QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#611712 | #7030. Ultraman vs. Aodzilla and Bodzilla | ucup-team4906# | TL | 0ms | 0kb | C++20 | 5.9kb | 2024-10-04 22:17:23 | 2024-10-04 22:17:24 |
answer
#include<bits/stdc++.h>
#define F(i, a, b) for(int i = a; i <= b; i ++)
#define pii pair<int, int>
#define fi first
#define se second
#define int long long
using namespace std;
const int N = 3e5 + 10;
int n, m, is_q[N];
pii a[N];
pii res[N];
int ans[N], Y[N];
typedef array<int, 3> a3;
int f1[N], f2[N];
vector<a3> op1[N], op2[N];
int lowbit(int x) {return x & -x;}
void add1(int x, int d) {for(; x <= n; x += lowbit(x)) f1[x] += d;}
int cal1(int l, int r) {
if(l > r) return 0;
int res = 0; l --;
for(; r; r -= lowbit(r)) res += f1[r];
for(; l; l -= lowbit(l)) res -= f1[l];
return res;
}
void add2(int x, int d) {for(; x <= n; x += lowbit(x)) f2[x] += d;}
int cal2(int l, int r) {
if(l > r) return 0;
int res = 0; l --;
for(; r; r -= lowbit(r)) res += f2[r];
for(; l; l -= lowbit(l)) res -= f2[l];
return res;
}
void sol() {
cin >> n >> m;
F(i, 1, n) {
int x, y; cin >> x >> y;
a[i] = {x, y};
Y[x] = y;
}
F(i, 1, m) ans[i] = 0, is_q[i] = 0;
F(i, 1, n) op1[i].clear(), op2[i].clear(), f1[i] = f2[i] = 0;
if(n == 1) {
F(i, 1, m) {
char op; cin >> op;
if(op == '!') cout << "0\n";
else if(op == '?') {
int x; cin >> x;
cout << 1 << " " << 1 << "\n";
}
}
return;
}
int x1 = 1, y1 = 1, x2 = n, y2 = n, fx1 = 1, fx2 = n, fy1 = 1, fy2 = n;
int f1 = 0, f2 = 0;
// x1: 包含了哪些位置 fx1: 目前的实际位置
// f1 是否合并上下的 只保留fx1
int tot1 = 0, tot2 = 0;
F(i, 1, m) {
char op; cin >> op;
if(op == '!') {
is_q[i] = 0;
if(f1 && f2) ans[i] = n * (n - 1) / 2;
else if(f1) {
ans[i] = y1 + n - y2 + 1;
} else if(f2) {
ans[i] = x1 + n - x2 + 1;
} else {
// op1 表示小于等于这个x的限制
op1[fx1].push_back({fy1, 0, i});
op1[fx1].push_back({fy2, 1, i});
op2[fx2].push_back({fy1, 0, i});
op2[fx2].push_back({fy2, 1, i});
}
} else if(op == '?') {
is_q[i] = 2;
int p; cin >> p;
auto [x, y] = a[p];
if(f1) x = fx1;
else if(x <= x1) x = fx1;
else if(x >= x2) x = fx2;
else x -= tot1;
if(f2) x = fy1;
else if(y <= y1) y = fy1;
else if(y >= y2) y = fy2;
else y -= tot2;
res[i] = {x, y};
} else {
is_q[i] = 1;
int x; cin >> x;
if(op == 'U') {
tot1 += x;
if(f1) { // 当前已经合并了
fx1 = max(1ll, fx1 - x);
continue;
}
// 挪到1的位置
int nd = min(fx1 - 1, x);
x -= nd;
fx1 -= nd, fx2 -= nd;
while(x) {
x1 ++, fx2 --;
if(x1 == x2) {
assert(fx2 == 1);
f1 = 1;
break;
}
x --;
}
} else if(op == 'D') {
tot1 -= x;
if(f1) {
fx1 = min(n, fx1 + x);
continue;
}
int nd = min(n - fx2, x);
x -= nd;
fx1 += nd, fx2 += nd;
while(x) {
x2 --, fx1 ++;
if(x1 == x2) {
assert(fx1 == n);
assert(fx1 == fx2);
f1 = 1;
break;
}
x --;
}
} else if(op == 'L') {
tot2 += x;
if(f2) {
fy1 = max(1ll, fy1 - x);
continue;
}
int nd = min(fy1 - 1, x);
x -= nd;
fy1 -= nd, fy2 -= nd;
while(x) {
y1 ++, fy2 --;
if(y1 == y2) {
assert(fy2 == 1);
f2 = 1;
break;
}
x --;
}
} else {
tot2 -= x;
if(f2) {
fy1 = min(n, fy1 + x);
continue;
}
int nd = min(n - fy2, x);
x -= nd;
fy1 += nd, fy2 += nd;
while(x) {
y2 --, fy1 ++;
if(x1 == x2) {
assert(fy1 == n);
assert(fy1 == fy2);
f2 = 1;
break;
}
x --;
}
}
}
}
F(x, 1, n) {
add1(Y[x], 1);
for(auto [y, flag, id] : op1[x]) {
int sum = 0;
if(flag == 0) sum = cal1(1, y);
else sum = cal1(y, n);
ans[id] += 1ll * sum * (sum - 1) / 2;
}
}
for(int x = n; x >= 1; x --) {
add2(Y[x], 1);
for(auto [y, flag, id] : op2[x]) {
int sum = 0;
if(flag == 0) sum = cal2(1, y);
else sum = cal2(y, n);
ans[id] += 1ll * sum * (sum - 1) / 2;
}
}
F(i, 1, m) {
if(is_q[i] == 2) cout << res[i].fi << " " << res[i].se << "\n";
if(is_q[i] == 0) cout << ans[i] << "\n";
}
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t; cin >> t;
F(i, 1, t) sol();
return 0;
}
Details
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Test #1:
score: 0
Time Limit Exceeded
input:
2 5 15 5 25 5 15 25 5