题目:
这题似乎BZOJ上数据强一些?
分析:
这题真的是……一言难尽
发现题面里没说权值的范围,怕出锅就写了离散化。后来经过面向数据编程(以及膜神犇代码)知道最大权值(1e5)(下文用(M)表示最大权值。注意如果没有这个限制,把所有数的权值和询问中提到的权值一起离散化后(M)也可以达到(n+2m=2.1e6)),如果没这个限制我很想知道正解应该怎么写……下面再细说
首先看到这种询问某个区间内在某个区域内值的数量的题,显然能想到把询问离线下来,用个什么数据结构维护莫队。
然后我大概脑补了一下,这个“数据结构”似乎可以用权值线段树?每个区间维护(sum)和(num)对应两种询问。对于叶子节点,若(sum)为(0)则(num)为(0),否则(num)为(1)。对于非叶子节点,(sum)和(num)都是分别是两儿子的(sum)/(num)之和。这样做正确性是显然的,复杂度(O(msqrt n imes log_2M)),大概是(1e6 imes sqrt{1e5} imes 21approx 6.64e9),过个鬼哦。
树状数组做法类似。维护两个树状数组(sum)和(num),同样对应两个询问。给(a)加的时候如果原本(sum[a])是(0)就给(num[a]),给(a)减的时候如果原本(sum[a])是(1)就给(num[a])减。复杂度和线段树是一样的,所以也过不去。
然后没想到竟然分块能过……把权值分块维护,(blsum)和(blnum)表示块内两种询问的答案,(num[a])表示(a)的出现次数,维护和查询方式详见代码。
在我僵化的思维里(log_2n)的树状数组和线段树比(sqrt n)的分块要优,但是这道题里分块的修改是(O(1)),查询是(O(sqrt M))。而查询不需要乘上莫队的(O(sqrt n)),所以最终复杂度是(O(m (sqrt n+sqrt M))),大概是(1e6 imes(sqrt{1e5} +sqrt{2.1e6})=1.77e9),还是悬。如果权值最大是(1e5)还是能跑过去的。
代码:
(线段树和树状数组没有AC)
线段树:(161)行,本地(83)s,内存(100.4)MB,BZOJ上TLE。下方代码中省略部分和树状数组基本一致
namespace Segment_Tree
{
struct node
{
int sum, num;
}tree[(N + (M << 1)) << 2];
pii operator + (const pii &a, const pii &b)
{
return make_pair(a.first + b.first, a.second + b.second);
}
pii operator += (pii &a, const pii &b)
{
return a = a + b;
}
void update(const int rot)
{
tree[rot].sum = tree[rot << 1].sum + tree[rot << 1 | 1].sum;
tree[rot].num = tree[rot << 1].num + tree[rot << 1 | 1].num;
}
void change(const int rot, const int lt, const int rt, const int pos, const int x)
{
if (lt == rt)
{
tree[rot].sum += x;
tree[rot].num = (tree[rot].sum ? 1 : 0);
return;
}
int mid = (lt + rt) >> 1;
if (pos <= mid)
change(rot << 1, lt, mid, pos, x);
else
change(rot << 1 | 1, mid + 1, rt, pos, x);
update(rot);
}
pii query(const int rot, const int lt, const int rt, const int ls, const int rs)
{
if (ls <= lt && rt <= rs)
return make_pair(tree[rot].sum, tree[rot].num);
int mid = (lt + rt) >> 1;
pii ans = make_pair(0, 0);
if (ls <= mid)
ans += query(rot << 1, lt, mid, ls, rs);
if (rs > mid)
ans += query(rot << 1 | 1, mid + 1, rt, ls, rs);
return ans;
}
}
树状数组:(153)行,本地(23)s(一样的复杂度比线段树快近(3)倍,结果BZOJ上还是TLE什么鬼),内存(37.0)MB
#include <cstdio>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
namespace zyt
{
template<typename T>
inline void read(T &x)
{
char c;
bool f = false;
x = 0;
do
c = getchar();
while (c != '-' && !isdigit(c));
if (c == '-')
f = true, c = getchar();
do
x = x * 10 + c - '0', c = getchar();
while (isdigit(c));
if (f)
x = -x;
}
template<typename T>
inline void write(T x)
{
static char buf[20];
char *pos = buf;
if (x < 0)
putchar('-'), x = -x;
do
*pos++ = x % 10 + '0';
while (x /= 10);
while (pos > buf)
putchar(*--pos);
}
inline void write(const pair<int, int> &a)
{
write(a.first), putchar(' '), write(a.second);
}
const int N = 1e5 + 10, M = 1e6 + 10;
int n, m, arr[N], belong[N];
typedef pair<int, int> pii;
pii ans[M];
struct _ask
{
int l, r, a, b, id;
bool operator < (const _ask &b) const
{
return belong[l] == belong[b.l] ? r < b.r : l < b.l;
}
}ask[M];
int cnt, tmp[N + (M << 1)];
class Tree_Array
{
private:
int data[N];
inline int lowbit(const int x)
{
return x & (-x);
}
public:
inline void add(int a, const int x)
{
while (a <= cnt)
data[a] += x, a += lowbit(a);
}
inline int query(int a)
{
int ans = 0;
while (a)
ans += data[a], a -= lowbit(a);
return ans;
}
inline int query(const int l, const int r)
{
return query(r) - query(l - 1);
}
}sum, num;
void update(const int pos, const int x)
{
if (x == 1)
{
if (!sum.query(pos, pos))
num.add(pos, 1);
sum.add(pos, 1);
}
if (x == -1)
{
sum.add(pos, -1);
if (!sum.query(pos, pos))
num.add(pos, -1);
}
}
void Mo_Algorithm()
{
int l = 1, r = 1;
update(arr[1], 1);
for (int i = 1; i <= m; i++)
{
while (l < ask[i].l)
update(arr[l++], -1);
while (l > ask[i].l)
update(arr[--l], 1);
while (r < ask[i].r)
update(arr[++r], 1);
while (r > ask[i].r)
update(arr[r--], -1);
if (ask[i].a > ask[i].b)
ans[ask[i].id] = make_pair(0, 0);
else
ans[ask[i].id] = make_pair(sum.query(ask[i].a, ask[i].b), num.query(ask[i].a, ask[i].b));
}
}
int work()
{
int block;
read(n), read(m);
block = sqrt(n);
for (int i = 1; i <= n; i++)
{
read(arr[i]), tmp[cnt++] = arr[i];
belong[i] = (i - 1) / block + 1;
}
for (int i = 1; i <= m; i++)
{
read(ask[i].l), read(ask[i].r), read(ask[i].a), read(ask[i].b);
ask[i].id = i;
tmp[cnt++] = ask[i].a, tmp[cnt++] = ask[i].b;
}
sort(tmp, tmp + cnt);
cnt = unique(tmp, tmp + cnt) - tmp;
for (int i = 1; i <= n; i++)
arr[i] = upper_bound(tmp, tmp + cnt, arr[i]) - tmp;
for (int i = 1; i <= m; i++)
{
ask[i].a = upper_bound(tmp, tmp + cnt, ask[i].a) - tmp;
ask[i].b = upper_bound(tmp, tmp + cnt, ask[i].b) - tmp;
}
sort(ask + 1, ask + m + 1);
Mo_Algorithm();
for (int i = 1; i <= m; i++)
write(ans[i]), putchar('
');
return 0;
}
}
int main()
{
return zyt::work();
}
分块:(136)行,本地(8)s,内存(28.5)MB,BZOJ上AC。这份代码是我知道最大权值(1e5)后写的,所以没有离散化。
#include <cstdio>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
namespace zyt
{
template<typename T>
inline void read(T &x)
{
char c;
bool f = false;
x = 0;
do
c = getchar();
while (c != '-' && !isdigit(c));
if (c == '-')
f = true, c = getchar();
do
x = x * 10 + c - '0', c = getchar();
while (isdigit(c));
if (f)
x = -x;
}
template<typename T>
inline void write(T x)
{
static char buf[20];
char *pos = buf;
if (x < 0)
putchar('-'), x = -x;
do
*pos++ = x % 10 + '0';
while (x /= 10);
while (pos > buf)
putchar(*--pos);
}
inline void write(const pair<int, int> &a)
{
write(a.first), putchar(' '), write(a.second);
}
const int N = 1e5 + 10, M = 1e6 + 10, BL = 510;
int n, m, arr[N], belong[N];
typedef pair<int, int> pii;
pii ans[M];
struct _ask
{
int l, r, a, b, id;
bool operator < (const _ask &b) const
{
return belong[l] == belong[b.l] ? r < b.r : l < b.l;
}
}ask[M];
int blsum[BL], blnum[BL], num[N], begin[BL];
inline void update(const int pos, const int x)
{
blsum[belong[pos]] += x;
if (x == -1 && !--num[pos])
--blnum[belong[pos]];
if (x == 1 && !num[pos]++)
++blnum[belong[pos]];
}
inline pii query(const int l, const int r)
{
pii ans = make_pair(0, 0);
if (belong[r] == belong[l])
{
for (int i = l; i <= r; i++)
if (num[i])
ans.first += num[i], ++ans.second;
}
else
{
for (int i = belong[l] + 1; i < belong[r]; i++)
ans.first += blsum[i], ans.second += blnum[i];
for (int i = l; i < begin[belong[l] + 1]; i++)
if (num[i])
ans.first += num[i], ++ans.second;
for (int i = begin[belong[r]]; i <= r; i++)
if (num[i])
ans.first += num[i], ++ans.second;
}
return ans;
}
void Mo_Algorithm()
{
int l = 1, r = 1, block = sqrt(N), blnum = ceil(N / (double)block);
for (int i = 1; i < N; i++)
belong[i] = (i - 1) / block + 1;
for (int i = 1; i <= blnum; i++)
begin[i] = (i - 1) * block + 1;
update(arr[1], 1);
for (int i = 1; i <= m; i++)
{
while (l < ask[i].l)
update(arr[l++], -1);
while (l > ask[i].l)
update(arr[--l], 1);
while (r < ask[i].r)
update(arr[++r], 1);
while (r > ask[i].r)
update(arr[r--], -1);
if (ask[i].a > ask[i].b)
ans[ask[i].id] = make_pair(0, 0);
else
ans[ask[i].id] = query(ask[i].a, ask[i].b);
}
}
int work()
{
int block;
read(n), read(m);
block = sqrt(n);
for (int i = 1; i <= n; i++)
{
read(arr[i]);
belong[i] = (i - 1) / block + 1;
}
for (int i = 1; i <= m; i++)
{
read(ask[i].l), read(ask[i].r), read(ask[i].a), read(ask[i].b);
ask[i].id = i;
}
sort(ask + 1, ask + m + 1);
Mo_Algorithm();
for (int i = 1; i <= m; i++)
write(ans[i]), putchar('
');
return 0;
}
}
int main()
{
return zyt::work();
}