2018 Multi-University Training Contest 3 Solution
A - Problem A. Ascending Rating
题意：给出n个数，给出区间长度m。对于每个区间，初始值的max为0，cnt为0.遇到一个a[i] > ans, 更新ans并且cnt++。计算
$A = \sum_{i = 1}^{i = n - m +1} (max \oplus i)$
$B = \sum_{i = 1}^{i = n - m +1} (cnt \oplus i)$
思路：单调队列，倒着扫一遍，对于每个区间的cnt就是队列的长度，扫一遍即可。
#include<bits/stdc++.h> using namespace std; const int maxn = 1e7 + ;
typedef long long ll; int t;
int n,m,k;
ll p, q, r, mod;
ll arr[maxn];
ll brr[maxn]; int main()
{
scanf("%d", &t);
while(t--)
{
scanf("%d %d %d %lld %lld %lld %lld", &n, &m, &k, &p, &q, &r, &mod);
for(int i = ; i <= n; ++i)
{
if(i <= k) scanf("%lld", arr + i);
else arr[i] = (p * arr[i - ] + q * i + r) % mod;
}
ll A = , B = ;
int head = , tail = -;
for(int i = n; i > (n - m + ); --i)
{
while(head <= tail && arr[i] >= arr[brr[tail]]) tail--;
brr[++tail] = i;
}
for(int i = n - m + ; i >= ; --i)
{
while(head <= tail && arr[i] >= arr[brr[tail]]) tail--;
brr[++tail] = i;
while(brr[head] - i >= m) head++;
A += arr[brr[head]] ^ i;
B += (tail - head + ) ^ i;
}
printf("%lld %lld\n", A, B);
}
return ;
}
B - Problem B. Cut The String
留坑。
C - Problem C. Dynamic Graph Matching
题意：给出n个点，两种操作，一种是加边，一种是减边，每次加一条或者减一条，每次操作后输出1, 2, ..., n/2 表示对应的数量的边，有几种取法，并且任意两个边不能相邻
思路：每加一条边，都可以转移状态 比如说 110100 到 111110 就是 dp[111110] += dp[110100]
#include<bits/stdc++.h> using namespace std; typedef long long ll;
const int maxn = << ;
const int MOD = 1e9 + ; int t, n, m;
ll dp[maxn];
ll ans[]; inline int cal(int x)
{
int cnt = ;
while(x)
{
if(x & ) cnt++;
x >>= ;
}
return cnt;
} int main()
{
scanf("%d", &t);
while(t--)
{
memset(ans, , sizeof ans);
memset(dp, , sizeof dp);
dp[] = ;
scanf("%d %d", &n, &m);
while(m--)
{
char op[];
int u, v;
scanf("%s %d %d", op, &u, &v);
--u,--v;
for(int s = ; s < ( << n); ++s)
{
if((s & ( << u)) || (s & ( << v))) continue;
int S = s | ( << u);
S |= ( << v);
if(op[] == '+')
{
dp[S] = (dp[S] + dp[s]) % MOD;
ans[cal(S)] = (ans[cal(S)] + dp[s]) % MOD;
}
else if(op[] == '-')
{
dp[S] = (dp[S] - dp[s] + MOD) % MOD;
ans[cal(S)] = (ans[cal(S)] - dp[s] + MOD) % MOD;
}
}
for(int i = ; i <= n; i += )
{
printf("%lld%c", ans[i], " \n"[i == n]);
}
}
}
return ;
}
D - Problem D. Euler Function
打表找规律
#include <bits/stdc++.h> using namespace std; #define N 100010 inline int gcd(int a, int b)
{
return b == ? a : gcd(b, a % b);
} inline void Init()
{
for (int i = ; i <= ; ++i)
{
int res = ;
for (int j = ; j < i; ++j)
res += (gcd(i, j) == );
printf("%d %d\n", i, res);
}
} int t, n; int main()
{
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
if (n == ) puts("");
else printf("%d\n", n + );
}
return ;
}
E - Problem E. Find The Submatrix
留坑，
F - Problem F. Grab The Tree
题意：一棵树，每个点有点权，小Q可以选择若干个点，并且任意两个点之间没有边，小T就是剩下的所有点，然后两个人的值就是拥有的点异或起来，求小Q赢还是输还是平局
思路：显然，只有赢或者平局的情况，只要考虑是否有平局情况就可以，当所有点异或起来为0便是平局
#include <bits/stdc++.h>
using namespace std; #define N 100010 int t, n;
int arr[N]; int main()
{
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
int res = ;
for (int i = ; i <= n; ++i)
{
scanf("%d", arr + i);
res ^= arr[i];
}
for (int i = , u, v; i < n; ++i) scanf("%d%d", &u, &v);
if (res == )
{
puts("D");
continue;
}
puts("Q");
}
return ;
}
G - Problem G. Interstellar Travel
题意：给出n个点，要从$i -> j$ 当且仅当 $x_i < x_j$ 花费为 $x_i \cdot y_j - x_j \cdot y_i$ 求从$1 -> n$ 求权值最小如果多解输出字典序最小的解
思路：可以发现权值就是叉积，求个凸包，然后考虑在一条线上的情况
#include <bits/stdc++.h>
using namespace std; #define N 200010 const double eps = 1e-; inline int sgn(double x)
{
if (fabs(x) < eps) return ;
if (x < ) return -;
else return ;
} struct Point
{
double x, y; int id;
inline Point () {}
inline Point(double x, double y) : x(x), y(y) {}
inline void scan(int _id)
{
id = _id;
scanf("%lf%lf", &x, &y);
}
inline bool operator == (const Point &r) const { return sgn(x - r.x) == && sgn(y - r.y) == ; }
inline bool operator < (const Point &r) const { return x < r.x || x == r.x && y < r.y || x == r.x && y == r.y && id < r.id; }
inline Point operator + (const Point &r) const { return Point(x + r.x, y + r.y); }
inline Point operator - (const Point &r) const { return Point(x - r.x, y - r.y); }
inline double operator ^ (const Point &r) const { return x * r.y - y * r.x; }
inline double distance(const Point &r) const { return hypot(x - r.x, y - r.y); }
}; struct Polygon
{
int n;
Point p[N];
struct cmp
{
Point p;
inline cmp(const Point &p0) { p = p0; }
inline bool operator () (const Point &aa, const Point &bb)
{
Point a = aa, b = bb;
int d = sgn((a - p) ^ (b - p));
if (d == )
{
return sgn(a.distance(p) - b.distance(p)) < ;
}
return d < ;
}
};
inline void norm()
{
Point mi = p[];
for (int i = ; i < n; ++i) mi = min(mi, p[i]);
sort(p, p + n, cmp(mi));
}
inline void Graham(Polygon &convex)
{
sort(p + , p + n - );
int cnt = ;
for (int i = ; i < n; ++i)
{
if (!(p[i] == p[i - ]))
p[cnt++] = p[i];
}
n = cnt;
norm();
int &top = convex.n;
top = ;
if (n == )
{
top = ;
convex.p[] = p[];
convex.p[] = p[];
return;
}
if(n == )
{
top = ;
convex.p[] = p[];
convex.p[] = p[];
convex.p[] = p[];
return ;
}
convex.p[] = p[];
convex.p[] = p[];
top = ;
for (int i = ; i < n; ++i)
{
while (top > && sgn((convex.p[top - ] - convex.p[top - ]) ^ (p[i] - convex.p[top - ])) >= )
{
if(sgn((convex.p[top - ] - convex.p[top - ]) ^ (p[i] - convex.p[top - ])) == )
{
if(p[i].id < convex.p[top - ].id) --top;
else break;
}
else
{
--top;
}
}
convex.p[top++] = p[i];
}
if (convex.n == && (convex.p[] == convex.p[])) --convex.n;
}
}arr, ans; int t, n; int main()
{
scanf("%d", &t);
while (t--)
{
scanf("%d", &n); arr.n = n;
// cout << n << endl;
for (int i = ; i < n; ++i) arr.p[i].scan(i + );
arr.Graham(ans);
for (int i = ; i < ans.n; ++i) printf("%d%c", ans.p[i].id, " \n"[i == ans.n - ]);
}
return ;
}
H - Problem H. Monster Hunter
留坑。
I - Problem I. Random Sequence
留坑。
J - Problem J. Rectangle Radar Scanner
留坑。
K - Problem K. Transport Construction
留坑。
L - Problem L. Visual Cube
按题意模拟即可，分块解决
#include <bits/stdc++.h>
using namespace std; int t, a, b, c, n, m;
char ans[][]; int main()
{
scanf("%d", &t);
while (t--)
{
memset(ans, , sizeof ans);
scanf("%d%d%d", &a, &b, &c);
n = (b + c) * + ;
m = (a + b) * + ;
for (int i = n, cnt = ; cnt <= c; ++cnt, i -= )
{
for (int j = ; j <= * a; j += )
ans[i][j] = '+', ans[i][j + ] = '-';
}
for (int i = n - , cnt = ; cnt <= c; ++cnt, i -= )
{
for (int j = ; j <= * a; j += )
ans[i][j] = '|';
}
for (int i = * b + , tmp = , cnt = ; cnt <= b + ; ++cnt, i -=, tmp += )
{
for (int j = + tmp, cntt = ; cntt <= a; ++cntt, j += )
ans[i][j] = '+', ans[i][j + ] = '-';
}
for (int i = * b, tmp = , cnt = ; cnt <= b; ++cnt, tmp += , i -= )
{
for (int j = + tmp, cntt = ; cntt <= a; ++cntt, j += )
ans[i][j] = '/';
}
for (int j = m, cntt = , tmp = ; cntt <= b + ; ++cntt, j -= , tmp += )
{
for (int i = + tmp, cnt = ; cnt <= c + ; ++cnt, i += )
{
ans[i][j] = '+';
if (cnt <= c) ans[i + ][j] = '|';
}
}
for (int j = m - , tmp = , cntt = ; cntt <= b; ++cntt, tmp += , j -= )
{
for (int i = + tmp, cnt = ; cnt <= c + ; ++cnt, i += )
ans[i][j] = '/';
}
for (int i = ; i <= n; ++i)
{
for (int j = ; j <= m; ++j)
if (!ans[i][j]) ans[i][j] = '.';
ans[i][m + ] = ;
printf("%s\n", ans[i] + );
}
}
return ;
}
M - Problem M. Walking Plan
题意：给出一张图，询问从$u -> v 至少经过k条路的最少花费$
思路：因为$k <= 10000$ 那么我们可以拆成 $k = x * 100 + y$ 然后考虑分块
$G[k][i][j] 表示 i -> j 至少经过k条路$
$dp[k][i][j] 表示 i -> j 至少经过 k * 100 条路$
然后查询的时候枚举中间点
#include <bits/stdc++.h>
using namespace std; #define N 210
#define INFLL 0x3f3f3f3f3f3f3f3f
#define ll long long int t, n, m, q;
ll G[N][][], dp[N][][]; inline void Init()
{
memset(G, 0x3f, sizeof G);
memset(dp, 0x3f, sizeof dp);
} inline void Floyd()
{
for (int l = ; l <= ; ++l)
for (int k = ; k <= n; ++k)
for (int i = ; i <= n; ++i)
for (int j = ; j <= n; ++j)
G[l][i][j] = min(G[l][i][j], G[l - ][i][k] + G[][k][j]);
for (int i = ; i <= n; ++i)
for (int j = ; j <= n; ++j)
for (int l = ; l >= ; --l)
G[l][i][j] = min(G[l][i][j], G[l + ][i][j]); // at least K roads;
for (int i = ; i <= n; ++i)
for (int j = ; j <= n; ++j)
dp[][i][j] = G[][i][j];
for (int l = ; l <= ; ++l)
for (int k = ; k <= n; ++k)
for (int i = ; i <= n; ++i)
for (int j = ; j <= n; ++j)
dp[l][i][j] = min(dp[l][i][j], dp[l - ][i][k] + dp[][k][j]);
} inline void Run()
{
scanf("%d", &t);
while (Init(), t--)
{
scanf("%d%d", &n, &m);
for (int i = , u, v, w; i <= m; ++i)
{
scanf("%d%d%d", &u, &v, &w);
G[][u][v] = min(G[][u][v], (ll)w);
}
Floyd(); scanf("%d", &q);
for (int i = , u, v, k; i <= q; ++i)
{
scanf("%d%d%d", &u, &v, &k);
int unit = floor (k * 1.0 / ), remind = k - unit * ;
ll ans = INFLL;
if (k <= ) ans = G[k][u][v];
else
{
for (int j = ; j <= n; ++j)
ans = min(ans, dp[unit][u][j] + G[remind][j][v]);
}
if (k > && remind == ) ans = min(ans, dp[unit][u][v]);
printf("%lld\n", ans >= INFLL ? - : ans);
}
}
} int main()
{
#ifdef LOCAL
freopen("Test.in", "r", stdin);
#endif Run();
return ;
}
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