E. Warp (DP)
定义 $dp[n][i][j]$: the number of paths that end up in $(x,y)$ after $n$ teleportations
复杂度: $O(N^3(\log{N} + \log{M}))$
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#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, m;
cin >> n >> m;
vector<long long> dx(3), dy(3);
for (int i = 0; i < 3; i++)
{
cin >> dx[i] >> dy[i];
}
set<pair<long long, long long>> obstacles;
for (int i = 0; i < m; i++)
{
long long x, y;
cin >> x >> y;
obstacles.insert({x, y});
}
long long mod = 998244353;
// dp[n][i][j]: the number of paths that end up in (x,y) after n teleportations
map<pair<long long, long long>, long long> dp;
dp[{0, 0}] = 1;
for (int i = 0; i < n; i++)
{
map<pair<long long, long long>, long long> newdp;
for (auto & p : dp)
{
long long x = p.first.first, y = p.first.second;
long long val = p.second;
for (int j = 0; j < 3; j++)
{
long long nx = x + dx[j], ny = y + dy[j];
if (!obstacles.count({nx, ny}))
{
newdp[{nx, ny}] += val;
newdp[{nx, ny}] %= mod;
}
}
}
swap(dp, newdp);
}
long long ans = 0;
for (auto & p : dp)
{
ans += p.second;
ans %= mod;
}
cout << ans << endl;
return 0;
}
- 方法2
定义 $dp[i][j][k]$: the number of paths in which you make each of the moves $x$, $y$, and $z$ times, during the $n$ teleportations
复杂度: $O(N^3\log{M})$
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#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, m;
cin >> n >> m;
vector<long long> dx(3), dy(3);
for (int i = 0; i < 3; i++)
{
cin >> dx[i] >> dy[i];
}
set<pair<long long, long long>> obstacles;
for (int i = 0; i < m; i++)
{
long long x, y;
cin >> x >> y;
obstacles.insert({x, y});
}
long long mod = 998244353;
// dp[i][j][k]: the number of paths in which you make each of the moves x, y, and z times, during the n teleportations
vector<vector<long long>> dp(1, vector<long long>(1));
dp[0][0] = 1;
for (int i = 0; i < n; i++)
{
vector<vector<long long>> newdp(i + 2, vector<long long>(i + 2));
for (long long a = 0; a <= i; a++)
{
for (long long b = 0; b <= i; b++)
{
long long x = a * dx[0] + b * dx[1] + (i - a - b) * dx[2];
long long y = a * dy[0] + b * dy[1] + (i - a - b) * dy[2];
for (long long k = 0; k < 3; k++)
{
if (!obstacles.count({x + dx[k], y + dy[k]}))
{
newdp[a + (k == 0)][b + (k == 1)] += dp[a][b];
newdp[a + (k == 0)][b + (k == 1)] %= mod;
}
}
}
}
swap(dp, newdp);
}
long long ans = 0;
for (int a = 0; a <= n; a++)
{
for (int b = 0; b <= n; b++)
{
ans += dp[a][b];
ans %= mod;
}
}
cout << ans << endl;
return 0;
}