LUOGU P4119 Ynoi2018 未来日记
题意
有一个长度为
1 l r x y，把区间内所有 变成 ； 2 l r k，查询区间内第 小值.
题解
考虑分块
对序列和值域同时分块，维护
对于散块每次查询构造
查询时对整块差分再加上散块的贡献，即可
先枚举
考虑修改
对于散块直接暴力修改
对于每个整块维护一个类似并查集的东西
给块内每一个值一个标号
把
- 块内没有
修改两个映射 - 块内有
暴力修改，通过势能分析证明总的暴力修改复杂度为
总的时间复杂度为
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using namespace std;
const int maxn = 100000, maxvs = 316;
const int maxs = 360, maxc = 278;
int cnt;
int a[maxn + 5];
int lp[maxc + 5], rp[maxvs + 5];
int bl[maxn + 5];
int vlp[maxvs + 5];
int vbl[maxn + 5];
int sum[maxc + 5][maxn + 5];
int blsum[maxc + 5][maxvs + 5];
int rep[maxc + 5][maxn + 5];
int col[maxc + 5][maxs + 5];
int id[maxn + 5];
int ccnt[maxc + 5];
void buildUnion(int bid) {
for (int i = 1; i <= ccnt[bid]; i++)
rep[bid][col[bid][i]] = 0;
ccnt[bid] = 0;
for (int i = lp[bid]; i <= rp[bid]; i++) {
if (rep[bid][a[i]] == 0) {
rep[bid][a[i]] = ++ccnt[bid];
col[bid][ccnt[bid]] = a[i];
}
id[i] = rep[bid][a[i]];
}
}
void reset(int bid) {
for (int i = lp[bid]; i <= rp[bid]; i++)
a[i] = col[bid][id[i]];
}
void build(int n) {
int siz = min(n, int(sqrt(1.3 * n)));
cnt = ceil(1. * n / siz);
for (int i = 1; i <= n; i++)
bl[i] = (i - 1) / siz + 1;
for (int i = 1; i <= cnt; i++) {
lp[i] = (i - 1) * siz + 1;
rp[i] = min(n, i * siz);
}
for (int i = 1; i <= maxn; i++)
vbl[i] = (i - 1) / maxvs + 1;
for (int i = 1; i <= maxvs + 1; i++)
vlp[i] = (i - 1) * maxvs + 1;
for (int i = 1; i <= cnt; i++) {
for (int j = 1; j <= maxn; j++)
sum[i][j] = sum[i - 1][j];
for (int j = 1; j <= maxvs + 1; j++)
blsum[i][j] = blsum[i - 1][j];
for (int j = lp[i]; j <= rp[i]; j++) {
sum[i][a[j]]++;
blsum[i][vbl[a[j]]]++;
}
buildUnion(i);
}
}
int tmp[maxn + 5], bltmp[maxvs + 5];
void add(int l, int r, int delta) {
for (int i = l; i <= r; i++) {
tmp[a[i]] += delta;
bltmp[vbl[a[i]]] += delta;
}
}
int query(int l, int r, int k) {
int lb = bl[l], rb = bl[r];
int res;
if (lb == rb) {
reset(lb);
copy(a + l, a + r + 1, tmp + l);
nth_element(tmp + l, tmp + l + k - 1, tmp + r + 1);
res = tmp[l + k - 1];
fill(tmp + l, tmp + r + 1, 0);
} else {
reset(lb);
reset(rb);
add(l, rp[lb], 1);
add(lp[rb], r, 1);
int i = 1, delta;
for (; k - (delta = bltmp[i] + blsum[rb - 1][i] - blsum[lb][i]) > 0;
i++)
k -= delta;
int j = vlp[i];
for (; k - (delta = tmp[j] + sum[rb - 1][j] - sum[lb][j]) > 0; j++)
k -= delta;
res = j;
add(l, rp[lb], -1);
add(lp[rb], r, -1);
}
return res;
}
void change(int l, int r, int x, int y, int bid) {
int chcnt = 0;
for (int i = l; i <= r; i++) {
if (a[i] == x) {
a[i] = y;
chcnt++;
}
}
sum[bid][x] -= chcnt;
sum[bid][y] += chcnt;
blsum[bid][vbl[x]] -= chcnt;
blsum[bid][vbl[y]] += chcnt;
}
void changeBlock(int l, int r, int x, int y, int bid) {
reset(bid);
change(l, r, x, y, bid);
buildUnion(bid);
}
void modify(int l, int r, int x, int y) {
int lb = bl[l], rb = bl[r];
if (x == y || sum[rb][x] - sum[lb - 1][x] == 0)
return;
for (int i = cnt; i >= lb; i--) {
sum[i][x] -= sum[i - 1][x];
sum[i][y] -= sum[i - 1][y];
blsum[i][vbl[x]] -= blsum[i - 1][vbl[x]];
blsum[i][vbl[y]] -= blsum[i - 1][vbl[y]];
}
if (lb == rb)
changeBlock(l, r, x, y, lb);
else {
changeBlock(l, rp[lb], x, y, lb);
changeBlock(lp[rb], r, x, y, rb);
for (int i = lb + 1; i < rb; i++) {
if (sum[i][x] == 0)
continue;
if (sum[i][y] == 0) {
col[i][rep[i][x]] = y;
swap(rep[i][y], rep[i][x]);
blsum[i][vbl[y]] += sum[i][x];
blsum[i][vbl[x]] -= sum[i][x];
sum[i][y] = sum[i][x];
sum[i][x] = 0;
} else
changeBlock(lp[i], rp[i], x, y, i);
}
}
for (int i = lb; i <= cnt; i++) {
sum[i][x] += sum[i - 1][x];
sum[i][y] += sum[i - 1][y];
blsum[i][vbl[x]] += blsum[i - 1][vbl[x]];
blsum[i][vbl[y]] += blsum[i - 1][vbl[y]];
}
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
build(n);
for (int i = 1; i <= m; i++) {
int op, l, r;
scanf("%d%d%d", &op, &l, &r);
if (op == 1) {
int x, y;
scanf("%d%d", &x, &y);
modify(l, r, x, y);
} else {
int k;
scanf("%d", &k);
printf("%d\n", query(l, r, k));
}
}
}
LUOGU P4119 Ynoi2018 未来日记