hdu_3804_Query on a tree(树链剖分)
题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=3804
题意:给你一棵树,然后给出树上边的价值,然后给出x,y,问从1到x的边上不超过y的最大值为多少
题解:这题如果直接写裸线段树来在线更新会T飞,正解是把边的值存下来,把询问也存下来,然后都按照价值从小到大排序,然后从最小的开始找,找之前把边的值插进线段树,这样就成了查询区间最大值了,很巧妙的方法。还有就是这题的内存卡的很紧,用前向星爆内存了,我没想通,用vector就过了。很尴尬
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<vector> 3 #include<algorithm> 4 #define F(i,a,b) for(int i=a;i<=b;++i) 5 #define root 1,n,1 6 #define ls l,m,rt<<1 7 #define rs m+1,r,rt<<1|1 8 using namespace std; 9 const int N=1e5+5; 10 vector<int>g[N]; 11 struct vs{ 12 int val,u,v; 13 bool operator<(const vs &b)const{return val<b.val;} 14 }va[N]; 15 struct ser{ 16 int x,k,id; 17 bool operator<(const ser &b)const{return k<b.k;} 18 }ask[N]; 19 int t,n,m,x,y,c,tot,ans[N],sum[N*4],sz[N],fa[N],dep[N],hs[N],top[N],tid[N],idx; 20 //线段树部分 21 void update(int x,int k,int l,int r,int rt){ 22 if(l==r){sum[rt]=k;return;} 23 int m=(l+r)>>1; 24 if(x<=m)update(x,k,ls); 25 else update(x,k,rs); 26 sum[rt]=max(sum[rt<<1],sum[rt<<1|1]); 27 } 28 29 int query(int L,int R,int l,int r,int rt){ 30 if(L<=l&&r<=R)return sum[rt]; 31 int m=(l+r)>>1,maxs=0; 32 if(L<=m)maxs=max(query(L,R,ls),maxs); 33 if(m<R)maxs=max(query(L,R,rs),maxs); 34 return maxs; 35 } 36 //树链部分 37 void dfs1(int u,int pre){ 38 sz[u]=1,fa[u]=pre,dep[u]=dep[pre]+1,hs[u]=0; 39 for(int i=0;i<g[u].size();i++){ 40 int vv=g[u][i]; 41 if(vv!=pre){ 42 dfs1(vv,u); 43 if(sz[vv]>sz[hs[u]])hs[u]=vv; 44 sz[u]+=sz[vv]; 45 } 46 } 47 } 48 49 void dfs2(int u,int tp){ 50 tid[u]=++idx,top[u]=tp; 51 if(hs[u])dfs2(hs[u],tp); 52 for(int i=0;i<g[u].size();i++){ 53 int vv=g[u][i]; 54 if(vv!=fa[u]&&vv!=hs[u])dfs2(vv,vv); 55 } 56 } 57 58 int findmax(int x){ 59 int fx=top[x],an=0; 60 while(fx!=1)an=max(query(tid[fx],tid[x],root),an),x=fa[fx],fx=top[x]; 61 if(x==1)return an; 62 return max(query(2,tid[x],root),an); 63 } 64 65 int main(){ 66 scanf("%d",&t); 67 while(t--){ 68 scanf("%d",&n); 69 F(i,1,n)g[i].clear(); 70 dep[0]=0,sz[0]=0; 71 F(i,0,4*N-1)sum[i]=0; 72 F(i,1,n-1)scanf("%d%d%d",&va[i].u,&va[i].v,&va[i].val),g[va[i].u].push_back(va[i].v),g[va[i].v].push_back(va[i].u); 73 dfs1(1,0),idx=0,dfs2(1,1); 74 scanf("%d",&m),tot=1; 75 F(i,1,m)scanf("%d%d",&ask[i].x,&ask[i].k),ask[i].id=i; 76 sort(va+1,va+n),sort(ask+1,ask+1+m); 77 F(i,1,m){ 78 while(tot<n&&va[tot].val<=ask[i].k){ 79 if(dep[va[tot].u]>dep[va[tot].v])c=va[tot].u;else c=va[tot].v; 80 update(tid[c],va[tot].val,root),tot++; 81 } 82 ans[ask[i].id]=findmax(ask[i].x); 83 } 84 F(i,1,m)printf("%d\n",ans[i]==0?-1:ans[i]); 85 } 86 return 0; 87 }