树链剖分-链的剖分（线段树维护+离线操作）
Query on a tree
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 622 Accepted Submission(s): 146
Problem Description
There are some queries on a tree which has n nodes. Every query is described as two integers (X, Y).For each query, you should find the maximum weight of the edges in set E, which satisfies the following two conditions.
1) The edge must on the path from node X to node 1.
2) The edge’s weight should be lower or equal to Y.
Now give you the tree and queries. Can you find out the answer for each query?
1) The edge must on the path from node X to node 1.
2) The edge’s weight should be lower or equal to Y.
Now give you the tree and queries. Can you find out the answer for each query?
Input
The first line of the input is an integer T, indicating the number of test cases. For each case, the first line contains an integer n indicates the number of nodes in the tree. Then n-1 lines follows, each line contains three integers
X, Y, W indicate an edge between node X and node Y whose value is W. Then one line has one integer Q indicates the number of queries. In the next Q lines, each line contains two integers X and Y as said above.
Output
For each test case, you should output Q lines. If no edge satisfy the conditions described above，just output “-1” for this query. Otherwise output the answer for this query.
Sample Input
1 3 1 2 7 2 3 5 4 3 10 3 7 3 6 3 4
Sample Output
7 7 5 -1Hint2<=n<=10^5 2<=Q<=10^5 1<=W,Y<=10^9 The data is guaranteed that your program will overflow if you use recursion.
分析：首先把边的权值从小到达排列，然后把询问的y值也从小到达排列，然后对于每一个询问，把小于y值的边更新到线段树中，接着求出该次询问的最大值，把答案记录到原来顺序的数组里面，最后输出即可；
程序:
#pragma comment(linker, "/STACK:102400000,102400000")
#include"stdio.h"
#include"string.h"
#include"iostream"
#include"map"
#include"string"
#include"queue"
#include"stdlib.h"
#include"math.h"
#define M 100009
#define inf 100000000
using namespace std;
struct node
{
int u,v,next;
}edge[M*2];
int t,head[M],son[M],fa[M],num[M],p[M],fp[M],deep[M],cnt[M],pos,top[M],Max;
void init()
{
t=pos=0;
memset(head,-1,sizeof(head));
memset(son,-1,sizeof(son));
}
void add(int u,int v)
{
edge[t].u=u;
edge[t].v=v;
edge[t].next=head[u];
head[u]=t++;
}
void dfs(int u,int f,int d)
{
deep[u]=d;
fa[u]=f;
num[u]=1;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(v!=f)
{
dfs(v,u,d+1);
num[u]+=num[v];
if(son[u]==-1||num[son[u]]<num[v])
son[u]=v;
}
}
}
void getpos(int u,int sp)
{
top[u]=sp;
p[u]=pos++;
fp[p[u]]=u;
if(son[u]==-1)return;
getpos(son[u],sp);
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(v!=fa[u]&&v!=son[u])
getpos(v,v);
}
}
struct Node
{
int l,r,maxi;
}tree[M*5];
void pushup(int i)
{
tree[i].maxi=max(tree[i*2].maxi,tree[i*2+1].maxi);
}
void make(int l,int r,int i)
{
tree[i].l=l;
tree[i].r=r;
if(l==r)
{
tree[i].maxi=-inf;
return;
}
int mid=(l+r)/2;
make(l,mid,i*2);
make(mid+1,r,i*2+1);
pushup(i);
}
void updata(int p,int q,int i)
{
if(tree[i].r==p&&tree[i].l==p)
{
tree[i].maxi=q;
return;
}
int mid=(tree[i].l+tree[i].r)/2;
if(p<=mid)
updata(p,q,i*2);
else
updata(p,q,i*2+1);
pushup(i);
}
void query(int l,int r,int i)
{
if(tree[i].l==l&&tree[i].r==r)
{
Max=max(Max,tree[i].maxi);
return;
}
int mid=(tree[i].r+tree[i].l)/2;
if(r<=mid)
query(l,r,i*2);
else if(l>mid)
query(l,r,i*2+1);
else
{
query(l,mid,i*2);
query(mid+1,r,i*2+1);
}
pushup(i);
}
int findmax(int u,int v)
{
int f1=top[u];
int f2=top[v];
int ans=-inf;
while(f1!=f2)
{
if(deep[f1]<deep[f2])
{
swap(f1,f2);
swap(u,v);
}
Max=-inf;
query(p[f1],p[u],1);
ans=max(ans,Max);
u=fa[f1];
f1=top[u];
}
if(v==u)
return ans;
if(deep[u]>deep[v])
swap(u,v);
Max=-inf;
query(p[son[u]],p[v],1);
ans=max(ans,Max);
return ans;
}
struct Edge
{
int u,v,w,x,y,kk;
}e[M],Q[M];
int cmp(const void *a,const void *b)
{
return (*(struct Edge*)a).w-(*(struct Edge*)b).w;
}
int cmpy(const void *a,const void *b)
{
return (*(struct Edge*)a).y-(*(struct Edge*)b).y;
}
int main()
{
int T;
cin>>T;
while(T--)
{
int n,i;
init();
scanf("%d",&n);
for(i=0;i<n-1;i++)
{
scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
add(e[i].u,e[i].v);
add(e[i].v,e[i].u);
}
dfs(1,1,0);
getpos(1,1);
make(1,pos-1,1);
qsort(e,n-1,sizeof(e[0]),cmp);
int que;
scanf("%d",&que);
for(i=0;i<que;i++)
{
scanf("%d%d",&Q[i].x,&Q[i].y);
Q[i].kk=i;
}
qsort(Q,que,sizeof(Q[0]),cmpy);
int j=0;
for(i=0;i<que;i++)
{
while(j<n-1&&e[j].w<=Q[i].y)
{
if(deep[e[j].u]<deep[e[j].v])
swap(e[j].u,e[j].v);
updata(p[e[j].u],e[j].w,1);
j++;
}
cnt[Q[i].kk]=findmax(1,Q[i].x);
if(cnt[Q[i].kk]<=-inf)
cnt[Q[i].kk]=-1;
}
for(i=0;i<que;i++)
printf("%d\n",cnt[i]);
}
return 0;
}
