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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 |
---|---|---|---|---|---|---|---|---|---|
#384377 | #6533. Traveling in Cells | mai123 | WA | 386ms | 5904kb | C++14 | 5.5kb | 2024-04-09 22:27:52 | 2024-04-09 22:27:53 |
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) {
if (T[node].val == T[node].r - T[node].l + 1) {
return T[node].l;
}
// 该区间未完全填充
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
}
// 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 + 1)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
}
signed main() {
ios;
int t;
cin >> t;
bool f = t > 100;
int cnt = 0;
while (t--) {
pos.clear();
int n, q;
cin >> n >> q;
if(f && cnt>=15){
cout<<n<<" "<<q<<endl;
}
for (int i = 1; i <= n; i++)tr[i] = 0;
for (int i = 1; i <= n; i++) {
cin >> c[i];
if(f && cnt>=15){
cout<<c[i]<<" ";
}
pos[c[i]].insert(i);
}
if(f && cnt>=15){
cout<<endl;
}
for (int i = 1; i <= n; i++) {
cin >> v[i];
if(f && cnt>=15){
cout<<v[i]<<" ";
}
add(i, v[i]);
}
if(f && cnt>=15){
cout<<endl;
}
while (q--) {
int op, p, x, k;
cnt++;
cin >> op;
if(f && cnt>=15)cout << op << " ";
if (op == 1) {
cin >> p >> x;
if(f && cnt>=15)cout << p << " " << x << endl;
//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) {
cin >> p >> x;
if(f && cnt>=15)cout << p << " " << x << endl;
int diff = x - v[p];
add(p, diff);
v[p] = x;
} else {
for (int i = 1; i <= (n << 2); i++)T[i] = {0, 0, 0};
build(1, 1, n);
cin >> x >> k;
if(f && cnt>=15)cout << x << " " << k << " ";
for (int i = 0; i < k; i++) {
int a;
cin >> a;
if(f && cnt>=15)cout << 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 = sum(right) - sum(left - 1);
if(f && cnt>=15)cout << endl;
else cout << ans << endl;
}
}
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 5904kb
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: -100
Wrong Answer
time: 386ms
memory: 5632kb
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 3 5 3 2 7 9 15 15 7 7 4 2 7 6 3 3 2 3 6 2 6 6 7 408865502 287523250 596421172 101361844 65318028 274851676 627677588 184057757 136346642 408433898 692051585 371882216 18263263 708799845 991435717 3 1...
result:
wrong answer 10th numbers differ - expected: '831633445', found: '3'