CODECHEF Nov. Challenge 2014 Chef & Churu
@(XSY)[分塊]
Hint: 題目原文是英文的, 寫得很難看, 因此翻譯為中文.
Input Format
First Line is the size of the array i.e. \(N\)
Next Line contains N space separated numbers \(A_i\) denoting the array
Next N line follows denoting \(Li\) and \(Ri\) for each functions.
Next Line contains an integer \(Q\) , number of queries to follow.
Next \(Q\) line follows , each line containing a query of Type 1 or Type 2.
1 x y : denotes a type 1 query,where x and y are integers
2 m n : denotes a type 2 query where m and n are integers
Output Format
For each query of type 2 , output as asked above.
Constraints
\(1 ≤ N ≤ 10^5\)
\(1 ≤ A i ≤ 10^9\)
\(1 ≤ L i ≤ N\)
$L i ≤ R i ≤ N $
\(1 ≤ Q ≤ 10^5\)
$1 ≤ x ≤ N $
\(1 ≤ y ≤ 10^9\)
$1 ≤ m ≤ N \(
\)m ≤ n ≤ N$
Subtask
Subtask \(1\): \(N ≤ 1000 , Q ≤ 1000\) , \(10\) points
Subtask \(2\): \(R-L ≤ 10\) , all x will be distinct ,\(10\) points
Subtask \(3\): Refer to constraints above , \(80\) points
Sample Input
5
1 2 3 4 5
1 3
2 5
4 5
3 5
1 2
4
2 1 4
1 3 7
2 1 4
2 3 5
Sample Output
41
53
28
Explanation
Functions values initially :
$F[1] = 1+ 2 + 3 = 6 \(
\)F[2] = 2 + 3 + 4 + 5 = 14 \(
\)F[3] = 4+5 = 9 \(
\)F[4] = 3+4+5 = 12 \(
\)F[5] = 1+2 = 3 $
Query \(1\): $F[1] + F[2] + F[3] + F[4] = 41 \(
After Update , the Functions are :
\)F[1] = 10 , F[2] = 18 , F[3] = 9 , F[4] = 16 , F[5] = 3 $
Query \(3\): $F[1] + F[2] + F[3] + F[4] = 53 $
Query \(4\): \(F[3]+F[4]+F[5] = 28\)
Solution
对\(a\)数组建立树状数组维护前缀和;
对函数进行分块处理. 维护两个数组, 其中\(sum[i]\)表示第\(i\)个块中的函数值的总和; \(cnt[i][j]\)表示第\(i\)个块中\(a[j]\)被累加的次数. \(cnt\)数组在预处理时可以通过累加前缀和的方法, \(O \left( n * sqrt(n) \right)\)完成. 而对于每次修改\(a[i]\)的值, 也可以在\(O \left(sqrt(n) * log(n) \right)\)的时间复杂度内完成维护.
#include<cstdio>
#include<cctype>
#include<cstring>
#include<cmath>
using namespace std;
inline long long read()
{
long long x = 0, flag = 1;
char c;
while(! isdigit(c = getchar()))
if(c == '-')
flag *= - 1;
while(isdigit(c))
x = x * 10 + c - '0', c = getchar();
return x * flag;
}
void println(long long x)
{
if(x < 0)
putchar('-'), x *= - 1;
if(x == 0)
putchar('0');
long long ans[1 << 5], top = 0;
while(x)
ans[top ++] = x % 10, x /= 10;
for(; top; top --)
putchar(ans[top - 1] + '0');
putchar('\n');
}
const long long N = 1 << 17;
long long n;
long long a[N];
long long L[N], R[N];
long long T[N];
inline void modify(long long u, long long x)
{
for(; u <= n; u += u & - u)
T[u] += (long long)x;
}
long long unit, num;
long long cnt[1 << 9][N];
long long sum[1 << 9];
void update(long long x, long long y)
{
for(long long i = 0; i < num; i ++)
sum[i] += (long long)cnt[i][x] * (y - a[x]);
modify(x, y - a[x]);
a[x] = y;
}
inline long long query(long long u)
{
long long ret = 0;
for(; u; u -= u & - u)
ret += T[u];
return ret;
}
long long ask(long long _L, long long _R)
{
long long lBlock = _L / unit, rBlock = _R / unit;
long long ret = 0;
if(lBlock == rBlock)
for(long long i = _L; i <= _R; i ++)
ret += query(R[i]) - query(L[i] - 1);
else
{
for(long long i = lBlock + 1; i < rBlock; i ++)
ret += sum[i];
for(long long i = _L; i < (lBlock + 1) * unit; i ++)
ret += query(R[i]) - query(L[i] - 1);
for(long long i = unit * rBlock; i <= _R; i ++)
ret += query(R[i]) - query(L[i] - 1);
}
return ret;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("chefAndChurus.in", "r", stdin);
freopen("chefAndChurus.out", "w", stdout);
#endif
n = read();
memset(T, 0, sizeof(T));
for(long long i = 1; i <= n; i ++)
modify(i, a[i] = read());
for(long long i = 0; i < n; i ++)
L[i] = read(), R[i] = read();
unit = (long long)sqrt(n);
long long cur = - 1;
memset(cnt, 0, sizeof(cnt));
for(long long i = 0; i < n; i ++)
{
if(i % unit == 0)
cur ++;
cnt[cur][L[i]] ++, cnt[cur][R[i] + 1] --;
}
num = cur + 1;
memset(sum, 0, sizeof(sum));
for(long long i = 0; i < num; i ++)
for(long long j = 1; j <= n; j ++)
{
cnt[i][j] += cnt[i][j - 1];
sum[i] += (long long)cnt[i][j] * a[j];
}
long long m = read();
for(long long i = 0; i < m; i ++)
{
long long opt = read(), x = read(), y = read();
if(opt == 1)
update(x, y);
else
println(ask(x - 1, y - 1));
}
}
