Problem T2. 作业
考虑状压 DP
\(f[mask]\) 表示当前已经排好序的颜色状态为 \(mask\),最少的交换次数。
预处理移动的最小代价即可。
时间复杂度 \(\mathcal O(400\times 2^{20}+20n\log n)\)
#include
using namespace std;
const int N = 300005, M = 21;
int n;
vector pos[M];
long long _move[M][M];
long long dp[1 << M];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n;
for (int i = 1; i <= n; ++i) {
int x;
cin >> x;
pos[x].push_back(i);
}
for (int i = 1; i <= 20; ++i) {
for (int j = 1; j <= 20; ++j) {
if (i == j)
continue;
_move[i][j] = 0;
for (int k = 0; k < (int)pos[i].size(); ++k) {
if (pos[j].size() == 0 || pos[j][0] > pos[i][k])
continue;
_move[i][j] += lower_bound(pos[j].begin(), pos[j].end(), pos[i][k]) - pos[j].begin();
}
}
}
memset(dp, 0x3f, sizeof dp);
dp[0] = 0;
for (int mask = 0; mask < (1 << 20); ++mask) {
for (int j = 0; j < 20; ++j) {
if (!(mask >> j & 1)) {
long long res = 0;
for (int k = 0; k < 20; ++k) {
if (mask >> k & 1) {
res += _move[k + 1][j + 1];
}
}
dp[mask | (1 << j)] = min(dp[mask | (1 << j)], dp[mask] + res);
}
}
}
cout << dp[(1 << 20) - 1] << '\n';
return 0;
}
Problem T3. Five-seventeen
将两个路径拆成向上和向下的路径,考虑树上倍增维护答案
为了减少边界的处理,令 \(prep_{i,j}\) 表示从 \(i\) 的父亲开始,向上倍增 \(j\) 次的答案
时间复杂度 \(\mathcal O(10\times n\log n)\)
#include
using namespace std;
const int N = 1e5 + 1;
struct edge {
int to, nxt;
} e[N << 1];
int head[N];
int edge_cnt;
void add_edge(int u, int v) {
e[++edge_cnt] = (edge){v, head[u]};
head[u] = edge_cnt;
}
const int LOG = 17;
int n, m, q;
int bz[LOG + 1][N][11];
vector tag[N];
int father[LOG + 1][N];
int dep[N];
void Dfs(int u, int Father) {
father[0][u] = Father;
dep[u] = dep[Father] + 1;
for (int i = 1; i <= LOG; ++i)
father[i][u] = father[i - 1][father[i - 1][u]];
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (v == Father)
continue;
Dfs(v, u);
}
}
int lca(int u, int v) {
if (dep[u] < dep[v])
swap(u, v);
for (int i = LOG; i >= 0; --i)
if (dep[father[i][u]] >= dep[v])
u = father[i][u];
if (u == v)
return u;
for (int i = LOG; i >= 0; --i)
if (father[i][u] != father[i][v])
u = father[i][u], v = father[i][v];
return father[0][u];
}
int c[11];
void merge(int* a, int* b) {
int x = 1, y = 1;
c[0] = 0;
for (int i = 1; i <= 10; ++i) {
if (x == a[0] + 1 && y == b[0] + 1)
break;
else if (x == a[0] + 1)
c[++c[0]] = b[y++];
else if (y == b[0] + 1)
c[++c[0]] = a[x++];
else
c[++c[0]] = a[x] < b[y] ? a[x++] : b[y++];
}
for (int i = 0; i <= c[0]; ++i)
a[i] = c[i];
}
void merge(int* a, vector b) {
int x = 1, y = 0;
c[0] = 0;
for (int i = 1; i <= 10; ++i) {
if (x == a[0] + 1 && y >= (int)b.size())
break;
else if (x == a[0] + 1)
c[++c[0]] = b[y++];
else if (y >= (int)b.size())
c[++c[0]] = a[x++];
else
c[++c[0]] = a[x] < b[y] ? a[x++] : b[y++];
}
for (int i = 0; i <= c[0]; ++i)
a[i] = c[i];
}
void Dfs_1(int u) {
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (v == father[0][u])
continue;
Dfs_1(v);
merge(bz[0][v], tag[u]);
}
}
void climb_up(int u, int v, int* ans) {
for (int i = LOG; i >= 0; --i) {
if (dep[father[i][u]] > dep[v]) {
merge(ans, bz[i][u]);
u = father[i][u];
}
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q;
for (int i = 1; i < n; ++i) {
int u, v;
cin >> u >> v;
add_edge(u, v);
add_edge(v, u);
}
Dfs(1, 0);
for (int i = 1; i <= n; ++i)
tag[i].clear();
for (int i = 1; i <= m; ++i) {
int x;
cin >> x;
tag[x].push_back(i);
}
Dfs_1(1);
for (int i = 1; i <= n; ++i) {
sort(bz[0][i] + 1, bz[0][i] + bz[0][i][0] + 1);
sort(tag[i].begin(), tag[i].end());
tag[i].resize(min((int)tag[i].size(), 10));
}
for (int i = 1; i <= LOG; ++i)
for (int j = 1; j <= n; ++j) {
for (int k = 0; k <= bz[i][j][0]; ++k)
bz[i][j][k] = bz[i - 1][j][k];
merge(bz[i][j], bz[i - 1][father[i - 1][j]]);
}
int ans[11];
for (int qt = 1; qt <= q; ++qt) {
int u, v, k;
cin >> u >> v >> k;
int _lca = lca(u, v);
ans[0] = 0;
merge(ans, tag[u]);
if (u != v)
merge(ans, tag[v]);
if (u == _lca) {
climb_up(v, u, ans);
} else if (v == _lca) {
climb_up(u, v, ans);
} else {
merge(ans, tag[_lca]);
climb_up(u, _lca, ans);
climb_up(v, _lca, ans);
}
cout << min(k, ans[0]) << ' ';
for (int i = 1; i <= min(k, ans[0]); ++i)
cout << ans[i] << ' ';
cout << '\n';
}
return 0;
}