SPOJ COT(树上k大，主席树＋LCA)

题目链接

给出一颗含有n个节点的树，每个节点有个权值。问u到v的路径上第k小的权值是多大。
做法就是主席树＋LCA，每个节点建立一颗从根（默认为1）到它的线断树，那么u->v路径的线断树就等于T[u] + T[v] - 2*LCA(u, v) + (l <= a[LCA] && a[LCA] <= r)。

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2016File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;#define MEM(x,y) memset(x, y,sizeof x)#define pk push_back#define lson rt << 1#define rson rt << 1 | 1#define bug cout << "BUG HERE\n"#define debug(x) cout << #x << " = " << x << endl#define ALL(v) (v).begin(), (v).end()#define lowbit(x) ((x)&(-x))#define Unique(x) sort(ALL(x)); (x).resize(unique(ALL(x)) - (x).begin())#define BitOne(x) __builtin_popcount(x)#define showtime printf("time = %.15f\n",clock() / (double)CLOCKS_PER_SEC)#define Rep(i, l, r) for (int i = l;i <= r;++i)#define Rrep(i, r, l) for (int i = r;i >= l;--i)typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;typedef pair<ii,int> iii;const double eps = 1e-8;const double pi = 4 * atan(1);const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;int nCase = 0;//精度正负、0的判断int dcmp(double x){if (fabs(x) < eps) return 0;return x < 0?-1:1;}template<class T> inline bool read(T &n){    T x = 0, tmp = 1;    char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T> inline void write(T n){    if(n < 0){putchar('-');n = -n;}    int len = 0,data[20];    while(n){data[len++] = n%10;n /= 10;}    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}LL QMOD(LL x, LL k) {    LL res = 1LL;    while(k) {if (k & 1) res = res * x % MOD;k >>= 1;x = x * x % MOD;}    return res;}const int maxn = 1e5 + 123;int n, m, q;int a[maxn], Hash[maxn];int head[maxn], pnt[maxn*2], nxt[maxn*2], ecnt;struct _LCA {    int dep[maxn];    const static int LOGN = 20;    int fa[LOGN][maxn];    void init() {        // memset(fa, -1, sizeof fa);    }    void dfs(int u,int pre, int depth) {        fa[0][u] = pre; dep[u] = depth;        for (int i = head[u];~i;i = nxt[i]) {            if (pnt[i] == pre) continue;            dfs(pnt[i], u, depth + 1);        }    }    inline void Build(int n) {        for (int k = 0;k < LOGN - 1;++k) {            for (int u = 1;u <= n;++u) {                if (fa[k][u] == -1) fa[k + 1][u] = -1;                else fa[k + 1][u] = fa[k][ fa[k][u] ];            }        }    }    inline int upslope(int u,int p) {        for (int k = 0;k < LOGN - 1;++k)            if ((p>>k) & 1) u = fa[k][u];        return u;    }    inline int LCA(int u,int v) {        if (dep[u] < dep[v]) swap(u, v);        u = upslope(u, dep[u] - dep[v]);        if (u == v) return u;        for (int k = LOGN - 1;k >= 0;--k) {            if (fa[k][u] != fa[k][v])                 u = fa[k][u], v = fa[k][v];        }        return fa[0][u];    }}lca;int root[maxn], ls[maxn*20], rs[maxn*20], sum[maxn*20], tot;void build(int&rt, int l, int r) {    rt = ++tot;    sum[rt] = 0;    if (l == r) return ;    int mid = (l + r) >> 1;    build(ls[rt], l, mid);    build(rs[rt], mid + 1, r);}void updata(int last,int&rt,int l,int r,int pos, int v) {    rt = ++tot;    ls[rt] = ls[last];    rs[rt] = rs[last];    sum[rt] = sum[last] + v;    if (l == r) return ;    int mid = (l + r) >> 1;    if (pos <= mid) updata(ls[last], ls[rt], l, mid, pos, v);    else updata(rs[last], rs[rt], mid + 1, r, pos, v);}void dfs_build(int u,int pre) {    updata(root[pre], root[u], 1, m, a[u], 1);    for (int i = head[u];~i;i = nxt[i]) {        if (pnt[i] == pre) continue;        dfs_build(pnt[i], u);    }}int find(int u_rt, int v_rt, int lca,int k) {    int lca_rt = root[lca];    int pos = a[lca];    int l = 1, r = m;    while(l < r) {        int mid = (l + r) >> 1;        int temp = sum[ls[u_rt]] + sum[ls[v_rt]] - 2*sum[ls[lca_rt]] + (l <= pos && pos <= mid);        if (temp >= k) {            u_rt = ls[u_rt];            v_rt = ls[v_rt];            lca_rt = ls[lca_rt];            r = mid;        }else {            k -= temp;            u_rt = rs[u_rt];            v_rt = rs[v_rt];            lca_rt = rs[lca_rt];            l = mid + 1;        }    }    return l;}void init() {    memset(head, -1, sizeof head), ecnt = 0, tot = 0;}int main(int argc, const char * argv[]){        // freopen("in.txt","r",stdin);    // freopen("out.txt","w",stdout);    // ios::sync_with_stdio(false);    // cout.sync_with_stdio(false);    // cin.sync_with_stdio(false);    // cout << (1 << 20) - 1 << endl;    while(~scanf("%d%d", &n, &q)) {        init();        Rep(i, 1, n) {scanf("%d", &a[i]); Hash[i] = a[i];}        sort(Hash + 1, Hash + 1 + n);        m = unique(Hash + 1, Hash + 1 + n) - Hash - 1;        Rep(i, 1, n) a[i] = lower_bound(Hash + 1, Hash + 1 + m, a[i]) - Hash;        Rep(i, 1, n - 1) {            int u, v;scanf("%d%d", &u, &v);            pnt[ecnt] = v, nxt[ecnt] = head[u], head[u] = ecnt++;            pnt[ecnt] = u, nxt[ecnt] = head[v], head[v] = ecnt++;        }        lca.dfs(1, -1, 0);        lca.Build(n);        build(root[0], 1, m);//m种数        dfs_build(1, 0);        while(q--) {            int u, v, k;scanf("%d%d%d", &u, &v, &k);            // printf("[u = %d, v = %d, lca = %d]\n", u, v, lca.LCA(u, v));            printf("%d\n", Hash[find(root[u], root[v], lca.LCA(u, v), k)]);        }    }    // showtime;    return 0;}