24-25 Winter Training
有些计算几何没写，有些字符串没写，有些题没补。
Long Round 1
原比赛链接：https://codeforces.com/gym/102770/
A
提前预处理日期的前缀和，模拟即可。
>#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#define MAXN 100010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
int ans;
bool leap_year(int x)
{
if (x % 100 != 0)
return (x % 4 == 0);
return (x % 400 == 0);
}
int yDay(int x, int y, int z)
{
if (y == 1 || y == 3 || y == 5 || y == 7 || y == 8 || y == 10 || y == 12)
return 31;
if (y == 2)
{
if (leap_year(x))
return 29;
return 28;
}
return 30;
}
int pd(int x, int y, int z)
{
// printf("x=%d y=%d z=%d\n", x, y, z);
if (x / 10 == 202)
return 1;
if (x % 1000 == 202)
return 1;
if (x % 100 == 20 && y / 10 == 2)
return 1;
if (x % 10 == 2 && y == 2)
return 1;
if (y == 20 && z / 10 == 2)
return 1;
if (y % 10 == 2 && z == 2)
return 1;
return 0;
}
int nextX(int x, int y, int z)
{
if (y == 12 && z == 31)
return x + 1;
return x;
}
int nextY(int x, int y, int z)
{
z++;
int day = yDay(x, y, z);
if (z > day)
{
y++;
if (y == 13)
y = 1;
}
return y;
}
int nextZ(int x, int y, int z)
{
z++;
int day = yDay(x, y, z);
if (z <= day)
return z;
return 1;
}
int sum[10000][13][32];
void solve()
{
ans = 0;
int X1 = read(), Y1 = read(), Z1 = read();
int X2 = read(), Y2 = read(), Z2 = read();
printf("%d\n", sum[X2][Y2][Z2] - sum[X1][Y1][Z1] + pd(X1, Y1, Z1));
}
void work()
{
int X1 = 2000, Y1 = 1, Z1 = 1;
int X2 = 9999, Y2 = 12, Z2 = 31;
while (X1 != X2 || Y1 != Y2 || Z1 != Z2)
{
int nx = nextX(X1, Y1, Z1);
int ny = nextY(X1, Y1, Z1);
int nz = nextZ(X1, Y1, Z1);
sum[nx][ny][nz] = sum[X1][Y1][Z1] + pd(nx, ny, nz);
X1 = nx, Y1 = ny, Z1 = nz;
// printf("nx=%d ny=%d nz=%d\n", nx, ny, nz);
}
}
int main()
{
// freopen("data.in", "r", stdin);
// freopen("A.out", "w", stdout);
work();
T = read();
while (T--)
solve();
return 0;
}
B
前者线段树，后者用set即可。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#define MAXN 1000010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
int n, C, a[MAXN], ans1, ans2;
int mx[MAXN << 2];
#define lson (now << 1)
#define rson ((now << 1) + 1)
void push_up(int now)
{
mx[now] = max(mx[lson], mx[rson]);
}
void ini(int now, int l, int r)
{
if (l == r)
mx[now] = C;
else
{
int mid = (l + r) >> 1;
ini(lson, l, mid);
ini(rson, mid + 1, r);
push_up(now);
}
}
void update(int now, int l, int r, int w)
{
if (l == r)
{
if (mx[now] == C)
ans1++;
mx[now] -= w;
}
else
{
int mid = (l + r) >> 1;
if (mx[lson] >= w)
update(lson, l, mid, w);
else
update(rson, mid + 1, r, w);
push_up(now);
}
}
map<int, int> mp;
void solve()
{
ans1 = 0, ans2 = 0;
mp.clear();
n = read(), C = read();
FUP(i, 1, n)
a[i] = read();
ini(1, 1, n);
FUP(i, 1, n)
{
update(1, 1, n, a[i]);
}
mp[C - a[1]] = 1;
ans2 = 1;
FUP(i, 2, n)
{
map<int, int>::iterator it = mp.lower_bound(a[i]);
if (it == mp.end())
{
mp[C - a[i]]++;
ans2++;
}
else
{
mp[it->first]--;
mp[it->first - a[i]]++;
if (mp[it->first] == 0)
mp.erase(it);
}
}
printf("%d %d\n", ans1, ans2);
}
int main()
{
T = read();
while (T--)
solve();
return 0;
}
C
大卡常题，卡吐了。反正就是拿trie去模拟，或者用unordered_map大概也行？我猜的。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#define MAXN 1010
#define MAXM 4000010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
int n, m, cnt;
ll sum[MAXM], ans;
char s[MAXN][MAXN];
int son[MAXM][26];
bool ok;
char tmps[MAXN];
void ins(string ss, int weight)
{
int now = 0;
int len = ss.length();
FUP(i, 0, len - 1)
{
if (!son[now][ss[i] - 'a'])
{
son[now][ss[i] - 'a'] = ++cnt;
}
now = son[now][ss[i] - 'a'];
}
// printf("now=%d\n", now);
sum[now] += weight;
}
void count(int len)
{
// printf("tmps:%s\n", tmps + 1);
int now = 0;
if (!len)
return;
FUP(i, 1, len)
{
if (!son[now][tmps[i] - 'a'])
{
if (T == 2)
puts("-1");
ok = false;
return;
}
now = son[now][tmps[i] - 'a'];
}
// printf("now=%d tot=%d sum=%lld\n", now, tot[now], sum[now]);
ans += sum[now];
}
void work1(int num)
{
int now = 1;
while (now <= n)
{
// printf("now=%d ns=", now);
int tmp = 0;
FUP(i, now, n)
{
if (s[num][i] == '#')
break;
tmps[++tmp] = s[num][i];
}
tmps[tmp + 1] = 0;
now += tmp + 1;
// cout << ns << endl;
count(tmp);
if (!ok)
return;
}
}
void work2(int num)
{
int now = 1;
while (now <= n)
{
int tmp = 0;
FUP(i, now, n)
{
if (s[i][num] == '#')
break;
tmps[++tmp] = s[i][num];
}
tmps[tmp + 1] = 0;
now += tmp + 1;
count(tmp);
if (!ok)
return;
}
}
void solve()
{
ok = true;
FUP(i, 0, cnt)
{
sum[i] = 0;
FUP(j, 0, 25)
son[i][j] = 0;
}
cnt = 0;
ans = 0;
n = read(), m = read();
FUP(i, 1, n)
scanf("%s", s[i] + 1);
FUP(i, 1, m)
{
string str;
int w = 0;
cin >> str >> w;
ins(str, w);
}
// FUP(i, 1, cnt)
// printf("i=%d tot=%d sum=%d\n", i, tot[i], sum[i]);
// line
FUP(i, 1, n)
{
// printf("line %d\n", i);
work1(i);
if (!ok)
return;
}
if (!ok)
return;
// printf("ans=%d\n", ans);
// column
FUP(i, 1, n)
{
// printf("column %d\n", i);
work2(i);
if (!ok)
return;
}
if (!ok)
return;
if (T == 2)
printf("%lld\n", ans);
}
int main()
{
// freopen("data.in", "r", stdin);
T = read();
FUP(tt, 1, T)
{
if (tt > 33700)
printf("tt=%d\n", tt);
solve();
}
return 0;
}
E
容易发现就是区间内前 \(k\) 大和一个平方和。于是使用可持久化平衡树即可。然而我忘记这个怎么写了，贺了以前自己的板子。但修改又怎么也修改不出来。很神奇啊！所以我选择了求前 \(len-k\) 小，然后用总和减去。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
#define MAXN 200010
#define INF 0x7fffffff
#define MOD 1000000007
#define ll long long
#define db double
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0')
{
w = w * 10 - '0' + ch, ch = getchar();
}
return w * flg;
}
int max(int x, int y) { return x > y ? x : y; }
int min(int x, int y) { return x < y ? x : y; }
int abs(int x) { return x < 0 ? (-x) : x; }
int num;
struct mpto
{
int data, id;
} a[MAXN];
struct mpbc
{
int d, ys;
} c[MAXN];
bool cmp(mpto x, mpto y)
{
return x.data < y.data;
}
int n, m, rt[MAXN];
struct zhuxishu
{
int ls[MAXN * 20], rs[MAXN * 20], siz[MAXN * 20];
ll sum[MAXN * 20];
int tot;
void build(int &p, int l, int r)
{
tot++;
p = tot;
if (l == r)
{
return;
}
int mid = (l + r) >> 1;
build(ls[p], l, mid);
build(rs[p], mid + 1, r);
}
void Insert(int &p, int pre, int l, int r, int x, int y)
{
tot++;
p = tot;
ls[p] = ls[pre];
rs[p] = rs[pre];
siz[p] = siz[pre] + 1;
// printf("%d,%d,%d,%d\n",l,r,p,siz[p]);
if (l == r)
{
sum[p] = sum[pre] + y;
return;
}
int mid = (l + r) >> 1;
if (mid >= x)
{
Insert(ls[p], ls[pre], l, mid, x, y);
}
else
{
Insert(rs[p], rs[pre], mid + 1, r, x, y);
}
sum[p] = sum[ls[p]] + sum[rs[p]];
}
int query(int p, int pre, int l, int r, int k)
{
if (l == r)
{
return c[l].ys * k;
}
int mid = (l + r) >> 1;
// printf("%d,%d,%d,%d,%d\n",p,l,r,ls[p],siz[ls[p]]);
if (siz[rs[p]] - siz[rs[pre]] >= k)
{
return query(rs[p], rs[pre], mid + 1, r, k);
}
return sum[rs[p]] - sum[rs[pre]] + query(ls[p], ls[pre], 1, mid, k - siz[rs[p]] + siz[rs[pre]]);
}
} T;
int w[MAXN];
int main()
{
n = read(), m = read();
FUP(i, 1, n)
{
a[i].data = read();
a[i].id = i;
w[i] = a[i].data;
}
// 离散化
sort(a + 1, a + n + 1, cmp);
a[0].data = INF;
FUP(i, 1, n)
{
if (a[i].data != a[i - 1].data)
{
num++;
}
c[a[i].id].d = num;
c[num].ys = a[i].data;
}
// cout<<num<<endl;
// 建主席树
T.build(rt[0], 1, num);
FUP(i, 1, n)
{
T.Insert(rt[i], rt[i - 1], 1, num, c[i].d, w[i]);
printf("i=%d sum=%lld\n", i, T.sum[rt[i]]);
// printf("%d,%d\n",rt[i],T.siz[rt[i]]);
}
// printf("%d",T.query(rt[1],rt[1],))
// 查询
FUP(i, 1, m)
{
int ql = read(), qr = read(), loc = read();
// loc+=T.siz[rt[ql-1]];
printf("%d\n", T.query(rt[qr], rt[ql - 1], 1, num, loc));
}
return 0;
}
F
诈骗题，如果两个 \(b\) 异号直接全 \(0\) 即可。否则你会发现我们很容易通过去修改某个向量的值去完成。要不然就是这个向量有 \(0\) 一类的，这时候再找个向量去配合一下就可以了。当且仅当这两个向量方向一致并且对应成比例的时候无解。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#define MAXN 100010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
int n;
double a[MAXN], b[MAXN];
double b1, b2;
void solve()
{
n = read();
FUP(i, 1, n)
a[i] = read();
b1 = read();
FUP(i, 1, n)
b[i] = read();
b2 = read();
if (b1 * b2 < 0)
{
FUP(i, 1, n)
printf("0 ");
puts("");
return;
}
FUP(i, 1, n)
{
if (a[i] * b[i] < 0)
{
FUP(j, 1, i - 1)
printf("0 ");
printf("10000 ");
FUP(j, i + 1, n)
printf("0 ");
puts("");
return;
}
if (a[i] == 0 && b[i] == 0)
continue;
if (a[i] != 0 && b[i] != 0 && a[i] * b2 != b[i] * b1)
{
FUP(j, 1, i - 1)
printf("0 ");
printf("%lf ", (-b1 / a[i] - b2 / b[i]) / 2);
FUP(j, i + 1, n)
printf("0 ");
puts("");
return;
}
// 1
if (a[i] == 0 && b1 > 0)
{
FUP(j, 1, i - 1)
printf("0 ");
if (b[i] > 0)
printf("-10000 ");
else
printf("10000 ");
FUP(j, i + 1, n)
printf("0 ");
puts("");
return;
}
// 2
if (b[i] == 0 && b2 > 0)
{
FUP(j, 1, i - 1)
printf("0 ");
if (a[i] > 0)
printf("-10000 ");
else
printf("10000 ");
FUP(j, i + 1, n)
printf("0 ");
puts("");
return;
}
// 3
if (a[i] == 0 && b1 < 0)
{
FUP(j, 1, i - 1)
printf("0 ");
if (b[i] > 0)
printf("10000 ");
else
printf("-10000 ");
FUP(j, i + 1, n)
printf("0 ");
puts("");
return;
}
// 4
if (b[i] == 0 && b2 < 0)
{
FUP(j, 1, i - 1)
printf("0 ");
if (a[i] > 0)
printf("10000 ");
else
printf("-10000 ");
FUP(j, i + 1, n)
printf("0 ");
puts("");
return;
}
}
puts("No");
}
int main()
{
T = read();
while (T--)
solve();
return 0;
}
G
dp题。考虑每次选择的向上吹的速度必须逐次增大，要不然你走了条劣于直线的路，中间经过的点还更菜，这不就亏麻了。而且因为你下一个点比现在更好，所以只要自己的高度正好飞到下个点，即到了之后速度为零就可以。所以按照向上速度建 \(DAG\) 然后跑dp就好了。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#define MAXN 4010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
db sx, sy, tx, ty;
db vf, vp, vh;
int n;
int deg[MAXN];
bool ari[MAXN];
db nx[MAXN], ny[MAXN], vu[MAXN];
vector<pair<int, db>> to[MAXN];
db dis(db a, db b, db c, db d)
{
return sqrt((c - a) * (c - a) + (d - b) * (d - b));
}
db re[MAXN];
void dfs(int u)
{
for (auto p : to[u])
{
int v = p.first;
re[v] = min(re[v], re[u] + p.second);
deg[v]--;
if (!deg[v])
dfs(v);
}
}
void cledfs(int u)
{
ari[u] = false;
for (auto p : to[u])
{
int v = p.first;
deg[v]--;
if (!deg[v])
cledfs(v);
}
}
void solve()
{
sx = read(), sy = read(), tx = read(), ty = read();
vf = read(), vp = read(), vh = read();
n = read();
FUP(i, 1, n + 2)
{
to[i].clear();
deg[i] = 0;
re[i] = LLINF;
ari[i] = true;
}
re[1] = 0;
FUP(i, 1, n + 1)
{
nx[i] = read(), ny[i] = read(), vu[i] = read();
}
nx[n + 2] = tx, ny[n + 2] = ty, vu[n + 2] = INF;
FUP(i, 1, n + 2)
{
if (vu[i] <= vp)
continue;
FUP(j, 1, n + 2)
{
if (i == j || vu[j] <= vp || vu[j] <= vu[i])
continue;
deg[j]++;
db t = dis(nx[i], ny[i], nx[j], ny[j]) / vh;
// printf("i=%d j=%d w=%.5lf\n", i, j, t + t * vp / (vu[i] - vp));
to[i].push_back(make_pair(j, t + t * vp / (vu[i] - vp)));
}
}
FUP(i, 2, n + 2)
{
if (deg[i] == 0)
cledfs(i);
}
FUP(i, 1, n + 2)
// printf("i=%d ari=%d\n", i, ari[i]);
dfs(1);
printf("%.10lf\n", re[n + 2]);
}
int main()
{
T = read();
while (T--)
solve();
return 0;
}
H
直接将每个点和每条线段去投影一下，然后把投影出的线段端点按坐标排序，维护每个阶段内有多少个点照不到自己就行了。注意考虑某个端点在自己这个点上方的情况，直接搞到2e9就行了。然后如果完全在自己上面，就不用考虑。如果覆盖住了这个点，就全图覆盖一下。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#define MAXN 510
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
int n, m;
db ans;
int cnt[MAXN];
int num;
struct Point
{
db x, y;
} p[MAXN];
struct Segment
{
Point u, v;
} s[MAXN];
int tot;
struct Shadow
{
db x;
int id, typ;
} sha[MAXN * MAXN * 2];
bool cmp(Shadow x, Shadow y)
{
return x.x < y.x;
}
void make_shadow(Point pf, Segment sf, int id)
{
Shadow s1, s2;
if ((pf.x - sf.u.x) * (sf.v.y - sf.u.y) == (pf.y - sf.u.y) * (sf.v.x - sf.u.x) && (pf.x - sf.u.x) * (pf.x - sf.v.x) <= 0)
{
s1.x = -2e9, s2.x = 2e9;
s1.id = id, s2.id = id;
s1.typ = 1, s2.typ = -1;
}
else
{
if (sf.u.y > sf.v.y)
swap(sf.u, sf.v);
if (sf.u.y > pf.y)
return;
s1.x = pf.x - pf.y * (pf.x - sf.u.x) / (pf.y - sf.u.y);
s1.id = id;
if (sf.v.y > pf.y)
{
db nx = sf.u.x + (pf.y - sf.u.y) * (sf.v.x - sf.u.x) / (sf.v.y - sf.u.y);
if (nx > pf.x)
s2.x = 2e9;
else
s2.x = -2e9;
}
else
{
s2.x = pf.x - pf.y * (pf.x - sf.v.x) / (pf.y - sf.v.y);
}
s2.id = id;
if (s1.x > s2.x)
swap(s1, s2);
s1.typ = 1, s2.typ = -1;
// printf("pf.x=%lf pf.y=%lf sf.u.x=%lf sf.u.y=%lf s1.x=%lf s2.x=%lf\n", pf.x, pf.y, sf.u.x, sf.u.y, s1.x, s2.x);
}
sha[++tot] = s1;
sha[++tot] = s2;
}
void solve()
{
ans = 0;
tot = 0;
num = 0;
FUP(i, 1, n)
cnt[i] = 0;
n = read(), m = read();
FUP(i, 1, n)
{
p[i].x = read();
p[i].y = read();
}
FUP(i, 1, m)
{
s[i].u.x = read(), s[i].u.y = read();
s[i].v.x = read(), s[i].v.y = read();
}
FUP(i, 1, n)
{
FUP(j, 1, m)
{
make_shadow(p[i], s[j], i);
}
}
sort(sha + 1, sha + tot + 1, cmp);
db lst = 0;
bool exi = 0;
FUP(i, 1, tot)
{
if (sha[i].typ == 1)
{
cnt[sha[i].id]++;
if (cnt[sha[i].id] == 1)
{
num++;
if (num == n)
{
exi = 1;
lst = sha[i].x;
}
}
}
else
{
cnt[sha[i].id]--;
if (cnt[sha[i].id] == 0)
{
num--;
if (num == n - 1)
{
if (exi)
{
exi = 0;
ans += sha[i].x - lst;
lst = 0;
}
}
}
}
// printf("i=%d x=%lf id=%d typ=%d num=%d lst=%lf ans=%lf\n", i, sha[i].x, sha[i].id, sha[i].typ, num, lst, ans);
}
if (ans > 1e9)
puts("-1");
else
printf("%.10lf\n", ans);
}
int main()
{
T = read();
while (T--)
solve();
return 0;
}
I
最后必定形成好几个环。只需考虑最大环即可。用 unordered_map 即可避免离散化。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <unordered_map>
#include <vector>
#define MAXN 100010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
int n, ans;
pair<int, int> pr[MAXN];
unordered_map<int, int> fa, cnt;
int find(int u)
{
if (fa[u] == 0)
return u;
return fa[u] = find(fa[u]);
}
void solve()
{
ans = 0;
fa.clear(), cnt.clear();
n = read();
FUP(i, 1, n)
{
int u = read(), v = read();
pr[i].first = u;
pr[i].second = v;
int ufa = find(u);
int vfa = find(v);
// printf("u=%d v=%d ufa=%d vfa=%d\n", u, v, ufa, vfa);
if (ufa == vfa)
continue;
fa[ufa] = vfa;
}
FUP(i, 1, n)
{
int ufa = find(pr[i].first);
int vfa = find(pr[i].second);
cnt[ufa]++;
cnt[vfa]++;
ans = max(ans, max(cnt[ufa], cnt[vfa]));
}
printf("%d\n", ans / 2);
}
int main()
{
// freopen("data.in", "r", stdin);
T = read();
while (T--)
solve();
return 0;
}
K
模拟判断即可。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#define MAXN 100010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
int n;
int a[MAXN], b[MAXN];
void solve()
{
n = read();
FUP(i, 1, n)
a[i] = read();
FUP(i, 1, n)
b[i] = read();
FUP(i, 1, n)
{
if (b[i] > a[i] * 3)
{
printf("%d\n", i);
return;
}
}
puts("-1");
}
int main()
{
T = read();
while (T--)
solve();
return 0;
}
Short Round 1
A
link:https://codeforces.com/problemset/problem/2000/B
模拟判断即可
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#define MAXN 200010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
int n, a[MAXN];
bool vis[MAXN];
void solve()
{
n = read();
FUP(i, 0, n + 1)
vis[i] = false;
FUP(i, 1, n)
a[i] = read();
vis[a[1]] = true;
FUP(i, 2, n)
{
if (!vis[a[i] - 1] && !vis[a[i] + 1])
{
puts("NO");
return;
}
vis[a[i]] = true;
}
puts("YES");
}
int main()
{
T = read();
while (T--)
solve();
return 0;
}
B
link:https://codeforces.com/problemset/problem/2028/B
忘了啥题了，光记得考虑一下 \(b=0\) 的情况了。别的情况显然都可以。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#define MAXN 100010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
ll read()
{
ll w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
ll n, b, c;
void solve()
{
n = read(), b = read(), c = read();
if (b == 0)
{
if (c < n - 2)
puts("-1");
else
{
if (c >= n)
printf("%lld\n", n);
else
printf("%lld\n", n - 1);
}
}
else
{
if (c >= n)
printf("%lld\n", n);
else
{
ll cnt = (n - c - 1) / b + 1;
printf("%lld\n", n - cnt);
}
}
}
int main()
{
T = read();
while (T--)
solve();
return 0;
}
C
link:https://codeforces.com/problemset/problem/1983/D
操作不改变逆序对奇偶性。奇偶性相同时，直接大力冒泡排序就可以证明一定可以。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#define MAXN 100010
#define MAXV 200010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int T;
int n, a[MAXN], b[MAXN], c[MAXN], d[MAXN];
int sum1[MAXV], sum2[MAXV];
int lbt(int x)
{
return x & (-x);
}
void add1(int x, int d)
{
for (int i = x; i <= 2e5; i += lbt(i))
{
sum1[i] += d;
}
}
void add2(int x, int d)
{
for (int i = x; i <= 2e5; i += lbt(i))
{
sum2[i] += d;
}
}
int query1(int x)
{
int re = 0;
for (int i = x; i > 0; i -= lbt(i))
{
re += sum1[i];
}
return re;
}
int query2(int x)
{
int re = 0;
for (int i = x; i > 0; i -= lbt(i))
{
re += sum2[i];
}
return re;
}
void solve()
{
n = read();
FUP(i, 1, n)
a[i] = c[i] = read();
FUP(i, 1, n)
b[i] = d[i] = read();
sort(c + 1, c + n + 1);
sort(d + 1, d + n + 1);
FUP(i, 1, n)
{
if (c[i] != d[i])
{
puts("NO");
return;
}
}
int cnt1 = 0, cnt2 = 0;
FUP(i, 1, n)
{
cnt1 = (cnt1 + i - 1 - query1(a[i])) % 2;
cnt2 = (cnt2 + i - 1 - query2(b[i])) % 2;
// printf("i=%d query1=%d query2=%d\n", i, i - 1 - query1(a[i]), i - 1 - query2(b[i]));
add1(a[i], 1);
add2(b[i], 1);
}
if (cnt1 == cnt2)
puts("YES");
else
puts("NO");
FUP(i, 1, n)
{
add1(a[i], -1);
add2(b[i], -1);
}
}
int main()
{
T = read();
while (T--)
solve();
return 0;
}
D
link:https://codeforces.com/problemset/problem/1635/E
只要有关系就一定方向相反。于是可以二分图染色确定方案。然后按照关系的种类确定谁在前谁在后。然后根据先后顺序建图，如果建出来个 \(DAG\) ，那么一定有解，否则无解。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#define MAXN 200010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
int n, m;
int uu[MAXN], vv[MAXN], typ[MAXN];
vector<int> to[MAXN];
int col[MAXN];
bool no;
void color(int u, int c)
{
col[u] = c;
for (auto v : to[u])
{
if (col[v] == 0)
color(v, -c);
else
{
if (col[v] * col[u] == 1)
{
no = true;
return;
}
}
}
}
vector<int> adj[MAXN];
int deg[MAXN];
int dfn[MAXN];
int cnt;
void dfs(int u)
{
dfn[u] = ++cnt;
for (auto v : adj[u])
{
deg[v]--;
if (deg[v] == 0)
dfs(v);
}
}
int main()
{
n = read(), m = read();
FUP(i, 1, m)
{
typ[i] = read(), uu[i] = read(), vv[i] = read();
to[uu[i]].push_back(vv[i]);
to[vv[i]].push_back(uu[i]);
}
FUP(i, 1, n)
{
if (col[i] != 0)
continue;
color(i, 1);
if (no)
{
puts("NO");
return 0;
}
}
FUP(i, 1, m)
{
if (col[uu[i]] == 1 && col[vv[i]] == -1)
{
if (typ[i] == 1)
{
adj[vv[i]].push_back(uu[i]);
deg[uu[i]]++;
}
else
{
adj[uu[i]].push_back(vv[i]);
deg[vv[i]]++;
}
}
else
{
if (typ[i] == 1)
{
adj[uu[i]].push_back(vv[i]);
deg[vv[i]]++;
}
else
{
adj[vv[i]].push_back(uu[i]);
deg[uu[i]]++;
}
}
}
FUP(i, 1, n)
{
if (dfn[i])
continue;
if (deg[i])
continue;
dfs(i);
}
if (cnt < n)
{
puts("NO");
return 0;
}
puts("YES");
FUP(i, 1, n)
{
if (col[i] == 1)
printf("R %d\n", dfn[i]);
else
printf("L %d\n", dfn[i]);
}
return 0;
}
E
links:https://codeforces.com/problemset/problem/1981/D
很有意思的题。首先发现我们应该选择质数，因为质数可以保证乘积分解的唯一性，不会出现类似于 \(6\times 2=3\times 4\) 一类的神经病情况。选了质数之后，就建图，找欧拉路径。注意这里可以有自环，但自环这个度数不属于欧拉路径存在的度数要求里的那个度数。然后注意需要奇数个质数的时候，每个点度数都是偶数，完全ok。偶数边就不大行了，所以需要去掉 \(2-3、4-5\cdot\cdot\cdot\) ，当然 \(1-n\) 需要留着，因为我们想要的是欧拉路径而非欧拉回路。然后跑那个 \(h\) 开头的字母的算法就好了。
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#define MAXN 300010
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define db double
#define FUP(i, x, y) for (int i = (x); i <= (y); i++)
#define FDW(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
int read()
{
int w = 0, flg = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
flg = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
w = w * 10 + ch - '0', ch = getchar();
return w * flg;
}
const int VN = 300000;
int pn, pri[MAXN];
bool vis[MAXN];
void init()
{
FUP(i, 2, VN)
{
if (!vis[i])
{
pri[++pn] = i;
}
FUP(j, 1, pn)
{
if (i * pri[j] > VN)
break;
vis[i * pri[j]] = true;
if (i % pri[j] == 0)
break;
}
}
}
int T;
bool ck(int mid, int n)
{
if (mid & 1)
{
if (1ll * mid * (mid + 1) / 2 >= n - 1)
return true;
return false;
}
else
{
if (1ll * mid * (mid + 1) / 2 - mid / 2 + 1 >= n - 1)
return true;
return false;
}
}
int head[MAXN], ednum;
int edid;
bool edvis[2100010];
struct edge
{
int nxt, to, id;
} ed[2100010];
void add_Edge(int u, int v, int mat)
{
// printf("u=%d v=%d mat=%d\n", u, v, mat);
ednum++;
ed[ednum].nxt = head[u];
ed[ednum].to = v;
ed[ednum].id = edid;
if (mat)
{
edid++;
ed[ednum].id++;
}
head[u] = ednum;
}
vector<int> ans;
int n;
void dfs(int u)
{
// printf("u=%d\n", u);
for (int &i = head[u]; i; i = ed[i].nxt)
{
int v = ed[i].to;
int id = ed[i].id;
if (edvis[id])
continue;
// printf("u=%d v=%d id=%d\n", u, v, id);
edvis[id] = true;
dfs(v);
if ((int)ans.size() == n)
return;
}
if ((int)ans.size() == n)
return;
ans.push_back(u);
}
void solve(int num)
{
n = read();
if (n == 2)
{
puts("1 1");
return;
}
if (n == 3)
{
puts("1 1 2");
return;
}
if (n == 4)
{
puts("1 1 2 2");
return;
}
int l = 1, r = n;
int re = -1;
while (l <= r)
{
int mid = (l + r) >> 1;
if (ck(mid, n))
re = mid, r = mid - 1;
else
l = mid + 1;
}
// printf("re=%d\n", re);
if (re & 1)
{
FUP(i, 1, re)
add_Edge(i, i, 1);
FUP(i, 1, re)
{
FUP(j, i + 1, re)
{
add_Edge(i, j, 1);
add_Edge(j, i, 0);
}
}
}
else
{
FUP(i, 1, re)
add_Edge(i, i, 1);
FUP(i, 1, re)
{
if (i % 2 == 1)
{
add_Edge(i, i + 1, 1);
add_Edge(i + 1, i, 0);
}
FUP(j, i + 2, re)
{
add_Edge(i, j, 1);
add_Edge(j, i, 0);
}
}
}
dfs(1);
FUP(i, 1, re)
{
head[i] = 0;
}
FUP(i, 1, ednum)
{
ed[i].nxt = ed[i].to = ed[i].id = 0;
}
for (auto x : ans)
printf("%d ", pri[x]);
puts("");
FUP(i, 1, edid)
edvis[i] = 0;
ednum = edid = 0;
ans.clear();
}
int main()
{
init();
T = read();
FUP(i, 1, T)
solve(i);
return 0;
}
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