[BZOJ2588]Spoj 10628. Count on a tree 可持久化线段树

离散化
对每个节点，用线段树表示该节点到根的路径上的值域
然后可持久化

询问的时候查询u,v,LCA(u,v),fa[ LCA(u,v)]结点上的四棵线段树
做一做减法就好

notice:
1、初始的权值已经超过了int，需要long long
2、最后一组询问的换行符不要输出！（我已经邮件了一发管理员希望他改题面）

写起来很好写

#include 
#include 
#include 
#include 

#define mid ((l+r)>>1)
#define LOG 20

#define N 200050

using namespace std;
typedef long long LL;
int tr[40*N],ls[40*N],rs[40*N];
LL a[N],d[N],ans;
int h[N],D[N],cur[N];
int Jump[N][LOG+5];
int n,m,cnt,v,x;
vector<int> e[N];

bool cmp(int p1,int p2) { return a[p1] < a[p2]; }

void ins(int l,int r,int las,int &t) {
    if (!t) t = ++cnt;
    if (l == r) { tr[t] = tr[las] + v; return ; }
    if (x <= mid) 
        ins(l,mid,ls[las],ls[t]) , rs[t] = rs[las];
    else
        ins(mid+1,r,rs[las],rs[t]) , ls[t] = ls[las];
    tr[t] = tr[ ls[t] ] + tr[ rs[t] ];
}

void dfs(int u,int fa) {
    Jump[u][0] = fa; D[u] = D[ fa ] + 1;
    x = a[u]; v = 1;
    ins(1,n,h[fa],h[u]);
    for (int i=0;i<(int)e[u].size();i++) 
        if (e[u][i] != fa) dfs(e[u][i],u);
}

void _preLCA() {
    for (int i=1;i<=LOG;i++)
        for (int j=1;j<=n;j++)
            Jump[j][i] = Jump[ Jump[j][i-1] ][i-1];
}
int LCA(int x,int y) {
    if (D[x] < D[y]) swap(x,y);
    for (int i=LOG;i>=0;i--)
        if (D[ Jump[x][i] ] >= D[y]) x = Jump[x][i];
    if (x == y) return x;

    for (int i=LOG;i>=0;i--)
        if (Jump[x][i] != Jump[y][i]) 
            x = Jump[x][i] , y = Jump[y][i];

    return Jump[x][0];
}

//p1 --> u , p2 --> v , p3 --> LCA
void query(int l,int r,int p1,int p2,int p3,int p4,int k) {
    if (l == r) { ans = l; return ; }
    int cur = tr[ ls[p1] ] + tr[ ls[p2] ] - tr[ ls[p3] ] - tr[ ls[p4] ];
    if (cur >= k)
        query(l,mid,ls[p1],ls[p2],ls[p3],ls[p4],k);
    else
        query(mid+1,r,rs[p1],rs[p2],rs[p3],rs[p4],k-cur);
}

int main() {
    #ifndef ONLINE_JUDGE
//      freopen("1.in","r",stdin);
    #endif
    scanf("%d%d",&n,&m);

    for (int i=1;i<=n;i++) scanf("%lld",&a[i]);

    for (int i=1;i<=n;i++) cur[i] = i;
    sort(cur+1,cur+n+1,cmp);

    //离散化 
    for (int i=1;i<=n;i++) 
        d[i] = a[ cur[i] ] , a[ cur[i] ] = i;

    for (int _=1;_<=n-1;_++) {
        int x,y; scanf("%d%d",&x,&y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    //建树 
    dfs(1,0);
    _preLCA();

//   ans = 0;
    for (int _=1;_<=m;_++) {
        LL u,v,k;
        scanf("%lld%lld%lld",&u,&v,&k);
        u ^= ans;
        int p = LCA(u,v);
        query(1,n,h[u],h[v],h[p],h[ Jump[p][0] ],k);
        printf("%lld",d[ans]);
        if (_!=m) printf("\n");
        ans = d[ans];
    }
    return 0;
}