树的直径+rmq+（伪）单调队列 -HDU4123

给定一棵n个点并且有边权的树，每个点的权值为该点能走的最远长度，并输入m个询问，每次询问最多有多少个编号连续的点，他们的最大最小点权差小于等于Q。N<=50000 M<=500 Q<=10000000

　　我们知道一个点能走的最远端点一定是树的直径的端点，所以我们只需从树的直径两端点dfs，就可以求出每个点能到的最远长度。。然后rmq+尺取即可。

这道题如果用系统的log会tle，所以必须手打log2。原理会再放一篇博客。

 #include <cctype>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 using namespace std;

 const int maxn=, maxmi=, INF=1e9;
 //const double exp=1e-6;
 int n, m, cntedge, maxdist, maxdistpos;
 int fir[maxn], val[maxn];
 int fmaxm[maxn][maxmi], fminm[maxn][maxmi];
 struct Edge{
     int to, v, next;
 };
 Edge edge[maxn];

 void addedge(int x, int y, int z){
     ++cntedge;
     Edge &nowedge1=edge[cntedge];
     nowedge1.to=y, nowedge1.v=z, nowedge1.next=fir[x];
     fir[x]=cntedge;
     ++cntedge;
     Edge &nowedge2=edge[cntedge];
     nowedge2.to=x, nowedge2.v=z, nowedge2.next=fir[y];
     fir[y]=cntedge;
     return;
 }

 void dfs(int now, int par, int dist){
     int nowedge, nowson;
     nowedge=fir[now];
     while (nowedge){
         nowson=edge[nowedge].to;
         if (nowson==par){
             nowedge=edge[nowedge].next;
             continue;
         }
         dfs(nowson, now, dist+edge[nowedge].v);
         nowedge=edge[nowedge].next;
     }
     if (dist>val[now]) val[now]=dist;
     if (dist>maxdist){
         maxdist=dist;
         maxdistpos=now;
     }
     return;
 }

 int flog2(float x) {
     return ((unsigned&)x>>&)-;
 }
 int dvalue(int head, int tail){
     int maxm=, minm=1e9;
     int lognum=flog2(tail-head+);
     maxm=max(fmaxm[head][lognum], fmaxm[tail-(<<lognum)+][lognum]);
     minm=min(fminm[head][lognum], fminm[tail-(<<lognum)+][lognum]);
     return maxm-minm;
 }

 void init(){
     memset(fir, , sizeof(fir));
     memset(val, , sizeof(val));
     for (int i=; i<maxn; ++i){
         edge[i].next=edge[i].to=edge[i].v=;
     }
     cntedge=;
 }

 int ri(){
     char c;
     int flag=, r=;
     do{
         c=getchar();
         if (c=='-') flag=-;
     } while (!isgraph(c));
     do{
         r=r*+c-;
         c=getchar();
     } while (isgraph(c));
     return r*flag;
 }

 int main(){
     int x, y, z;
     while (~scanf("%d%d", &n, &m)){
         if (n==&&m==) break;
         init();
         for (int i=; i<n; ++i){
             x=ri(), y=ri(), z=ri();
             addedge(x, y, z);
         }
         int s=, far1, far2;
         maxdist=, dfs(s, , );
         far1=maxdistpos;
         maxdist=, dfs(far1, , );
         far2=maxdistpos;
         maxdist=, dfs(far2, , );
         //for (int i=1; i<=n; ++i)
         //printf("%d\n", val[i]);
         int q=;
         for (int i=; i<=n; ++i)
             fmaxm[i][]=fminm[i][]=val[i];
         for (int i=; i<maxmi; ++i){
             for (int j=; j<=n-(<<i)+; ++j){
                 fmaxm[j][i]=max(fmaxm[j][i-], fmaxm[j+(<<(i-))][i-]);
                 fminm[j][i]=min(fminm[j][i-], fminm[j+(<<(i-))][i-]);
             }
         }
         int h=, t=, maxm=;
         for (int i=; i<m; ++i){
             q=ri();
             h=, t=, maxm=;
             while (t<=n){
                 if (dvalue(h, t)<=q) ++t;
                 else ++h;
                 if ((t-h)>maxm) maxm=t-h;
             }
             printf("%d\n", maxm);
         }
     }
     return ;
 }