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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#384410 | #6533. Traveling in Cells | mai123 | TL | 846ms | 6112kb | C++14 | 5.0kb | 2024-04-09 23:09:24 | 2024-04-09 23:09:25 |
Judging History
answer
#include<bits/stdc++.h>
#define int long long
#define ios ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
using namespace std;
typedef pair<int, int> PII;
const int Maxn = 1e5 + 10;
int c[Maxn], v[Maxn];
unordered_map<int, unordered_set<int>> pos;//记录每个颜色的位置
// 树状数组维护区间和
int tr[Maxn];
int lowbit(int x) {
return x & -x;
}
void add(int x, int d) {
for (; x < Maxn; x += lowbit(x))tr[x] += d;
}
int sum(int x) {
int res = 0;
for (; x > 0; x -= lowbit(x))res += tr[x];
return res;
}
struct Node {
int l, r, val;
} T[Maxn << 2];
void pushUp(int node) {
T[node].val = T[node << 1].val + T[node << 1 | 1].val;
}
void build(int node, int l, int r) {
T[node] = {l, r, 0};
if (l == r)return;
int mid = (l + r) >> 1;
build(node << 1, l, mid);
build(node << 1 | 1, mid + 1, r);
pushUp(node);
}
void update(int node, int l, int r, int d) {
if (l <= T[node].l && T[node].r <= r) {
T[node].val = d;
return;
}
int mid = (T[node].l + T[node].r) >> 1;
if (l <= mid)update(node << 1, l, r, d);
if (mid < r)update(node << 1 | 1, l, r, d);
pushUp(node);
}
// 查询[l,r]区间内,最左边的完全填充区
int query_left(int node, int l, int r) {
int mid = (T[node].l + T[node].r) >> 1;
// 该节点被查询区间完全包括
if (l <= T[node].l && T[node].r <= r) {
// 区间和 等于 长度,代表该区间被全覆盖
if (T[node].val == T[node].r - T[node].l + 1)return T[node].l;
// 区间未覆盖
if (T[node].l == T[node].r)return T[node].r + 1;
// 先查询靠近x的区域
int left = query_left(node << 1 | 1, l, r);
if (left == mid + 1) left = query_left(node << 1, l, r);
return left;
} else if (l <= mid && mid < r) {
// 查询区间被当前节点包括,且中点被查询区间包括
int left = query_left(node << 1 | 1, l, r);
if (left == mid + 1)left = query_left(node << 1, l, r);
return left;
} else if (l <= mid)return query_left(node << 1, l, r);//查左子树
else if (mid < r)return query_left(node << 1 | 1, l, r);//查右子树
return INT_MAX;
}
// 查询[l,r]区间内,最右边的完全填充区
int query_right(int node, int l, int r) {
int mid = (T[node].l + T[node].r) >> 1;
// 该节点被查询区间完全包括
if (l <= T[node].l && T[node].r <= r) {
if (T[node].val == T[node].r - T[node].l + 1) {
return T[node].r;
}
// 该区间未完全填充
if (T[node].l == T[node].r)return T[node].l - 1;
int right = query_right(node << 1, l, r);
if (right == mid)right = query_right(node << 1 | 1, l, r);
return right;
} else if (l <= mid && mid < r) {
// 查询区间被当前节点包括,且中点被查询区间包括
int right = query_right(node << 1, l, r);
if (right == mid)right = query_right(node << 1 | 1, l, r);
return right;
} else if (l <= mid)return query_right(node << 1, l, r);//查左子树
else if (mid < r)return query_right(node << 1 | 1, l, r);//查右子树
return INT_MIN;
}
signed main() {
int t;
scanf("%lld", &t);
while (t--) {
pos.clear();
int n, q;
scanf("%lld%lld", &n, &q);
for (int i = 1; i <= n; i++)tr[i] = 0;
for (int i = 1; i <= n; i++) {
scanf("%lld", &c[i]);
pos[c[i]].insert(i);//记录每种颜色的所有位置
}
// 树状数组维护的是区间和
for (int i = 1; i <= n; i++) {
scanf("%lld", &v[i]);
add(i, v[i]);
}
while (q--) {
int op, p, x, k;
scanf("%lld", &op);
if (op == 1) {
scanf("%lld%lld", &p, &x);
// cp改成x
pos[c[p]].erase(p);
if (pos[c[p]].empty())pos.erase(c[p]);
pos[x].insert(p);
c[p] = x;
} else if (op == 2) {
scanf("%lld%lld",&p,&x);
// vp改成x
int diff = x - v[p];
add(p, diff);
v[p] = x;
} else {
build(1, 1, n);
scanf("%lld%lld", &x, &k);
for (int i = 0; i < k; i++) {
int a;
scanf("%lld", &a);
// 遍历颜色a的所有位置,并 +1
for (int c_pos: pos[a]) {
update(1, c_pos, c_pos, 1);
}
}
int left = query_left(1, 1, x);
int right = query_right(1, x, n);
int ans = 0;
if (left <= x)ans -= sum(left - 1);
if (x <= right)ans += sum(right);
printf("%lld\n", ans);
}
}
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 6112kb
input:
2 5 10 1 2 3 1 2 1 10 100 1000 10000 3 3 1 3 3 3 2 2 3 2 5 20000 2 3 200 3 3 2 1 3 3 3 3 1 2 3 1 3 4 2 1 100000 1 2 2 3 1 2 1 2 4 1 1 2 3 4 1000000 1000000 1000000 1000000 3 4 4 1 2 3 4
output:
100 110 1200 21211 100010 4000000
result:
ok 6 numbers
Test #2:
score: 0
Accepted
time: 235ms
memory: 6068kb
input:
20000 15 15 1 1 3 3 2 3 3 3 3 3 2 3 2 3 3 634593973 158136379 707704004 998369647 831633445 263092797 937841412 451774682 552617416 483763379 50360475 794662797 74247465 537217773 901809831 3 6 4 1 3 5 10 3 5 7 1 2 3 4 5 9 10 3 4 3 3 8 9 2 13 608033129 3 15 2 3 5 1 9 3 3 8 4 1 3 7 10 2 6 727472865 3...
output:
2689089686 8377825475 1706073651 1439027604 2689089686 792730352 8904867081 8904867081 8270273108 831633445 692051585 2782432626 697783016 883944422 184057757 287523250 184057757 696388752 184057757 1675459344 2667693025 2614711258 4006992373 1767091974 5348541057 5348541057 390631780 2290216252 942...
result:
ok 200062 numbers
Test #3:
score: 0
Accepted
time: 846ms
memory: 5900kb
input:
2000 150 150 8 3 8 8 8 6 8 4 2 7 6 8 8 5 8 7 7 8 5 6 8 8 6 8 8 8 8 7 8 6 6 8 8 8 6 2 3 4 8 8 7 8 5 8 2 6 8 7 8 8 6 8 6 8 3 8 8 8 8 4 7 8 7 3 7 6 7 5 5 8 6 8 8 6 3 8 6 7 6 8 8 7 4 8 6 7 8 7 7 7 7 8 8 8 8 2 5 2 8 8 6 7 6 3 8 8 7 8 8 8 6 6 8 6 6 7 5 8 8 8 7 8 7 7 6 8 8 8 8 8 8 6 5 7 5 5 8 7 8 7 7 7 6 5...
output:
4449391171 3290849667 852793841 5178673994 995994209 11431868919 4327723427 5071541023 3032743466 962345334 2997656427 4534278452 3851900075 3611231417 5071541023 1477584218 1299005818 1299005818 2145605244 854143763 886347565 2081234124 2333808475 2455955801 4179722063 2328504333 1473735464 4107685...
result:
ok 199987 numbers
Test #4:
score: -100
Time Limit Exceeded
input:
10 30000 30000 3 4 2 4 4 4 4 3 4 3 4 3 4 3 4 4 2 4 4 4 4 4 3 3 3 4 3 4 3 4 3 3 4 2 4 3 3 3 3 4 3 4 4 4 4 2 3 3 4 2 3 4 4 4 4 1 4 4 4 4 4 4 4 4 3 3 3 4 4 4 4 4 2 3 4 4 4 4 3 4 4 3 3 3 4 4 3 4 4 2 3 4 4 4 4 3 2 4 3 4 3 2 4 4 3 4 2 2 4 4 4 4 2 4 3 2 4 4 3 4 4 4 2 4 4 3 2 3 2 3 3 3 4 2 4 3 4 1 4 3 4 4 4...
output:
6959437173 934970676 72461245502 8365928740 8384151048 984567228 12482909122 1904927816 15134139942 3759040688 92670874909 332468911 5936663371 3562978848 1300592004 10314009201 5581540905 131246926443 15087184135864 4077066271 1124704817 1520626740 4388174158 744377942 2770411457 6231852240 1508724...