2019 ICPC Universidad Nacional de Colombia Programming Contest 题解
目录
2019 ICPC Universidad Nacional de Colombia Programming Contest 题解
最近比较忙，题解写的有点水
A. Amazon
题意：给定 \(n\) 条直线，两条相互垂直的直线交点处要修建一个十字路口，询问需要修建几个十字路口。
分析：这题题意相当坑，并不是求相互垂直的直线的不同交点数量，而是不同的“十字路口数量”，这有什么区别呢？参考下图：
左边虽然有两对相互垂直的直线，但只要建一个“十字路口”，但右边需要建两个。
看明白题意之后这题就不难了，跟 CCPC2019 秦皇岛站 A. Angle Beats 的做法是一样的，把所有直线用最简形式的向量存储，删除共线后计数即可。
#include <bits/stdc++.h>
#define mp make_pair
#define ll long long
using namespace std;
void io() { ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); }
int __gcd(int a, int b) {
return b == 0 ? a : __gcd(b, a % b);
}
int main() {
io(); int t;
cin >> t;
while (t--) {
int n; cin >> n;
vector<pair<pair<int, int>, int> > p;
map<pair<int, int>, int> MP;
for (int i = 1; i <= n; ++i) {
int xa, ya, xb, yb;
cin >> xa >> ya >> xb >> yb;
int x = xb - xa, y = yb - ya;
int g = __gcd(x, y);
x /= g, y /= g;
if (y < 0) x = -x, y = -y;
else if (y == 0) x = abs(x);
p.emplace_back(mp(mp(x, y), xa * y - ya * x));
}
sort(p.begin(), p.end());
p.erase(unique(p.begin(), p.end()), p.end());
for (auto it : p) MP[it.first]++;
ll ans = 0;
for (auto it : p) {
int x = it.first.second, y = -it.first.first;
if (y < 0) x = -x, y = -y;
else if (y == 0) x = abs(x);
ans += MP[mp(x, y)];
}
cout << ans / 2ll << '\n';
}
}
B. Boring Non-Palindrome
题意：略。
分析：水题。
#include <bits/stdc++.h>
#define ll long long
using namespace std;
void io() { ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); }
int main() {
io(); string x, y;
cin >> x;
y = x; reverse(y.begin(), y.end());
for (int i = 0; i <= x.length(); ++i) {
string p = y.substr(y.size() - i, i);
string tmp = x + p;
string tp = tmp;
reverse(tp.begin(), tp.end());
if (tp == tmp) {
cout << tmp;
return 0;
}
}
}
C. Common Subsequence
题意：给定两个由 \(A,C,G,T\) 构成的长度为 \(n\) 的字符串，求出他们的最长公共子序列，然后判断这个子序列的长度和 \(0.99n\) 的大小关系。
分析：\(dp\) ，队友秒了。
#include<bits/stdc++.h>
#define ll long long
#define maxn 100100
#define mod 998244353
using namespace std;
int dp[1010][1010];
char s1[maxn], s2[maxn];
int main() {
scanf("%s%s", s1, s2);
int n = strlen(s1);
int m = n / 100 + 1;
int ans = 0;
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= m; j++) {
while (i + dp[i][j] < n && j + dp[i][j] < n && s1[i + dp[i][j]] == s2[j + dp[i][j]]) dp[i][j]++;
dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
dp[i][j + 1] = max(dp[i][j + 1], dp[i][j]);
ans = max(ans, dp[i][j]);
}
}
if (ans * 100 >= n * 99) printf("Long lost brothers D:\n");
else printf("Not brothers :(\n");
return 0;
}
D. Do Not Try This Problem
题意：给定一个字符串 \(s\) ，\(q\) 次修改，每次将 \(i+ak\) 这些位置的字母修改，询问修改完后的字符串。
分析：数据水，暴力剪枝假算法过了。
#define _CRT_SECURE_NO_WARNINGS
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#include <bits/stdc++.h>
#define mp make_pair
#define SIZE 100010
#define ll long long
using namespace std;
void io() { ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); }
vector<pair<int, char> > in[SIZE];
vector<pair<int, char> > out[SIZE];
int main() {
io(); string s;
cin >> s;
s = " " + s;
int q; cin >> q;
vector<int> lazy(s.length(), 0);
for (int i = 1; i <= q; ++i) {
int p, a, k; char c;
cin >> p >> a >> k >> c;
if (a == 1) {
in[p].emplace_back(mp(i, c));
out[p + k + 1].emplace_back(mp(i, c));
}
else {
for (int j = 0; j <= k; ++j) {
s[p + a * j] = c;
lazy[p + a * j] = i;
}
}
}
set<pair<int, char> > st;
for (int i = 1; i < s.length(); ++i) {
for (auto it : in[i])
if (!st.count(it))
st.insert(it);
for (auto it : out[i])
if (st.count(it))
st.erase(it);
if (st.size()) {
if (st.rbegin()->first > lazy[i]) cout << st.rbegin()->second;
else cout << s[i];
}
else cout << s[i];
}
}
E. Extreme Image
题意：求一个扇形（剪掉一个小扇形）区域内最多能覆盖多少点。
分析：\(POJ2482\) 翻版，从水平扫描线变成旋转扫描线，不过还是一个标准的扫描线+线段树裸题。
#include<bits/stdc++.h>
#define ll long long
#define maxn 100100
#define mod 998244353
#define eps 1e-8
using namespace std;
struct cv {
int x, y;
}a[maxn * 2];
bool cmp(cv p, cv q) {
return p.y < q.y;
}
int tr[maxn * 4], lz[maxn * 4];
void up(int x) {
tr[x] = max(tr[x * 2], tr[x * 2 + 1]);
}
void pe(int x) {
if (lz[x]) {
tr[x * 2] += lz[x];
tr[x * 2 + 1] += lz[x];
lz[x * 2] += lz[x];
lz[x * 2 + 1] += lz[x];
lz[x] = 0;
}
}
void cg(int x, int l, int r, int s, int e, int y) {
if (l >= s && r <= e) {
tr[x] += y;
lz[x] += y;
return;
}
pe(x);
int mid = (l + r) / 2;
if (mid >= s) cg(x * 2, l, mid, s, e, y);
if (mid < e) cg(x * 2 + 1, mid + 1, r, s, e, y);
up(x);
}
int main() {
int n, d, w;
double wi;
scanf("%d%d%lf", &n, &d, &wi);
w = (int)(wi * 100 + eps);
for (int i = 0; i < n; i++) {
scanf("%d%lf", &a[i].x, &wi);
a[i].y = (int)(wi * 100 + eps);
}
sort(a, a + n, cmp);
for (int i = 0; i < n; i++) {
a[i + n] = a[i];
a[i + n].y += 36000;
}
n *= 2;
int l = 0, r = 0, ans = 0;
memset(tr, 0, sizeof(tr));
memset(lz, 0, sizeof(lz));
while (r < n) {
int cnt = 0;
for (int i = r; i < n; i++) {
if (a[i].y == a[r].y) {
cg(1, 0, 100000, max(0, a[i].x - d), a[i].x, 1);
cnt++;
}
else break;
}
while (l < r) {
if (a[r].y - a[l].y > w) {
cg(1, 0, 100000, max(0, a[l].x - d), a[l].x, -1);
l++;
}
else break;
}
r += cnt;
ans = max(ans, tr[1]);
}
printf("%d\n", ans);
return 0;
}
F. Fraction Formula
题意：给出一个式子，求结果。
分析：分数类大模拟，注意必定爆 \(int\) ，如果运算顺序不好还会爆 \(long\ long\) 。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define int LL
const int maxn = 2e5 + 5;
char s[maxn];
struct Fac {
int x;
int y;
Fac() {}
Fac(int x, int y) : x(x), y(y) {}
void change() {
x *= -1;
}
void trans() {
if (!y) return;
if (!x) {
y = 1;
return;
}
int gcd = __gcd(x, y);
x /= gcd;
y /= gcd;
if (y < 0) {
y *= -1;
x *= -1;
}
}
Fac friend operator+(const Fac& a, const Fac& b) {
int lcm = a.y / __gcd(a.y, b.y) * b.y;
int x1 = a.x * (lcm / a.y), x2 = b.x * (lcm / b.y);
Fac res(x1 + x2, lcm);
res.trans();
return res;
}
void print() {
cout << x << "/" << y << '\n';
}
};
vector<Fac> res;
vector<int> fres;
signed main() {
while (~scanf("%s", s + 1)) {
res.clear();
int g = 1;
fres.clear();
fres.push_back(1);
int len = strlen(s + 1);
int f = g;
for (int i = 1; i <= len; i++) {
if (s[i] == '-') f *= -1;
else if (s[i] == '+') f *= 1;
else if (s[i] >= '0' && s[i] <= '9') {
int x = 0, y = 0, ff = 1;
while (s[i] != '/')
x = x * 10 + s[i] - '0',
i++;
i++;
if (s[i] == '-')
ff *= -1, i++;
while (s[i] >= '0' && s[i] <= '9')
y = y * 10 + s[i] - '0',
i++;
i--;
Fac a = Fac(x, y);
a.trans();
if (f * ff == -1) a.change();
f = g;
if (res.empty()) res.push_back(a);
else res.back() = res.back() + a;
} else if (s[i] == '(') {
fres.push_back(f);
g = fres.back();
f = g;
} else if (s[i] == ')') {
fres.pop_back();
g = fres.back();
f = g;
}
}
res.back().trans();
res.back().print();
}
return 0;
}
G. Graduation
我也不知道是啥，队友秒了。
#include<bits/stdc++.h>
#define ll long long
#define maxn 10010
#define mod 998244353
using namespace std;
vector<int>b[maxn];
int ans[maxn];
void sol(int x, int d) {
ans[d]++;
for (int i = 0; i < b[x].size(); i++) {
sol(b[x][i], d + 1);
}
}
int main() {
int n, k;
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; i++) {
int x;
scanf("%d", &x);
b[x].push_back(i);
}
sol(0, 0);
int r = maxn - 1;
while (!ans[r]) r--;
int sum = 0, num = 0;
for (int i = r; i > 0; i--) {
sum += ans[i];
num++;
while (num * k < sum) num++;
}
printf("%d\n", num);
return 0;
}
H. Hardest Challenge
题意：略。
分析：折半搜索。
#define _CRT_SECURE_NO_WARNINGS
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#include <bits/stdc++.h>
#define SIZE 5000100
#define ll long long
using namespace std;
void io() { ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); }
const ll mod = 1e15 + 37;
vector<string> s(3);
ll a[SIZE], b[SIZE], p[30];
int pa, pb, n, m;
void dfs(int pos, ll sum, ll a[], int& cnt, int m) {
if (pos == m) {
a[cnt++] = sum;
return;
}
for (int i = 0; i < 3; ++i) {
ll tmp = (sum + 1ll * s[i][pos] * p[n - 1 - pos] % mod) % mod;
dfs(pos + 1, tmp, a, cnt, m);
}
}
ll solve() {
for (int i = 0; i < 3; ++i) cin >> s[i];
n = s[0].length();
pa = pb = 0;
m = n >> 1;
dfs(0, 0, a, pa, m);
dfs(m, 0, b, pb, n);
sort(a, a + pa);
sort(b, b + pb);
a[pa] = a[0] + mod;
b[pb] = b[0] + mod;
int pos = pa;
ll ans = mod;
for (int i = 0; i < pb; ++i) {
while (pos > 0 && b[i] + a[pos - 1] >= mod) --pos;
ans = min(ans, a[pos] + b[i] - mod);
}
return ans;
}
int main() {
io(); p[0] = 1;
for (int i = 1; i < 30; ++i) p[i] = p[i - 1] * 127ll % mod;
int x; cin >> x >> x;
ll A = solve();
ll B = solve();
if (A < B) cout << "Owls";
else if (A > B) cout << "Goats";
else cout << "Tie";
}
I. Integer Prefix
题意：找最长的前缀数字。
分析：签到。
#include <bits/stdc++.h>
#define ll long long
using namespace std;
void io() { ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); }
int main() {
io(); string s;
cin >> s;
bool f = false;
for (auto i : s) {
if (i >= '0' && i <= '9') {
f = true;
cout << i;
}
else break;
}
if (!f) cout << "-1";
}
J. Jail Destruction
队友说是简单的线段树。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
typedef long long LL;
int a[maxn];
namespace Segement {
const LL inf = 0X3f3f3f3f3f3f3f3f;
LL tree[maxn << 2], vis[maxn << 2], h[maxn << 2];
LL lazy[maxn << 2];
void build(int root, int left, int right)
{
if (left == right) {
vis[root] = 1;
h[root] = tree[root] = a[left];
return;
}
int mid = (left + right) >> 1;
build(root << 1, left, mid);
build(root << 1 | 1, mid + 1, right);
h[root] = min(h[root << 1], h[root << 1 | 1]);
vis[root] = vis[root << 1] + vis[root << 1 | 1];
tree[root] = tree[root << 1] + tree[root << 1 | 1];
}
void pushdown(int root)
{
if (!lazy[root])
return;
h[root << 1] -= lazy[root];
h[root << 1 | 1] -= lazy[root];
tree[root << 1] -= vis[root << 1] * lazy[root];
tree[root << 1 | 1] -= vis[root << 1 | 1] * lazy[root];
lazy[root << 1] += lazy[root];
lazy[root << 1 | 1] += lazy[root];
lazy[root] = 0;
}
void update(int root, int left, int right, int stdl, int stdr, LL val)
{
if (left >= stdl && right <= stdr && val <= h[root]) {
h[root] -= val;
tree[root] -= vis[root] * val;
lazy[root] += val;
return;
}
if (left == right && h[root] < val) {
tree[root] = vis[root] = 0;
h[root] = inf;
return;
}
pushdown(root);
int mid = (left + right) >> 1;
if (stdl <= mid)
update(root << 1, left, mid, stdl, stdr, val);
if (stdr > mid)
update(root << 1 | 1, mid + 1, right, stdl, stdr, val);
h[root] = min(h[root << 1], h[root << 1 | 1]);
vis[root] = vis[root << 1] + vis[root << 1 | 1];
tree[root] = tree[root << 1] + tree[root << 1 | 1];
}
LL query(int root, int left, int right, int stdl, int stdr)
{
if (left >= stdl && right <= stdr) {
return tree[root];
}
pushdown(root);
LL res = 0;
int mid = (left + right) >> 1;
if (stdl <= mid)
res += query(root << 1, left, mid, stdl, stdr);
if (stdr > mid)
res += query(root << 1 | 1, mid + 1, right, stdl, stdr);
return res;
}
}
int main()
{
int n, q;
scanf("%d%d", &n, &q);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
Segement::build(1, 1, n);
int p, a, b, w;
while (q--) {
scanf("%d", &p);
if (p == 1) {
scanf("%d%d", &a, &b);
LL ans;
if (a <= b)
ans = Segement::query(1, 1, n, a, b);
else
ans = Segement::query(1, 1, n, a, n) + Segement::query(1, 1, n, 1, b);
printf("%lld\n", ans);
}
else {
scanf("%d%d%d", &a, &b, &w);
if (a <= b)
Segement::update(1, 1, n, a, b, w);
else
Segement::update(1, 1, n, a, n, w),
Segement::update(1, 1, n, 1, b, w);
}
}
return 0;
}
K. Kernel Of Love
队友说是签到。
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
scanf("%d", &t);
while (t--) {
int n;
scanf("%d", &n);
int ans = n / 3;
if (ans)
ans = (ans - 1) * 2 + 1;
if (n / 3 && n % 3)
ans++;
if (n >= 3)
ans++;
printf("%d\n", ans);
}
return 0;
}
L. Liquid X
题意：交互题，给定一堆试管，每次能够用这些试管加入试剂，加多了会变红，少了会变绿，正好会变黄，询问应该加多少试剂，或者判断不能求出应该加入多少试剂。
分析：用背包把所有可以表示的求出来，然后二分。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
bool dp[maxn];
int pre[maxn], p[maxn];
int n, a[105], vis[105];
int cnt[maxn];
void check(int mid)
{
int now = cnt[mid];
memset(vis, 0, sizeof(vis));
while (now) {
vis[p[now - pre[now]]]++;
now = pre[now];
}
printf("1\n");
for (int i = 1; i <= n; i++)
printf("%d ", vis[i]);
printf("\n");
fflush(stdout);
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int limit = 1e6;
dp[0] = true;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= limit - a[i]; j++)
if (dp[j])
dp[j + a[i]] = true,
pre[j + a[i]] = j;
}
int k = 0;
for (int i = 1; i <= limit; i++)
if (dp[i])
cnt[++k] = i;
for (int i = 1; i <= n; i++)
p[a[i]] = i;
char s[20];
int left = 1, right = k, mid, ans = -1;
while (left <= right) {
mid = (left + right) >> 1;
check(mid);
scanf("%s", s + 1);
if (s[1] == 'y') {
ans = cnt[mid];
break;
} else if (s[1] == 'g')
left = mid + 1;
else
right = mid - 1;
}
if (ans == -1) {
if (cnt[mid] == 2 && s[1] == 'r')
ans = 1;
if (cnt[mid] == limit - 1 && s[1] == 'g')
ans = limit;
if (abs(cnt[right] - cnt[left]) == 2)
ans = min(cnt[left], cnt[right]) + 1;
}
printf("2\n%d\n", ans);
fflush(stdout);
return 0;
}
