The 13th Shandong ICPC Provincial Collegiate Programming Contest
A. Orders
按照订单的结束时间排序，然后遍历一遍即可
#include<bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using i32 = int32_t;
void solve() {
int n, k;
cin >> n >> k;
vector<pii> p(n);
for (auto &[a, b]: p)
cin >> a >> b;
sort(p.begin(), p.end());
int cnt = 0, sum = 0;
for (int lst = 0; const auto &[a, b]: p) {
cnt += ( a - lst ) * k , sum += b , lst = a;
if( cnt < sum ) {
cout << "No\n";
return ;
}
}
cout << "Yes\n";
return;
}
i32 main() {
int TC;
for (cin >> TC; TC; TC--)
solve();
return 0;
}
B. Building Company
解法类似求拓扑序，只是约束关系比较复杂。
我们统计每一种员工不同的项目要求的数量。每次吸引到新的员工后就检查该种员工是否满足了新的项目要求。如果新的项目所有要求都被满足了，就把他加入到队列中。
#include<bits/stdc++.h>
using namespace std;
#define int long long
using i32 = int32_t;
using vi = vector<int>;
using pii = pair<int, int>;
i32 main() {
ios::sync_with_stdio(0), cin.tie(0);
int g;
cin >> g;
unordered_map<int, int> have;
for (int t, u; g; g--)
cin >> t >> u, have[t] += u;
int n;
cin >> n;
// 把所有的需求按照员工种类分类并排序
unordered_map<int, priority_queue<pii, vector<pii>, greater<pii>>> requirements;
// 每个项目还有多少个需求等待满足
vi cnt(n, 0);
// 完成每个项目可以吸引的员工数量
vector attract(n, vector<pii>());
queue<int> q;
for (int i = 0, k; i < n; i++) {
cin >> cnt[i];
for (int j = cnt[i], a, b; j; j--) {
cin >> a >> b;
if (have[a] >= b) {
cnt[i]--;
} else requirements[a].emplace(b, i);
}
if (cnt[i] == 0) q.push(i);
cin >> k, attract[i].resize(k);
for (auto &[c, d]: attract[i]) cin >> c >> d;
}
int res = 0;
for (int u; !q.empty();) {
u = q.front(), q.pop();
res++;
for (auto [c, d]: attract[u]) {
have[c] += d;
for (int v, id; !requirements[c].empty();) {
v = requirements[c].top().first, id = requirements[c].top().second;
if (v > have[c]) break;
requirements[c].pop(), cnt[id]--;
if (cnt[id] == 0) q.push(id);
}
}
}
cout << res << "\n";
return 0;
}
D. Fast and Fat
二分答案。
我们二分的最终的速度\(x\)，那么就可以把人分成两类，一类是速度大于\(x\)，另一类是速度小于\(x\)。
对于速度大于\(x\)的人，最大可以背负的人的重量为\(v+w-x\)。对于速度小于\(x\)的人，其速度是没有意义的，只需要记录下重量，然后贪心的优先满足重量大的选手，这样就可以判断出是否可以使得所有人的速度都大于\(x\)
#include<bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using i32 = int32_t;
const int inf = 1e9;
void solve() {
int n;
cin >> n;
vector<pii> p(n);
for (auto &[w, v]: p) cin >> v >> w;
int l = 0, r = inf, res = -1;
auto check = [p](int x) -> bool {
vector<int> a;
multiset<int> b;
for (const auto &[w, v]: p)
if (v < x) a.push_back(w);
else b.insert(v + w - x);
if (a.size() > b.size()) return false;
sort(a.begin(), a.end(), greater<int>());
for (const auto &i: a) {
if (i > *b.rbegin()) return false;
b.erase(b.lower_bound(i));
}
return true;
};
for (int mid; l <= r;) {
mid = (l + r) / 2;
if (check(mid)) res = mid, l = mid + 1;
else r = mid - 1;
}
cout << res << "\n";
return;
}
i32 main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int TC;
for (cin >> TC; TC; TC--)
solve();
return 0;
}
E. Math Problem
可以注意到的是先乘再除是没有意义的，所以最优解一定是先做除法。
可以枚举做除法的次数，然后计算出做乘法可以得到结果的区间，直到区间可以包含\(m\)时结束。
过程中会溢出，我这里使用了__int128
#include<bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using i32 = int32_t;
using i128 = __int128;
const int inf = 1e9;
istream &operator>>(istream &is, i128 &x) {
int y;
is >> y, x = y;
return is;
}
ostream &operator<<(ostream &os, i128 &x) {
int y = x;
os << y;
return os;
}
void solve() {
i128 n, k, m, a, b;
cin >> n >> k >> m >> a >> b;
if (n % m == 0) {
cout << "0\n";
return;
}
if (k == 1) {
cout << "-1\n";
return;
}
auto check = [m](i128 l, i128 r) {
i128 t = (l / m) * m;
if (l <= t and t <= r) return true;
t += m;
if (l <= t and t <= r) return true;
return false;
};
i128 res = LLONG_MAX;
for (i128 cnt = 0, sum = 0, l, r; true;) {
for (l = n, r = n, cnt = 0; check(l, r) == false and l <= r; l = l * k, r = r * k + k - 1, cnt += a);
res = min(res, cnt + sum);
if (n == i128(0)) break;
sum += b, n /= k;
}
cout << res << "\n";
return;
}
i32 main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int TC;
for (cin >> TC; TC; TC--)
solve();
return 0;
}
G. Matching
对于原始等式可以修改为\(i-a_i=j-a_j\)的形式，然后把点按照\(i-a_i\)的形式分类，贪心选择就好了
#include<bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using i32 = int32_t;
const int inf = 1e9;
void solve() {
int n;
cin >> n;
map<int, vector<int>> g;
for (int i = 1, a; i <= n; i++) {
cin >> a;
g[i - a].push_back(a);
}
int res = 0;
for (auto &[k, v]: g) {
sort(v.begin(), v.end(), greater<int>());
for (int i = 0; i < v.size(); i += 2)
if (i + 1 < v.size() and v[i] + v[i + 1] >= 0) res += v[i] + v[i + 1];
}
cout << res << "\n";
}
i32 main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int TC;
for (cin >> TC; TC; TC--)
solve();
return 0;
}
I. 三只骰子
暴力枚举
#include<bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using i32 = int32_t;
pii operator+(const pii &a, const pii &b) {
pii ans = {a.first + b.first, a.second + b.second};
return ans;
}
i32 main() {
vector<pii> T = {{1, 0},
{0, 2},
{0, 3},
{4, 0},
{0, 5},
{0, 6}};
pii t;
cin >> t.first >> t.second;
for (auto i: T)
for (auto j: T)
for (auto k: T)
if ((i + j + k) == t) {
cout << "Yes\n";
return 0;
}
cout << "No\n";
return 0;
}
L. Puzzle: Sashigane
首先我们可以一圈一圈的放置，直到目标点一定在边界上是，我们可以不断的包裹目标点，直到目标点变成一个在角落的正方形，然后再包裹这个正方形就好了。
#include<bits/stdc++.h>
using namespace std;
#define int long long
using pii = pair<int, int>;
using i32 = int32_t;
using i128 = __int128;
const int inf = 1e9;
using vi = vector<int>;
vector<array<int, 4>> res;
const int dx[] = {1, 1, -1, -1};
const int dy[] = {1, -1, 1, -1};
void op2(int sx, int sy, int tx, int ty, int x, int y) {
vi cnt(4);
for (int i = 0, px, py; i < 4; i++) {
px = x + dx[i], py = y + dy[i];
while (sx <= px and px <= tx and sy <= py and py <= ty)
cnt[i]++, px += dx[i], py += dy[i];
}
int t = -1;
for (int i = 0; i < 4; i++) {
if (cnt[i] == 0) continue;
if (t == -1 or cnt[i] < cnt[t]) t = i;
}
int d = tx - sx - cnt[t];
cnt[t] = 0;
for (int px = x + dx[t], py = y + dy[t], p = -1;
sx <= px and px <= tx and sy <= py and py <= ty; px += dx[t], py += dy[t], p--) {
res.push_back({px, py, p * dx[t], p * dy[t]});
}
t = -1;
for (int i = 0; i < 4; i++) {
if (cnt[i] == 0) continue;
if (t == -1) t = i;
}
if (t == -1 or d == 0 ) return;
for (int px = (dx[t] == -1 ? sx : tx), py = (dy[t] == -1 ? sy : ty), p = sx - tx;
sx <= px and px <= tx and sy <= py and py <= ty and d > 0;
px -= dx[t], py -= dy[t], p++, d--) {
res.push_back({px, py, p * dx[t], p * dy[t]});
}
return;
}
void op1(int sx, int sy, int tx, int ty, int x, int y) {
if (sx == x or sy == y or tx == x or ty == y) {
op2(sx, sy, tx, ty, x, y);
return;
}
int d = tx - sx;
res.push_back({sx, sy, d, d});
res.push_back({tx, ty, 1 - d, 1 - d});
op1(sx + 1, sy + 1, tx - 1, ty - 1, x, y);
return;
}
i32 main() {
ios::sync_with_stdio(false), cin.tie(nullptr);
int n, x, y;
cin >> n >> x >> y;
if (n == 1 and x == 1 and y == 1) {
cout << "Yes\n0\n";
return 0;
}
op1(1, 1, n, n, x, y);
cout << "Yes\n" << res.size() << "\n";
for (auto it: res) {
for (auto i: it) cout << i << " ";
cout << "\n";
}
return 0;
}
赛后看了官解，发现自己的做法太糟糕了
#include<bits/stdc++.h>
using namespace std;
#define int long long
using i32 = int32_t;
i32 main() {
ios::sync_with_stdio(0), cin.tie(0);
int n, l, r, u, d;
cin >> n >> u >> l, r = l, d = u;
cout << "Yes\n" << n - 1 << "\n";
for (int i = 1; i < n; i++) {
if (u > 1 and l > 1) {
u--, l--;
cout << u << " " << l << " " << d - u << " " << r - l << "\n";
} else if (u > 1 and r < n) {
u--, r++;
cout << u << " " << r << " " << d - u << " " << l - r << "\n";
} else if (l > 1) {
d++, l--;
cout << d << " " << l << " " << u - d << " " << r - l << "\n";
} else {
d++, r++;
cout << d << " " << r << " " << u - d << " " << l - r << "\n";
}
}
return 0;
}
J. Not Another Path Query Problem
思路类似数位 dp。
首先大于的\(V\)的整数\(v’\)在二进制下一定有一个前缀和\(V\)相同。我们枚举这个前缀，然后包含所有的使用包含这个前缀的所有边，这样每次再用 bfs 计算一下连通性就可以判断答案是否有解。
#include<bits/stdc++.h>
using namespace std;
#define int long long
using i32 = int32_t;
using vi = vector<int>;
using pii = pair<int, int>;
i32 main() {
ios::sync_with_stdio(0), cin.tie(0);
int n, m, q, V;
cin >> n >> m >> q >> V;
vector e(n + 1, vector<pii>());
for (int u, v, w; m; m--) {
cin >> u >> v >> w;
e[u].emplace_back(v, w);
e[v].emplace_back(u, w);
}
vector<pii> query(q);
for (auto &[a, b]: query) cin >> a >> b;
vi res(q);
auto calc = [&res, query, n, e, q](int t) {
vi col(n + 1);
auto bfs = [&col, e](int s, int t) {
queue<int> q;
q.push(s), col[s] = s;
for (int u; !q.empty();) {
u = q.front(), q.pop();
for (auto [v, w]: e[u]) {
if (col[v] or (w & t) != t) continue;
q.push(v), col[v] = s;
}
}
};
for (int i = 1; i <= n; i++)
if (!col[i]) bfs(i, t);
for (int i = 0; i < q; i++)
res[i] |= col[query[i].first] == col[query[i].second];
};
if (V == 0) calc(0);
else for (int t = V, T = (1ll << 60); t < T; t += (t & -t)) calc(t), cerr << t << "\n";
for (auto i: res)
cout << (i ? "Yes\n" : "No\n");
return 0;
}
