SPOJ10628--COT（树上第K大）主席树
You are given a tree with N nodes.The tree nodes are numbered from 1 to N.Each node has an integer weight.
We will ask you to perform the following operation:
- u v k : ask for the kth minimum weight on the path from node u to node v
Input
In the first line there are two integers N and M.(N,M<=100000)
In the second line there are N integers.The ith integer denotes the weight of the ith node.
In the next N-1 lines,each line contains two integers u v,which describes an edge (u,v).
In the next M lines,each line contains three integers u v k,which means an operation asking for the kth minimum weight on the path from node u to node v.
Output
For each operation,print its result.
Example
Input:
8 5
8 5 105 2 9 3 8 5 7 7 1 2 1 3 1 4 3 5 3 6 3 7 4 8 2 5 1 2 5 2 2 5 3 2 5 4 7 8 2
Output: 2 8 9 105 7
题意：求u到v路径上的第K大值
思路：和普通的区间第K大类似，不过普通第K大是每棵线段树根据后一棵线段树建立，而树上的主席树是每个节点根据父亲节点建立。
#include
#include
#include
#include
using namespace std;
#define maxn 208000
#define maxm 2500000
int T[maxn],first[100080],nxt[maxn],vv[maxn],dfn[maxn],dep[maxn],key[maxn],X[maxn];
int c[maxm],lson[maxm],rson[maxm];
int tot,e,n,m,fuck,cnt;
int fa[maxn][20];//fa[i][j]表示i的2^j祖先是谁。
void scanf_(int & num)
{
char in;
while((in=getchar())>'9'||in<'0');
num=in-'0';
while(in=getchar(),in>='0'&&in<='9')
num*=10,num+=in-'0';
}
void addedge(int u,int v)
{
vv[e] = v; nxt[e] = first[u]; first[u] = e++;
vv[e] = u; nxt[e] = first[v]; first[v] = e++;
}
int build(int l,int r)
{
int root = tot++;
c[root] = 0;
if(l != r)
{
int mid = (l+r) >> 1;
lson[root] = build(l,mid);
rson[root] = build(mid+1,r);
}
return root;
}
int Update(int root,int pos,int val)
{
int newnode = tot++,tmp = newnode;
c[newnode] = c[root] + val;
int l = 1,r = fuck;
while(l < r)
{
int mid = (l+r) >> 1;
if(pos <= mid)
{
r = mid;
lson[newnode] = tot++; rson[newnode] = rson[root];
newnode = lson[newnode]; root = lson[root];
}
else
{
l = mid + 1;
rson[newnode] = tot++; lson[newnode] = lson[root];
newnode = rson[newnode]; root = rson[root];
}
c[newnode] = c[root] + val;
}
return tmp;
}
int Query(int u,int v,int lca,int lca_root,int k)
{
int l = 1,r = fuck;
int pos = lower_bound(X+1,X+fuck,key[lca]) - X;
while(l < r)
{
int mid = (l+r) >> 1;
if(c[lson[u]] + c[lson[v]] - 2*c[lson[lca_root]] + (pos >= l && pos <= mid) >= k)
{
r = mid;
u = lson[u];
v = lson[v];
lca_root = lson[lca_root];
}
else
{
k -= c[lson[u]] + c[lson[v]] - 2*c[lson[lca_root]] + (pos >= l && pos <= mid);
u = rson[u];
v = rson[v];
lca_root = rson[lca_root];
l = mid + 1;
}
}
return l;
}
int LCA(int u,int v)
{
if(dfn[u] == dfn[v])
return u;
if(dfn[u] < dfn[v])
swap(u,v);
//if(dfn[fa[u][0]] <= dfn[v])
//return LCA(fa[u][0],v);
for(int i = 1;i >= 0;i++)
{
if(dfn[fa[u][i]] < dfn[v])
return LCA(fa[u][i-1],v);
if(dfn[fa[u][i]] == dfn[v])
return v;
}
}
void init()
{
memset(first,-1,sizeof(first));
memset(dfn,0,sizeof(dfn));
dfn[1] = dep[1] = 0;
tot = e = cnt = 0;
}
void dfs(int u,int pre)//建树同时求出每个点的2的若干次幂祖先
{
int pos = lower_bound(X+1,X+1+fuck,key[u]) - X;
T[u] = Update(T[pre],pos,1);
dfn[u] = ++cnt;
dep[u] = dep[pre] + 1;
fa[u][0] = pre;
for(int j = 1;dep[u] - (1<= 0;j++)
fa[u][j] = fa[fa[u][j-1]][j-1];
for(int i = first[u];i != -1;i = nxt[i])
{
int v = vv[i];
if(v == pre) continue;
dfs(v,u);
}
}
int main()
{
//freopen("in.txt","r",stdin);
memset(fa,0,sizeof(fa));
scanf_(n);scanf_(m);
init();
for(int i = 1;i <= n;i++)
{
scanf_(key[i]);
X[i] = key[i];
}
sort(X+1,X+1+n);
fuck = unique(X+1,X+1+n) - X - 1;
T[0] = build(1,fuck);
for(int i = 1;i < n;i++)
{
int u,v;
scanf_(u),scanf_(v);
addedge(u,v);
}
dfs(1,0);
while(m--)
{
int u,v,k;
scanf_(u),scanf_(v),scanf_(k);
int lca = LCA(u,v);
printf("%d\n",X[Query(T[u],T[v],lca,T[lca],k)]);
}
return 0;
}