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(N^1.5logN). 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
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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

By wangyisong1996加强数据