BZOJ 3236: [Ahoi2013]作业( 莫队 + BIT )

莫队..用两个树状数组计算.时间复杂度应该是O(N1.5logN). 估计我是写残了...跑得很慢...

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#include<bits/stdc++.h>
 
using namespace std;
 
#define lowbit(x) ((x) & -(x))
 
const int maxn = 100009;
const int maxm = 1000009;
 
int N, M, block[maxn], seq[maxn], ans[maxm][2], L, R;
 
inline int read() {
int ans = 0;
char c = getchar();
for(; !isdigit(c); c = getchar());
for(; isdigit(c); c = getchar())
   ans = ans * 10 + c - '0';
return ans;
}
 
int buf[12];
inline void write(int x) {
if(!x) {
putchar('0');
return;
}
int n = 0;
for(; x; x /= 10) buf[n++] = x % 10;
while(n--) putchar(buf[n] + '0');
}
 
struct BIT {
int b[maxn];
BIT() {
memset(b, 0, sizeof b);
}
inline void update(int x, int v) {
for(; x <= N; x += lowbit(x)) b[x] += v;
}
inline int sum(int x) {
int ret = 0;
for(; x; x -= lowbit(x)) ret += b[x];
return ret;
}
inline int query(int l, int r) {
return sum(r) - sum(l - 1);
}
} cnt, exist;
 
struct Q {
int l, r, a, b, p;
inline void Read(int t) {
l = read() - 1; r = read() - 1;
a = read(); b = read();
p = t;
}
bool operator < (const Q &o) const {
return block[l] < block[o.l] || (block[l] == block[o.l] && r < o.r);
}
} A[maxm];
 
void init() {
N = read(); M = read();
for(int i = 0; i < N; i++) seq[i] = read();
int blocks = (int) sqrt(N);
for(int i = 0; i < N; i++) block[i] = i / blocks;
for(int i = 0; i < M; i++) A[i].Read(i);
}
 
void work(int x) {
Q* c = A + x;
for(; L < c->l; L++) {
cnt.update(seq[L], -1);
if(!cnt.query(seq[L], seq[L])) exist.update(seq[L], -1);
}
while(L > c->l) {
L--;
if(!cnt.query(seq[L], seq[L])) exist.update(seq[L], 1);
cnt.update(seq[L], 1);
}
for(; R > c->r; R--) {
cnt.update(seq[R], -1);
if(!cnt.query(seq[R], seq[R])) exist.update(seq[R], -1);
}
while(R < c->r) {
R++;
if(!cnt.query(seq[R], seq[R])) exist.update(seq[R], 1);
cnt.update(seq[R], 1);
}
ans[c->p][0] = cnt.query(c->a, c->b);
ans[c->p][1] = exist.query(c->a, c->b);
}
 
int main() {
init();
sort(A, A + M);
cnt.update(seq[L = R = 0], 1);
exist.update(seq[0], 1);
for(int i = 0; i < M; i++) work(i);
for(int i = 0; i < M; i++) {
write(ans[i][0]);
putchar(' ');
write(ans[i][1]);
putchar('\n');
}
return 0;
}

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3236: [Ahoi2013]作业

Time Limit: 100 Sec  Memory Limit: 512 MB
Submit: 899  Solved: 345
[Submit][Status][Discuss]

Description

Input

Output

Sample Input

3 4
1 2 2
1 2 1 3
1 2 1 1
1 3 1 3
2 3 2 3

Sample Output

2 2
1 1
3 2
2 1

HINT


N=100000,M=1000000

Source

 

posted @ 2015-09-03 14:32  JSZX11556  阅读(206)  评论(0编辑  收藏  举报