(好题)树状数组+离散化+DFS序+离线/莫队 HDOJ 4358 Boring counting
题意：给你一棵树，树上的每个节点都有树值，给m个查询，问以每个点u为根的子树下有多少种权值恰好出现k次。
分析：首先要对权值离散化，然后要将树形转换为线形，配上图：
。然后按照右端点从小到大排序，离线操作：将每一个深度的权值分组到相同权值的cnt中，当sz == k时，用树状数组更新+1，表示在该深度已经存在k个相同的权值，如果>k，之前k个-2(-1是恢复原样，再-1是为下次做准备？)，然后一个点的子树的答案就是 sum (r) - sum (l-1)。
当然，区间离线问题用莫队算法是很好做的。
收获：1. 离散化 2. DFS序，树形变线形 3. 离线操作 4. 莫队算法
代码(树状数组)：
/************************************************
* Author :Running_Time
* Created Time :2015/9/10 星期四 19:15:22
* File Name :I.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Edge {
int v, nex;
}edge[N*2];
struct Query {
int l, r, id;
Query (int _l = 0, int _r = 0, int _id = 0) : l (_l), r (_r), id (_id) {}
bool operator < (const Query &x) const {
return r < x.r;
}
}q[N];
struct BIT {
int c[N];
void init(void) {
memset (c, 0, sizeof (c));
}
void updata(int i, int x) {
while (i < N) {
c[i] += x; i += i & (-i);
}
}
int query(int i) {
int ret = 0;
while (i) {
ret += c[i]; i -= i & (-i);
}
return ret;
}
}bit;
int head[N], dfn[N], low[N], w[N], p[N], val[N], ans[N];
vector<int> cnt[N];
int e, dep;
void init(void) {
memset (head, -1, sizeof (head));
e = 0; dep = 0;
}
void add_edge(int u, int v) {
edge[e].v = v; edge[e].nex = head[u];
head[u] = e++;
}
bool cmp(int i, int j) {
return w[i] < w[j];
}
void compress(int n) {
for (int i=1; i<=n; ++i) p[i] = i;
sort (p+1, p+1+n, cmp);
int rk = 0, pre = w[p[1]] - 1;
for (int i=1; i<=n; ++i) {
if (pre != w[p[i]]) {
pre = w[p[i]];
w[p[i]] = ++rk;
}
else {
w[p[i]] = rk;
}
}
}
void DFS(int u, int fa) {
dfn[u] = ++dep; val[dep] = w[u];
for (int i=head[u]; ~i; i=edge[i].nex) {
int v = edge[i].v;
if (v == fa) continue;
DFS (v, u);
}
low[u] = dep;
}
int main(void) {
int T, cas = 0; scanf ("%d", &T);
while (T--) {
if (cas) puts ("");
printf ("Case #%d:\n", ++cas);
int n, k; scanf ("%d%d", &n, &k);
init ();
for (int i=1; i<=n; ++i) {
scanf ("%d", &w[i]);
}
compress (n); //离散化，升序排序，相同的还是相同的
for (int u, v, i=1; i<n; ++i) {
scanf ("%d%d", &u, &v);
add_edge (u, v); add_edge (v, u);
}
DFS (1, -1); //先序遍历，得到DFS序，树形变线形
int m; scanf ("%d", &m);
for (int u, i=1; i<=m; ++i) {
scanf ("%d", &u);
q[i] = Query (dfn[u], low[u], i);
}
sort (q+1, q+1+m); //按照DFS序排序
for (int i=1; i<=n; ++i) cnt[i].clear ();
int qu = 1; bit.init ();
for (int i=1; i<=n; ++i) { //按照dep深度从小到大
int v = val[i];
cnt[v].push_back (i); //表示离散后的相同的权值的个数
int sz = cnt[v].size ();
if (sz >= k) {
if (sz == k) bit.updata (cnt[v][sz-k], 1);
else {
bit.updata (cnt[v][sz-k-1], -2); //?
bit.updata (cnt[v][sz-k], 1);
}
}
while (qu <= m && q[qu].r == i) {
ans[q[qu].id] = bit.query (q[qu].r) - bit.query (q[qu].l - 1);
qu++;
}
}
for (int i=1; i<=m; ++i) {
printf ("%d\n", ans[i]);
}
}
return 0;
}
代码(莫队算法)：
/************************************************
* Author :Running_Time
* Created Time :2015/9/10 星期四 19:15:22
* File Name :I.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Edge {
int v, nex;
}edge[N*2];
struct Data {
int l, r, id, b;
Data () {}
Data (int l, int r, int id, int b) : l (l), r (r), id (id), b (b) {}
bool operator < (const Data &x) const {
if (b == x.b) return r < x.r;
else return b < x.b;
}
}data[N];
int head[N], dfn[N], low[N], w[N], p[N], val[N], ans[N];
int cnt[N];
int n, k, m, e, dep, sum;
void init(void) {
memset (head, -1, sizeof (head));
e = 0; dep = 0;
}
void add_edge(int u, int v) {
edge[e].v = v; edge[e].nex = head[u];
head[u] = e++;
}
bool cmp(int i, int j) {
return w[i] < w[j];
}
void compress(int n) {
for (int i=1; i<=n; ++i) p[i] = i;
sort (p+1, p+1+n, cmp);
int rk = 0, pre = w[p[1]] - 1;
for (int i=1; i<=n; ++i) {
if (pre != w[p[i]]) {
pre = w[p[i]];
w[p[i]] = ++rk;
}
else {
w[p[i]] = rk;
}
}
}
void DFS(int u, int fa) {
dfn[u] = ++dep; val[dep] = w[u];
for (int i=head[u]; ~i; i=edge[i].nex) {
int v = edge[i].v;
if (v == fa) continue;
DFS (v, u);
}
low[u] = dep;
}
inline void updata(int x, int c) {
cnt[x] += c;
if (cnt[x] == k) sum++;
else if (c > 0 && cnt[x] == k + 1) sum--;
else if (c < 0 && cnt[x] == k - 1) sum--;
}
void Modui(void) {
memset (cnt, 0, sizeof (cnt));
sum = 0;
int l = 1, r = 0;
for (int i=1; i<=m; ++i) {
while (data[i].l < l) updata (val[--l], 1);
while (data[i].l > l) updata (val[l], -1), l++;
while (data[i].r > r) updata (val[++r], 1);
while (data[i].r < r) updata (val[r], -1), r--;
ans[data[i].id] = sum;
}
for (int i=1; i<=m; ++i) {
printf ("%d\n", ans[i]);
}
}
int main(void) {
int T, cas = 0; scanf ("%d", &T);
while (T--) {
if (cas) puts ("");
printf ("Case #%d:\n", ++cas);
scanf ("%d%d", &n, &k);
init ();
for (int i=1; i<=n; ++i) {
scanf ("%d", &w[i]);
}
compress (n); //离散化，升序排序，相同的还是相同的
for (int u, v, i=1; i<n; ++i) {
scanf ("%d%d", &u, &v);
add_edge (u, v); add_edge (v, u);
}
DFS (1, -1); //先序遍历，得到DFS序，树形变线形
int block = (int) sqrt (n + 0.5);
scanf ("%d", &m);
for (int u, i=1; i<=m; ++i) {
scanf ("%d", &u);
data[i] = Data (dfn[u], low[u], i, dfn[u] / block);
}
sort (data+1, data+1+m); //按照DFS序排序
Modui ();
}
return 0;
}
Codeforces Round #136 (Div. 1) B. Little Elephant and Array
分析：估计这是上面那道题的母题(弱化版)，直接套用就行了，现在稍微理解了树状数组的维护方法，配上图：
代码(树状数组)：
/************************************************
* Author :Running_Time
* Created Time :2015/9/11 星期五 18:39:17
* File Name :B.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int n, m;
struct BIT {
int c[N];
void init(void) {
memset (c, 0, sizeof (c));
}
void updata(int i, int x) {
while (i <= n) {
c[i] += x; i += i & (-i);
}
}
int query(int i) {
int ret = 0;
while (i) {
ret += c[i]; i -= i & (-i);
}
return ret;
}
}bit;
struct Query {
int l, r, id;
bool operator < (const Query &x) const {
return r < x.r;
}
}q[N];
int a[N], ans[N];
vector<int> cnt[N];
int main(void) {
scanf ("%d%d", &n, &m);
for (int i=1; i<=n; ++i) scanf ("%d", &a[i]);
for (int i=1; i<=m; ++i) {
scanf ("%d%d", &q[i].l, &q[i].r);
q[i].id = i;
}
sort (q+1, q+1+m);
int qu = 1;
for (int i=1; i<=n; ++i) cnt[i].clear ();
bit.init ();
for (int i=1; i<=n; ++i) {
if (a[i] > n) continue;
cnt[a[i]].push_back (i);
int sz = cnt[a[i]].size ();
if (sz >= a[i]) {
bit.updata (cnt[a[i]][sz-a[i]], 1);
if (sz > a[i]) bit.updata (cnt[a[i]][sz-a[i]-1], -2);
if (sz > a[i] + 1) bit.updata (cnt[a[i]][sz-a[i]-2], 1);
}
while (qu <= m && q[qu].r == i) {
ans[q[qu].id] = bit.query (q[qu].r) - bit.query (q[qu].l - 1);
qu++;
}
}
for (int i=1; i<=m; ++i) {
printf ("%d\n", ans[i]);
}
return 0;
}
代码(莫队)：
/************************************************
* Author :Running_Time
* Created Time :2015/9/11 星期五 20:05:22
* File Name :B_Modui.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Data {
int l, r, id, b;
bool operator < (const Data &x) const {
if (b == x.b) return r < x.r;
else return b < x.b;
}
}data[N];
int a[N], ans[N], cnt[N];
int n, m, sum;
inline void updata(int x, int c) {
if (x > n) return ;
cnt[x] += c;
if (cnt[x] == x) sum++;
else if (c > 0 && cnt[x] == x + 1) sum--;
else if (c < 0 && cnt[x] == x - 1) sum--;
}
void Modui(void) {
sum = 0;
memset (cnt, 0, sizeof (cnt));
int l = 1, r = 0;
for (int i=1; i<=m; ++i) {
while (data[i].l < l) updata (a[--l], 1);
while (data[i].l > l) updata (a[l], -1), l++;
while (data[i].r > r) updata (a[++r], 1);
while (data[i].r < r) updata (a[r], -1), r--;
ans[data[i].id] = sum;
}
for (int i=1; i<=m; ++i) {
printf ("%d\n", ans[i]);
}
}
int main(void) {
scanf ("%d%d", &n, &m);
int block = (int) sqrt (n + 0.5);
for (int i=1; i<=n; ++i) {
scanf ("%d", &a[i]);
}
for (int i=1; i<=m; ++i) {
scanf ("%d%d", &data[i].l, &data[i].r);
data[i].id = i; data[i].b = data[i].l / block;
}
sort (data+1, data+1+m);
Modui ();
return 0;
}
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