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【比赛链接】
【题解链接】
**【A】**Berzerk
【思路要点】
- 博弈搜索,将状态按先后手拆点,建出游戏图。
- 若一个点存在出边指向必败态,则该点为必胜态。
- 若一个点所有出边指向必胜态,则该点为必败态。
- 不满足上述两点的点为平局态。
- 用一个类似拓扑排序的过程实现即可。
- 时间复杂度 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 2e4 + 5; typedef long long ll; typedef long double ld; typedef unsigned long long ull; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } int n, tot, point[MAXN][2], d[MAXN]; bool vis[MAXN], win[MAXN]; vector <int> s, t, a[MAXN]; int main() { read(n); int k; read(k); while (k--) { int x; read(x); s.push_back(x); } read(k); while (k--) { int x; read(x); t.push_back(x); } for (int i = 1; i <= n; i++) { point[i][0] = ++tot; d[tot] = s.size(); point[i][1] = ++tot; d[tot] = t.size(); } int l = 0, r = 1; static int q[MAXN]; q[0] = point[1][0], q[1] = point[1][1]; vis[q[0]] = vis[q[1]] = true; win[q[0]] = win[q[1]] = false; while (l <= r) { int tmp = q[l++], type = tmp % 2; int pos = (tmp + 1) / 2; if (win[tmp]) { if (type) { for (unsigned i = 0; i < t.size(); i++) { int tnp = point[(pos - t[i] + n - 1) % n + 1][1]; if (!vis[tnp] && --d[tnp] == 0) { vis[tnp] = true; win[tnp] = false; q[++r] = tnp; } } } else { for (unsigned i = 0; i < s.size(); i++) { int tnp = point[(pos - s[i] + n - 1) % n + 1][0]; if (!vis[tnp] && --d[tnp] == 0) { vis[tnp] = true; win[tnp] = false; q[++r] = tnp; } } } } else { if (type) { for (unsigned i = 0; i < t.size(); i++) { int tnp = point[(pos - t[i] + n - 1) % n + 1][1]; if (!vis[tnp]) { vis[tnp] = true; win[tnp] = true; q[++r] = tnp; } } } else { for (unsigned i = 0; i < s.size(); i++) { int tnp = point[(pos - s[i] + n - 1) % n + 1][0]; if (!vis[tnp]) { vis[tnp] = true; win[tnp] = true; q[++r] = tnp; } } } } } for (int i = 2; i <= n; i++) if (vis[point[i][0]]) { if (win[point[i][0]]) printf("Win "); else printf("Lose "); } else printf("Loop "); printf("\n"); for (int i = 2; i <= n; i++) if (vis[point[i][1]]) { if (win[point[i][1]]) printf("Win "); else printf("Lose "); } else printf("Loop "); printf("\n"); return 0; }
**【B】**Legacy
【思路要点】
- 线段树建图,跑单源最短路。
- 时间复杂度 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 2e5 + 5; typedef long long ll; typedef long double ld; typedef unsigned long long ull; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } namespace ShortestPath { const ll INF = 1e18; const int MAXP = 1e6; struct edge {int dest, len; }; int n; ll dist[MAXP]; vector <edge> a[MAXP]; set <pair <ll, int> > st; void addedge(int x, int y, int z) { a[x].push_back((edge) {y, z}); } void init(int x) { n = x; st.clear(); for (int i = 1; i <= n; i++) { dist[i] = INF; a[i].clear(); } } void work(int s) { dist[s] = 0; st.insert(make_pair(0, s)); while (!st.empty()) { pair <ll, int> tmp = *st.begin(); st.erase(tmp); for (unsigned i = 0; i < a[tmp.second].size(); i++) { int dest = a[tmp.second][i].dest; ll newlen = tmp.first + a[tmp.second][i].len; if (newlen < dist[dest]) { st.erase(make_pair(dist[dest], dest)); dist[dest] = newlen; st.insert(make_pair(dist[dest], dest)); } } } } } int tot, n, q, s, totp, root; int lc[MAXN], rc[MAXN], point[MAXN][2]; void build(int &root, int l, int r) { root = ++totp; point[root][0] = ++tot; point[root][1] = ++tot; if (l == r) return; int mid = (l + r) / 2; build(lc[root], l, mid); build(rc[root], mid + 1, r); } void query(int from, int root, int l, int r, int ql, int qr, int len, int type) { if (l == ql && r == qr) { if (type == 0) ShortestPath :: addedge(from, point[root][0], len); else ShortestPath :: addedge(point[root][1], from, len); return; } int mid = (l + r) / 2; if (mid >= ql) query(from, lc[root], l, mid, ql, min(mid, qr), len, type); if (mid + 1 <= qr) query(from, rc[root], mid + 1, r, max(mid + 1, ql), qr, len, type); } int main() { read(n), read(q), read(s), tot = n; build(root, 1, n); ShortestPath :: init(tot); for (int i = 1, j = 0; i <= totp; i++) { if (lc[i]) { ShortestPath :: addedge(point[i][0], point[lc[i]][0], 0); ShortestPath :: addedge(point[i][0], point[rc[i]][0], 0); ShortestPath :: addedge(point[lc[i]][1], point[i][1], 0); ShortestPath :: addedge(point[rc[i]][1], point[i][1], 0); } else { j++; ShortestPath :: addedge(point[i][0], j, 0); ShortestPath :: addedge(j, point[i][1], 0); } } for (int i = 1; i <= q; i++) { int opt; read(opt); if (opt == 1) { int x, y, z; read(x), read(y), read(z); ShortestPath :: addedge(x, y, z); } if (opt == 2) { int x, l, r, z; read(x), read(l), read(r), read(z); query(x, root, 1, n, l, r, z, 0); } if (opt == 3) { int x, l, r, z; read(x), read(l), read(r), read(z); query(x, root, 1, n, l, r, z, 1); } } ShortestPath :: work(s); for (int i = 1; i <= n; i++) { ll tmp = ShortestPath :: dist[i]; if (tmp == 1e18) printf("-1 "); else write(tmp), putchar(' '); } return 0; }
**【C】**Till I Collapse
【思路要点】
- 显然所有答案的总和不超过 。
- 从左到右考虑所有左端点,用树状数组维护右端点对应的区间内不同数的个数,二分最远点实现贪心。
- 时间复杂度 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 2e5 + 5; typedef long long ll; typedef long double ld; typedef unsigned long long ull; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } struct BinaryIndexTree { int n, a[MAXN]; void init(int x) { n = x; memset(a, 0, sizeof(a)); } void modify(int x, int d) { for (int i = x; i <= n; i += i & -i) a[i] += d; } int query(int x) { int ans = 0; for (int i = 20; i >= 0; i--) { int tmp = 1 << i; if (ans + tmp <= n && a[ans + tmp] <= x) { x -= a[ans + tmp]; ans += tmp; } } return ans; } } BIT; int n, val[MAXN], last[MAXN], ans[MAXN]; vector <int> a[MAXN], home[MAXN]; int main() { read(n); for (int i = 1; i <= n; i++) { read(val[i]); a[last[val[i]]].push_back(i); last[val[i]] = i; } for (int i = 1; i <= n; i++) home[1].push_back(i); BIT.init(n); for (int i = 1; i <= n; i++) { if (i != 1) BIT.modify(i - 1, -1); for (unsigned j = 0; j < a[i - 1].size(); j++) BIT.modify(a[i - 1][j], 1); for (unsigned j = 0; j < home[i].size(); j++) { int tmp = home[i][j]; ans[tmp]++, home[BIT.query(tmp) + 1].push_back(tmp); } } for (int i = 1; i <= n; i++) write(ans[i]), putchar(' '); return 0; }
**【D】**Rap God
【思路要点】
- 标算是点分治。
- 我们用树上倍增 哈希实现快速比较字符串。
- 运用点分治,我们将问题转化到了询问 中的 在分治重心的情况。
- 将 所在分治子树的字符串通过哈希比较排序,询问时二分即可。
- 时间复杂度 。
但上面的做法比较难写,下面笔者的代码是暴力的精细实现,时间复杂度 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 2e5 + 5; typedef long long ll; typedef long double ld; typedef unsigned long long ull; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } char from[MAXN]; int n, m, father[MAXN], ans[MAXN]; vector <pair <int, int> > q[MAXN]; vector <pair <int, char> > a[MAXN]; void dfs(int pos, int fa) { for (unsigned i = 0; i < a[pos].size(); i++) if (a[pos][i].first != fa) { father[a[pos][i].first] = pos; from[a[pos][i].first] = a[pos][i].second; dfs(a[pos][i].first, pos); } } int solve(int pos, int to) { static int vis[MAXN], depth[MAXN], tag = 0; static char val[MAXN]; static int cmp[MAXN]; tag++; int tmp = n, ans = -1; for (int now = to; now != pos; now = father[now], tmp--) { val[tmp] = from[now]; depth[now] = tmp; vis[now] = tag; cmp[now] = 1; ans++; } vis[pos] = tag, cmp[pos] = 1, depth[pos] = tmp; for (int i = 1; i <= n; i++) if (vis[i] != tag) { int top = 0, now = i; static int st[MAXN]; while (vis[now] != tag) st[top++] = now, now = father[now]; for (int j = top - 1; j >= 0; j--) { int now = st[j]; depth[now] = depth[father[now]] + 1; vis[now] = tag; if (cmp[father[now]] != 1) cmp[now] = cmp[father[now]]; else if (val[depth[now]] > from[now]) cmp[now] = 0; else if (val[depth[now]] == from[now]) cmp[now] = 1; else cmp[now] = 2; ans += (cmp[now] == 0) || (cmp[now] == 1 && depth[now] < n); } } return ans; } void work(int pos, int fa) { for (unsigned i = 0; i < q[pos].size(); i++) ans[q[pos][i].second] = solve(pos, q[pos][i].first); for (unsigned i = 0; i < a[pos].size(); i++) if (a[pos][i].first != fa) { int dest = a[pos][i].first; father[pos] = dest; father[dest] = 0; from[pos] = a[pos][i].second; work(dest, pos); father[dest] = pos; father[pos] = 0; from[dest] = a[pos][i].second; } } int main() { read(n), read(m); for (int i = 1; i <= n - 1; i++) { int x, y; char c; scanf("%d%d %c", &x, &y, &c); a[x].push_back(make_pair(y, c)); a[y].push_back(make_pair(x, c)); } dfs(1, 0); for (int i = 1; i <= m; i++) { int x, y; read(x), read(y); q[x].push_back(make_pair(y, i)); } work(1, 0); for (int i = 1; i <= m; i++) writeln(ans[i]); return 0; }
**【E】**ALT
【思路要点】
- 最大权闭合子图 树上倍增建图。
- 时间复杂度 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXP = 4e5 + 5; const int MAXN = 2e4 + 5; const int MAXLOG = 16; const int INF = 2e9; typedef long long ll; typedef long double ld; typedef unsigned long long ull; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } struct edge {int dest, flow; unsigned pos; }; vector <edge> a[MAXP]; int n, m, s, t, dist[MAXP]; unsigned curr[MAXP]; vector <int> b[MAXN], home[MAXN]; int tot, point[MAXN][MAXLOG], pos[MAXN]; int depth[MAXN], father[MAXN][MAXLOG]; void addedge(int x, int y, int z) { a[x].push_back((edge) {y, z, a[y].size()}); a[y].push_back((edge) {x, 0, a[x].size() - 1}); } int dinic(int pos, int limit) { if (pos == t) return limit; int used = 0, tmp; for (unsigned &i = curr[pos]; i < a[pos].size(); i++) if (a[pos][i].flow != 0 && dist[pos] + 1 == dist[a[pos][i].dest] && (tmp = dinic(a[pos][i].dest, min(limit - used, a[pos][i].flow)))) { used += tmp; a[pos][i].flow -= tmp; a[a[pos][i].dest][a[pos][i].pos].flow += tmp; if (used == limit) return used; } return used; } bool bfs() { static int q[MAXP]; int l = 0, r = 0; memset(dist, 0, sizeof(dist)); dist[s] = 1, q[0] = s; while (l <= r) { int tmp = q[l]; for (unsigned i = 0; i < a[tmp].size(); i++) if (dist[a[tmp][i].dest] == 0 && a[tmp][i].flow != 0) { q[++r] = a[tmp][i].dest; dist[q[r]] = dist[tmp] + 1; } l++; } return dist[t] != 0; } void work(int pos, int fa) { depth[pos] = depth[fa] + 1; father[pos][0] = fa; for (int i = 1; i < MAXLOG; i++) { father[pos][i] = father[father[pos][i - 1]][i - 1]; if (father[pos][i]) { point[pos][i] = ++tot; addedge(point[pos][i], point[pos][i - 1], INF); addedge(point[pos][i], point[father[pos][i - 1]][i - 1], INF); } } for (unsigned i = 0; i < b[pos].size(); i++) if (b[pos][i] != fa) { point[b[pos][i]][0] = home[pos][i]; work(b[pos][i], pos); } } int lca(int x, int y) { if (depth[x] < depth[y]) swap(x, y); for (int i = MAXLOG - 1; i >= 0; i--) if (depth[father[x][i]] >= depth[y]) x = father[x][i]; if (x == y) return x; for (int i = MAXLOG - 1; i >= 0; i--) if (father[x][i] != father[y][i]) { x = father[x][i]; y = father[y][i]; } return father[x][0]; } int main() { read(n), read(m); for (int i = 1; i <= n - 1; i++) { int x, y; read(x), read(y); b[x].push_back(y); home[x].push_back(i); b[y].push_back(x); home[y].push_back(i); } s = 0, tot = n - 1; work(1, 0); for (int i = 1; i <= m; i++) { int x, y; read(x), read(y); int z = lca(x, y); pos[i] = ++tot; addedge(s, pos[i], 1); for (int j = MAXLOG - 1; j >= 0; j--) if (depth[father[x][j]] >= depth[z]) { addedge(pos[i], point[x][j], INF); x = father[x][j]; } for (int j = MAXLOG - 1; j >= 0; j--) if (depth[father[y][j]] >= depth[z]) { addedge(pos[i], point[y][j], INF); y = father[y][j]; } } t = ++tot; for (int i = 1; i <= n - 1; i++) addedge(i, t, 1); int ans = 0; while (bfs()) { memset(curr, 0, sizeof(curr)); ans += dinic(s, INF); } printf("%d\n", ans); bfs(); int size = 0; for (int i = 1; i <= m; i++) if (dist[pos[i]] == 0) size++; printf("%d ", size); for (int i = 1; i <= m; i++) if (dist[pos[i]] == 0) printf("%d ", i); printf("\n"); size = 0; for (int i = 1; i <= n - 1; i++) if (dist[i] != 0) size++; printf("%d ", size); for (int i = 1; i <= n - 1; i++) if (dist[i] != 0) printf("%d ", i); printf("\n"); return 0; }