將整個區間分爲sqrt(n)段小區間,每個段長度爲sqrt(n),並維護每一段的統計量,這樣可以在sqrt(n)時間內完成一次區間查詢或更新。平衡查詢和更新之間矛盾是數據結構設計時的重要考慮因素,而這種做法是經典的平衡做法,早就瞭解,但以前沒有遇到過只能用“分段”解決的題目,因此總覺着這是迫不得已的“賴招”,也從來沒有寫過,昨天第一次寫,小細節沒注意,結果悲劇的寫成每次查詢O(n)了。總結一下錯誤:
1.對於整個區間的更新,不需要實際處理每個元素的值,只需要用一個變量標記一下即可,當整段元素更新量不一致時再更新每個元素,這樣才能夠保證查詢時sqrt(n)的複雜度,否則就O(n)了。。。
2.如果使用map代替哈希表,如果元素不在map中不要直接查詢,因爲查詢不存在的元素會自動插入,空間複雜度會變爲n*n。
3.對於最後一段,長度可能與前邊的段不同,因此需要特殊處理最後一段的長度。
鑑於java中HashMap效率很客觀,自己懶得寫哈希表,使用map會增加複雜度,雖然oj支持不好而且不支持下標caoz,最終還是選擇了用java搞分段哈希
其它題目:
Spoj3261 Race Against Time修改某個值,查詢謀取區間內小於等於k的元素個數
解法:分段+排序/Treap
spoj 861 SWAPS動態維護逆序對
解法:分段/Treap+樹狀數組(http://blog.csdn.net/kksleric/article/details/7929903)
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.util.HashMap;
public class Hash {
int maxn = 100010, maxm = 335;
class Zone {
HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>();
int color[] = new int[maxm], len, all;
void init(int ln) {
mp.clear();
all = -1;
len = ln;
}
int get(int c) {
if (mp.containsKey(c))
return mp.get(c);
else
return 0;
}
void set(int c) {
all = c;
}
void update(int left, int right, int c) {
if (all == c)
return;
if (all != -1) {
for (int i = 1; i < left; i++)
color[i] = all;
for (int i = right + 1; i <= len; i++)
color[i] = all;
}
for (int i = left; i <= right; i++)
color[i] = c;
all = -1;
mp.clear();
for (int i = 1; i <= len; i++)
mp.put(color[i], get(color[i])+1);
}
int query(int left, int right, int c) {
if (all == -1) {
int sum = 0;
for (int i = left; i <= right; i++)
if (color[i] == c)
sum++;
return sum;
} else {
if (all == c)
return right - left + 1;
else
return 0;
}
}
}
Zone zz[] = new Zone[maxm];
int belong[] = new int[maxn], id[] = new int[maxn];
int a[] = new int[maxn], M;
void build(int n) {
for (int i = 1; i * i <= n; i++)
M = i;
int cnt = 0, len = M;
for (int i = 1; i <= n; i++) {
if (len == M) {
zz[++cnt].init(M);
len = 0;
}
belong[i] = cnt;
id[i] = ++len;
zz[cnt].color[len] = a[i];
zz[cnt].mp.put(a[i], zz[cnt].get(a[i]) + 1);
}
zz[cnt].len = len;
}
void update(int left, int right, int c) {
int l = belong[left], r = belong[right];
if (l == r) {
zz[r].update(id[left], id[right], c);
} else {
zz[l].update(id[left], zz[l].len, c);
for (int i = l + 1; i < r; i++)
zz[i].set(c);
zz[r].update(1, id[right], c);
}
}
int query(int left, int right, int c) {
int sum = 0;
int l = belong[left], r = belong[right];
if (l == r) {
sum += zz[l].query(id[left], id[right], c);
} else {
sum += zz[l].query(id[left], zz[l].len, c);
for (int i = l + 1; i < r; i++) {
if (zz[i].all == c)
sum += zz[i].len;
if(zz[i].all==-1)
sum += zz[i].get(c);
}
sum += zz[r].query(1, id[right], c);
}
return sum;
}
void run() throws IOException {
for (int i = 1; i < maxm; i++)
zz[i] = new Zone();
while (in.nextToken() != in.TT_EOF) {
int n = (int) in.nval;
int m = nextInt();
for (int i = 1; i <= n; i++)
a[i] = nextInt();
build(n);
int t, a, b, c;
while (m-- > 0) {
t = nextInt();
a = nextInt() + 1;
b = nextInt() + 1;
c = nextInt();
if (t == 1)
update(a, b, c);
else
System.out.println(query(a, b, c));
}
}
}
StreamTokenizer in = new StreamTokenizer(new BufferedReader(
new InputStreamReader(System.in)));
int nextInt() throws IOException {
in.nextToken();
return (int) in.nval;
}
public static void main(String[] args) throws IOException {
new Hash().run();
}
}