单点修改莫队 UVA - 12345【Dynamic len(set(a[L:R]))】
https://cn.vjudge.net/contest/304248#problem/L
题意
给定 n 个数和 m 次操作，数组下标 [0, n-1] ，操作分为两种：
Q x y —— 查询区间 [x, y] 内出现过几个值M x y —— 将 a[x] 修改成 y
分析
手贱全删没了...不想写了，下一篇里都差不多的...
代码
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e6+5;
int n, m;
int a[maxn];
int new_a[maxn];
int block;
int cnt, cnt_;
int ans[maxn];
int mark[maxn];
int sum[maxn];
int res;
struct node{
int l, r, id, t;
bool operator < (const node &a) const {
if(l/block == a.l/block && r/block == a.r/block) {
return t < a.t;
}
if(l/block == a.l/block) {
return r < a.r;
}
return l < a.l;
}
}Q[maxn];
struct Change {
int x, now, cur;
}C[maxn];
void init() {
block = (int)pow(n, 2.0/3);
memset(mark, 0, sizeof(mark));
memset(sum, 0, sizeof(sum));
cnt = cnt_ = 0;
res = 0;
}
void update(int x) {
if(mark[x]) {
if(sum[a[x]] == 1) {
res --;
}
sum[a[x]]--;
}
else {
if(sum[a[x]] == 0) {
res ++;
}
sum[a[x]] ++;
}
mark[x] ^= 1;
}
void change_time(int x, int val) {
if(mark[x]) {
update(x);
a[x] = val;
update(x);
}
else {
a[x] = val;
}
}
int main() {
// fopen("in.txt", "r", stdin);
// fopen("out.txt", "w", stdout);
while(~scanf("%d%d", &n, &m)) {
init();
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
new_a[i] = a[i];
}
for(int i = 1; i <= m; i++) {
char f[10];
int x, y;
scanf("%s%d%d", f, &x, &y);
x = x + 1;
if(f[0] == 'Q') {
Q[++cnt].l = x;
Q[cnt].r = y;
Q[cnt].id = cnt;
Q[cnt].t = cnt_;
}
else {
C[++cnt_].x = x;
C[cnt_].now = y;
C[cnt_].cur = new_a[x];
new_a[x] = y;
}
}
sort(Q+1, Q+1+cnt);
int l = 1, r = 0;
int Time = 0;
for(int i = 1; i <= cnt; i++) {
while(l < Q[i].l) {
update(l);
l++;
}
while(l > Q[i].l) {
l--;
update(l);
}
while(r < Q[i].r) {
r++;
update(r);
}
while(r > Q[i].r) {
update(r);
r--;
}
while(Time < Q[i].t) {
Time++;
change_time(C[Time].x, C[Time].now);
}
while(Time > Q[i].t) {
change_time(C[Time].x, C[Time].cur);
Time--;
}
ans[Q[i].id] = res;
}
for(int i = 1; i <= cnt; i++) {
printf("%d\n", ans[i]);
}
}
return 0;
}
转载于:https://www.cnblogs.com/Decray/p/10947035.html
