SPOJ PTO7J Query on a tree III(dfs序+主席树)

题意：

给定一棵树，节点带权， q次询问，每次询问一颗子树上第k大的权值。

思路：

求出树的dfs序列，子树的问题就变成了区间问题，也就是经典的区间第k大，这里用主席树水过了。。

#include<iostream>#include<string>#include<cstring>#include<cstdio>#include<cmath>#include<queue>#include<set>#include<algorithm>using namespace std;#define LL long long#define eps 1e-8#define MP make_pair#define N 200020#define M 400020#define NLGN 400002#pragma comment(linker, "/STACK:1024000000,1024000000")#define ls (i << 1)#define rs (ls | 1)#define md ((ll + rr) >> 1)#define lson ll, md, ls#define rson md + 1, rr, rs#define mod 258280327#define inf 0x3f3f3f3f#define pii pair<int, int>#define ULL unsigned long longint readint() {    char c;    while((c = getchar()) && !(c >= '0' && c <= '9') && c != '-');    int ret = c - '0', sgn = 0;    if(c == '-') sgn = 1, ret = 0;    while((c = getchar()) && c >= '0' && c <= '9')        ret = ret * 10 + c - '0';    if(sgn) ret = -ret;    return ret;}char buf[8000000],*pt = buf,*o = buf;int getint(){    int f = 1,x = 0;    while((*pt != '-') && (*pt < '0' || *pt > '9'))    pt ++;    if(*pt == '-')    f = -1,pt ++;    else    x = *pt++ - 48;    while(*pt >= '0' && *pt <= '9')    x = x * 10 + *pt ++ - 48;    return x * f;}char getch(){    char ch;    while(*pt < 'A' || *pt > 'Z')    pt ++;    ch=*pt;pt++;    return ch;}int n, fst[N], nxt[M], vv[M], e;void init() {    memset(fst, -1, sizeof fst);    e = 0;}void add(int u, int v) {    vv[e] = v, nxt[e] = fst[u], fst[u] = e++;}int val[N], san[N], cnt;int dc, c[N], d[N], lab[N];int root[N];int ch[N*20][2], sum[N*20], tot;int pos[N];int haxi(int x) {    return lower_bound(san + 1, san + cnt + 1, x) - san;}void dfs(int u, int p) {    c[u] = ++dc;    lab[dc] = u;    for(int i = fst[u]; ~i; i = nxt[i]) {        int v = vv[i];        if(v == p) continue;        dfs(v, u);    }    d[u] = dc;}int update(int i, int x, int ll, int rr) {    int k = ++tot;    ch[k][0] = ch[i][0];    ch[k][1] = ch[i][1];    sum[k] = sum[i] + 1;    if(ll == rr) return k;    if(x <= md) ch[k][0] = update(ch[i][0], x, ll, md);    else ch[k][1] = update(ch[i][1], x, md + 1, rr);    return k;}int query(int i, int o, int k, int ll, int rr) {    if(ll == rr) return pos[ll];    int x = sum[ch[i][0]] - sum[ch[o][0]];    if(k <= x) return query(ch[i][0], ch[o][0], k, ll, md);    return query(ch[i][1], ch[o][1], k - x, md + 1, rr);}int main() {    while(scanf("%d", &n) != EOF) {        init();        cnt = 0;        for(int i = 1; i <= n; ++i) {            scanf("%d", &val[i]);            san[++cnt] = val[i];        }        for(int i = 1; i < n; ++i) {            int u, v;            scanf("%d%d", &u, &v);            add(u, v);            add(v, u);        }        sort(san + 1, san + cnt + 1);        cnt = unique(san + 1, san + cnt + 1) - san - 1;        for(int i = 1; i <= n; ++i) val[i] = haxi(val[i]), pos[val[i]] = i;        dc = 0;        dfs(1, -1);        tot = 0;        for(int i = 1; i <= n; ++i) {            int u = lab[i];            root[i] = update(root[i-1], val[u], 1, cnt);        }        int q;        scanf("%d", &q);        while(q--) {            int x, k;            scanf("%d%d", &x, &k);            int t = query(root[d[x]], root[c[x]-1], k, 1, cnt);            printf("%d\n", t);        }    }    return 0;}