CodeForces 375D Tree and Queries 莫队算法

题目：http://codeforces.com/problemset/problem/375/D

题意：给定一个以1为根的树，树中每个节点有一个颜色，求以某个节点v为根的子树中颜色出现次数大于等于k的颜色有几种

思路：dfs序分块，然后莫队乱搞，统计每种颜色的次数，用树状数组维护次数的前缀和，查询大于等于k次的颜色便是sum(n) - sum(k-1)

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define debug() puts("here")using namespace std;const int N = 100010;struct edge{    int to, next;}g[N*2];struct node{    int v, l, r, k, id;}q[N];int n, m, unit, tmp;int cnt, head[N];int tot, a[N], in[N], out[N], num[N], val[N], res[N], bit[N];void add_edge(int v, int u){    g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;}void dfs(int v, int fa){    in[v] = ++tot;    val[tot] = a[v];    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(u != fa) dfs(u, v);    }    out[v] = tot;}bool cmp(node a, node b){    return a.l/unit != b.l/unit ? a.l/unit < b.l/unit : a.r < b.r;}void add(int i, int x){    while(i <= n) bit[i] += x, i += i & -i;}int sum(int i){    int res = 0;    while(i > 0) res += bit[i], i -= i & -i;    return res;}void _add(int i){    if(num[i] > 0) add(num[i], -1);    num[i]++;    add(num[i], 1);}void del(int i){    add(num[i], -1);    num[i]--;    if(num[i] > 0) add(num[i], 1);}void solve(){    dfs(1, -1);    unit = (int)sqrt(1.0*n);    for(int i = 1; i <= m; i++)        q[i].id = i, q[i].l = in[q[i].v], q[i].r = out[q[i].v];    sort(q+1, q+1+m, cmp);    int l = 1, r = 0;    tmp = 0;    for(int i = 1; i <= m; i++)    {        while(r < q[i].r) _add(val[++r]);        while(r > q[i].r) del(val[r--]);        while(l < q[i].l) del(val[l++]);        while(l > q[i].l) _add(val[--l]);        if(q[i].k > n) res[q[i].id] = 0;        else res[q[i].id] = sum(n) - sum(q[i].k - 1);    }    for(int i = 1; i <= m; i++) printf("%d\n", res[i]);}int main(){    scanf("%d%d", &n, &m);    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);    cnt = 0;    memset(head, -1, sizeof head);    int v, u;    for(int i = 1; i <= n - 1; i++)    {        scanf("%d%d", &v, &u);        add_edge(v, u), add_edge(u, v);    }    for(int i = 1; i <= m; i++) scanf("%d%d", &q[i].v, &q[i].k);    solve();    return 0;}

另外一种不用树状数组的写法，直接用一个数组维护

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define debug() puts("here")using namespace std;const int N = 100010;struct edge{    int to, next;}g[N*2];struct node{    int v, l, r, k, id;}q[N];int n, m, unit, tmp;int cnt, head[N];int tot, a[N], in[N], out[N], num[N], val[N], res[N], bit[N], rec[N];void add_edge(int v, int u){    g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;}void dfs(int v, int fa){    in[v] = ++tot;    val[tot] = a[v];    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(u != fa) dfs(u, v);    }    out[v] = tot;}bool cmp(node a, node b){    return a.l/unit != b.l/unit ? a.l/unit < b.l/unit : a.r < b.r;}void add(int i){    rec[++num[i]]++;}void del(int i){    rec[num[i]--]--;}void solve(){    tot = 0;    dfs(1, -1);    unit = (int)sqrt(1.0*n);    for(int i = 1; i <= m; i++)        q[i].id = i, q[i].l = in[q[i].v], q[i].r = out[q[i].v];    sort(q+1, q+1+m, cmp);    int l = 1, r = 0;    tmp = 0;    for(int i = 1; i <= m; i++)    {        while(r < q[i].r) add(val[++r]);        while(r > q[i].r) del(val[r--]);        while(l < q[i].l) del(val[l++]);        while(l > q[i].l) add(val[--l]);        res[q[i].id] = rec[q[i].k];    }    for(int i = 1; i <= m; i++) printf("%d\n", res[i]);}int main(){    scanf("%d%d", &n, &m);    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);    cnt = 0;    memset(head, -1, sizeof head);    int v, u;    for(int i = 1; i <= n - 1; i++)    {        scanf("%d%d", &v, &u);        add_edge(v, u), add_edge(u, v);    }    for(int i = 1; i <= m; i++) scanf("%d%d", &q[i].v, &q[i].k);    solve();    return 0;}