep)
 {
     vis[d]=;
     if (red[d]==)
         c=;
     cost[d]=c;
     len[d]=l;
     dep[++ind]=deep;
     p[ind]=d;
     lef[d]=ind;
     vector<pair<int,ll> >::iterator j;
     for (j=e[d].begin();j!=e[d].end();j++)
         if (!vis[j->first])
         {
             dfs(j->first,c+j->second,l+j->second,deep+);
             dep[++ind]=deep;
             p[ind]=d;
         }
 }

 int cmp(int a,int b)
 {
     return cost[a]>cost[b];
 }

 int main()
 {
     int t,n,m,q,u,v,g,d,deep,i,j,k,l;
     ll w,result,r;
     bool use;
     for (i=;i<=;i++)
         er[i]=<<i;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%d%d%d",&n,&m,&q);
         for (i=;i<=n;i++)
         {
             red[i]=;
             vis[i]=;
             e[i].clear();
         }
         for (i=;i<=m;i++)
             scanf("%d",&d),red[d]=;
         for (i=;i<n;i++)
         {
             scanf("%d%d%lld",&u,&v,&w);
             e[u].push_back({v,w});
             e[v].push_back({u,w});
         }
         ind=;
         dfs(,,,);
         for (j=;j<=ind;j++)
             f[j][]=j;
         deep=(int)(log(ind)/log()+minv);
         for (i=;i<=deep;i++)
             for (j=,k=er[i-]+;j<=ind-er[i]+;j++,k++)
                 f[j][i]=(dep[f[j][i-]]<dep[f[k][i-]])?f[j][i-]:f[k][i-];
         while (q--)
         {
             l=; r=;
             scanf("%d",&g);
             for (i=;i<=g;i++)
             {
                 scanf("%d",&point[i]);
                 r=max(r,cost[point[i]]);
             }
             while (l<=r)
             {
                 m=(l+r)>>;
                 use=;
                 for (i=;i<=g;i++)
                     if (cost[point[i]]>m)
                     {
                         if (!use)
                         {
                             use=;
                             d=point[i];
                             continue;
                         }
                         u=lef[d];
                         v=lef[point[i]];
                         if (u>v)
                             swap(u,v);
                         deep=(int)(log(v-u+)/log()+minv);
                         d=(dep[f[u][deep]]<dep[f[v-er[deep]+][deep]])?f[u][deep]:f[v-er[deep]+][deep];
                         d=p[d];
                         if (len[point[]]-len[d]>m)
                             break;
                     }
                 if (i==g+)
                     r=m-;
                 else
                     l=m+;
             }
             printf("%lld\n",l);
         }
     }
     return ;
 }
 /*
 10
 10 2 100
 7 10
 1 2 3
 1 3 4
 1 4 1
 2 5 1
 3 6 2
 5 7 3
 3 8 2
 4 9 3
 9 10 10
 5 1 2 3 4 5

 */