BZOJ 2724: [Violet 6]蒲公英( 分块 )

虽然AC了但是时间惨不忍睹...不科学....怎么会那么慢呢... 

无修改的区间众数..分块, 预处理出Mode[i][j]表示第i块到第j块的众数, sum[i][j]表示前i块j出现次数(前缀和,事实上我是写后缀和..因为下标从0开始..), cnt[i][j][k]表示第i块中的前j个数中,k出现次数。预处理O(N^1.5), 询问每次O(N^0.5), 总O((N+M)N^0.5)

-------------------------------------------------------------------

 #include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

#include<cctype>

 

using namespace std;

 

typedef pair<int, int> pii;

 

const int maxb = 209;

const int maxn = 40009;

 

struct HASH {

int h[maxn], n;

HASH() : n(0) {

}

void Add(int v) {

h[n++] = v;

}

void Work() {

sort(h, h + n);

n = unique(h, h + n) - h;

}

inline int Hash(int v) {

return lower_bound(h, h + n, v) - h;

}

inline int _Hash(int v) {

return h[v];

}

} H;

 

int read() {

char c = getchar();

int ret = 0;

for(; !isdigit(c); c = getchar()) c = getchar();

for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0';

return ret;

}

 

int seq[maxn];

int sum[maxb][maxn], cnt[maxb][maxb][maxb], c[maxn], Id[maxb][maxn];

int N, Q, B, n;

pii Mode[maxb][maxb]; // <numId, cnt>

 

void Init() {

N = read(); Q = read();

B = (int) sqrt(N);

n = N / B;

if(N % B) n++;

for(int i = 0; i < N; i++)

H.Add(seq[i] = read());

H.Work();

for(int i = 0; i < N; i++)

seq[i] = H.Hash(seq[i]);

memset(sum, 0, sizeof sum);

for(int i = 0; i < N; i++)

sum[i / B][seq[i]]++;

for(int i = n; i--; )

for(int j = 0; j < H.n; j++)

sum[i][j] += sum[i + 1][j];

for(int i = 0; i < n; i++) {

memset(c, 0, sizeof c);

int Max = 0, numId;

for(int j = i; j < n; j++) {

int p = j * B;

for(int k = 0; k < B; k++, p++) if(++c[seq[p]] > Max) {

numId = seq[p];

Max = c[seq[p]];

} else if(c[seq[p]] == Max) {

numId = min(numId, seq[p]);

}

Mode[i][j] = make_pair(numId, Max);

}

}

memset(cnt, 0, sizeof cnt);

memset(Id, -1, sizeof Id);

for(int i = 0; i < n; i++) {

Id[i][H.n] = 0;

int p = i * B;

for(int j = 0; j < B; j++, p++) {

int v = seq[p];

if(!~Id[i][v])

Id[i][v] = Id[i][H.n]++;

cnt[i][j][Id[i][v]]++;

}

for(int j = B; j--; )

for(int k = 0; k < Id[i][H.n]; k++)

cnt[i][j][k] += cnt[i][j + 1][k];

}

}

 

inline int getSum(int l, int r, int v) {

return sum[l][v] - sum[r + 1][v];

}

 

inline int getCnt(int Block, int l, int r, int numId) {

return cnt[Block][l][Id[Block][numId]] - cnt[Block][r + 1][Id[Block][numId]];

}

 

int solve(int l, int r) {

if(l > r) swap(l, r);

int lb = l / B, rb = r / B, ans, CNT = 0;

memset(c, 0, sizeof c);

if(rb - lb > 1) {

pii &t = Mode[lb + 1][rb - 1];

ans = t.first;

CNT = t.second;

for(int i = l; i / B == lb; i++) {

int v = seq[i], Cnt = getCnt(lb, l % B, B - 1, v) + getCnt(rb, 0, r % B, v);

Cnt += getSum(lb + 1, rb - 1, v);

if(Cnt > CNT)

ans = v, CNT = Cnt;

else if(Cnt == CNT)

ans = min(ans, v);

}

for(int i = rb * B; i <= r; i++) {

int v = seq[i], Cnt = getCnt(lb, l % B, B - 1, v) + getCnt(rb, 0, r % B, v);

Cnt += getSum(lb + 1, rb - 1, v);

if(Cnt > CNT)

ans = v, CNT = Cnt;

else if(Cnt == CNT)

ans = min(ans, v);

}

} else if(lb + 1 == rb) {

for(int i = l; i / B == lb; i++) {

int v = seq[i], Cnt = getCnt(lb, l % B, B - 1, v) + getCnt(rb, 0, r % B, v);

if(Cnt > CNT)

ans = v, CNT = Cnt;

else if(Cnt == CNT)

ans = min(ans, v);

}

for(int i = rb * B; i <= r; i++) {

int v = seq[i], Cnt = getCnt(lb, l % B, B - 1, v) + getCnt(rb, 0, r % B, v);

if(Cnt > CNT)

ans = v, CNT = Cnt;

else if(Cnt == CNT)

ans = min(ans, v);

}

} else if(lb == rb) {

for(; l <= r; l++) {

int v = seq[l], Cnt = getCnt(lb, l % B, r % B, v);

if(Cnt > CNT)

ans = v, CNT = Cnt;

else if(Cnt == CNT)

ans = min(ans, v);

}

}

return H._Hash(ans);

}

 

void Work() {

int ans = 0;

while(Q--) {

int l = read(), r = read();

printf("%d ", ans = solve((l + ans - 1) % N, (r + ans - 1) % N));

}

}

 

int main() {

Init();

Work();

return 0;

}

-------------------------------------------------------------------

2724: [Violet 6]蒲公英

Time Limit: 40 Sec  Memory Limit: 512 MB
Submit: 1140  Solved: 373
[Submit][Status][Discuss]

Description

Input

修正一下

l = (l_0 + x - 1) mod n + 1, r = (r_0 + x - 1) mod n + 1

Output

Sample Input

6 3
1 2 3 2 1 2
1 5
3 6
1 5

Sample Output

1
2
1

HINT

修正下：

n <= 40000, m <= 50000

Source

Vani原创