Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P8360 [SNOI2022] 军队

P8360 [SNOI2022] 军队

考虑分块,对于整块,操作时维护每个颜色的个数和答案和,并且将操作记录下来。对于散块,重构散块,将操作暴力实现。

考虑如何重构,按照倒序处理操作,记录每个颜色应该变为哪个颜色,以及每个颜色增加的军队。

查看代码
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#pragma GCC optimize ("O2")
#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...rest>
void read(Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write(Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
typedef long long LL;
const int N = 25e4 + 10, M = 510;
LL A[N], s[M], f[N];
int n, q, m, C[N];
int p[N];
int B, tot, id[N], L[N], R[N];
short c[M][N];
vector <int> op[M];
struct OP { int op, l, r, a, b; } o[N];
void rebuild (int x)
{
for (int i = L[x]; i <= R[x]; ++i) c[x][C[i]] = 0;
reverse(op[x].begin(), op[x].end());
for (int i : op[x])
if (o[i].op == 1)
p[o[i].a] = p[o[i].b], f[o[i].a] = f[o[i].b], c[x][o[i].b] = 0;
else if (o[i].op == 2) f[o[i].a] += o[i].b;
for (int i = L[x]; i <= R[x]; ++i)
A[i] += f[C[i]], ++c[x][C[i] = p[C[i]]];
for (int i : op[x])
p[o[i].a] = o[i].a, f[o[i].a] = 0;
op[x].clear();
}
void modify (int l, int r, int x, int y, int t)
{
if (id[l] == id[r])
{
rebuild(id[l]);
for (int i = l; i <= r; ++i) if (C[i] == x)
--c[id[l]][x], ++c[id[l]][C[i] = y];
return;
}
rebuild(id[l]), rebuild(id[r]);
for (int i = l; i <= R[id[l]]; ++i) if (C[i] == x)
--c[id[l]][x], ++c[id[l]][C[i] = y];
for (int i = id[l] + 1; i < id[r]; ++i)
{
c[i][y] += c[i][x], c[i][x] = 0;
op[i].push_back(t);
}
for (int i = L[id[r]]; i <= r; ++i) if (C[i] == x)
--c[id[r]][x], ++c[id[r]][C[i] = y];
}
void add (int l, int r, int x, int v, int t)
{
if (id[l] == id[r])
{
rebuild(id[l]);
for (int i = l; i <= r; ++i) if (C[i] == x)
A[i] += v, s[id[l]] += v;
return;
}
rebuild(id[l]), rebuild(id[r]);
for (int i = l; i <= R[id[l]]; ++i) if (C[i] == x)
A[i] += v, s[id[l]] += v;
for (int i = id[l] + 1; i < id[r]; ++i)
{
s[i] += (LL)v * c[i][x];
op[i].push_back(t);
}
for (int i = L[id[r]]; i <= r; ++i) if (C[i] == x)
A[i] += v, s[id[r]] += v;
}
LL query (int l, int r)
{
LL res = 0;
if (id[l] == id[r])
{
rebuild(id[l]);
for (int i = l; i <= r; ++i) res += A[i];
return res;
}
rebuild(id[l]), rebuild(id[r]);
for (int i = l; i <= R[id[l]]; ++i) res += A[i];
for (int i = id[l] + 1; i < id[r]; ++i) res += s[i];
for (int i = L[id[r]]; i <= r; ++i) res += A[i];
return res;
}
int main ()
{
read(n, q, m);
for (int i = 1; i <= n; ++i) read(A[i]);
for (int i = 1; i <= n; ++i) read(C[i]);
B = sqrt(n);
for (; R[tot] < n; ++tot)
L[tot + 1] = R[tot] + 1, R[tot + 1] = R[tot] + B;
R[tot] = n;
for (int i = 1; i <= tot; ++i)
for (int j = L[i]; j <= R[i]; id[j++] = i)
s[i] += A[j], ++c[i][C[j]];
for (int i = 1; i <= m; ++i) p[i] = i;
for (int i = 1; i <= q; ++i)
{
read(o[i].op, o[i].l, o[i].r);
if (o[i].op == 1)
{
read(o[i].a, o[i].b);
modify(o[i].l, o[i].r, o[i].a, o[i].b, i);
}
else if (o[i].op == 2)
{
read(o[i].a, o[i].b);
add(o[i].l, o[i].r, o[i].a, o[i].b, i);
}
else if (o[i].op == 3)
write(query(o[i].l, o[i].r)), puts("");
}
return 0;
}