题目链接
这题我很久之前用分块写过,没写出来。。
今天又看到了,于是下决心把这题做出来。
这次我用线段树写的,直接对每本书的编号Hash一下然后离散化然后各建一棵线段树,维护当前编号在某个位置有没有书,就行了。
为了卡空间,我用了(vector),同时指针建树,结构体里不保存当前节点维护的区间,区间作为参数递归,这样就能过了,空间复杂度应该是(O(N+M log N))。
另外Hash的边界搞大一点,第一次只弄了10W 80分,改成100W就A了。
#include <iostream>
#include <cstdio>
#include <vector>
#define re register
using namespace std;
inline int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') { s = s * 10 + ch - '0'; ch = getchar(); }
return s * w;
}
const int MAXN = 100010;
int num;
struct Seg_Tree{
int sum;
Seg_Tree *l, *r;
Seg_Tree() { l = NULL; r = NULL; sum = 0; }
void pushup(){
sum = 0;
if(l != NULL) sum += l->sum;
if(r != NULL) sum += r->sum;
}
void Add(int c, int p, int L, int R){
if(L == R){
sum += c;
return;
}
int mid = (L + R) >> 1;
if(p > mid){
if(r == NULL)
r = new Seg_Tree;
r->Add(c, p, mid + 1, R);
}
else{
if(l == NULL)
l = new Seg_Tree;
l->Add(c, p, L, mid);
}
pushup();
}
int Ask(int L, int R, int wl, int wr){
int ans = 0;
if(L >= wl && R <= wr) return sum;
if(L > wr || R < wl) return 0;
int mid = (L + R) >> 1;
if(l != NULL) ans += l->Ask(L, mid, wl, wr);
if(r != NULL) ans += r->Ask(mid + 1, R, wl, wr);
return ans;
}
};
vector <Seg_Tree> tree;
int n, m, v[MAXN * 10], id[MAXN * 10];
int getHash(int x){
int hash = x % 1000000;
if(!v[hash]) v[hash] = x;
else while(v[hash] != x && v[hash]) hash = (hash + 233) % 10000;
return hash;
}
int a, b, c;
int w[MAXN];
char ch;
int main(){
tree.push_back(Seg_Tree());
n = read(); m = read();
for(re int i = 1; i <= n; ++i){
w[i] = read();
re int hash = getHash(w[i]);
if(!id[hash]) id[hash] = ++num, tree.push_back(Seg_Tree());
tree[id[hash]].Add(1, i, 1, n);
}
for(re int i = 1; i <= m; ++i){
do{
ch = getchar();
}while(ch != 'C' && ch != 'Q');
if(ch == 'C'){
a = read(); b = read();
re int hash = getHash(w[a]);
tree[id[hash]].Add(-1, a, 1, n);
hash = getHash(b);
if(!id[hash]) id[hash] = ++num, tree.push_back(Seg_Tree());
tree[id[hash]].Add(1, a, 1, n);
w[a] = b;
}
else{
a = read(); b = read(); c = read();
re int hash = getHash(c);
if(!id[hash]) id[hash] = ++num, tree.push_back(Seg_Tree());
printf("%d
", tree[id[hash]].Ask(1, n, a, b));
}
}
return 0;
}