B.
Offline segment tree maintenance partition by merging the rank of disjoint-set bare title.
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int MAXN = 100000 + 10; 6 const int MAXV = 9; 7 8 typedef pair<int, int> pii; 9 10 struct Edge { 11 int u, v; 12 int w; 13 int st, ed; 14 Edge(int u=0, int v=0, int w=0, int st=0, int ed=0) 15 : u(u), v(v), w(w), st(st), ed(ed) {} 16 } e[MAXN]; 17 int ec; 18 19 pii q[MAXN]; 20 int qc; 21 map<pii, int> id; 22 int ans[MAXN]; 23 24 int par[MAXN]; 25 int sz[MAXN]; 26 int st[MAXN], top; 27 int find(int x) { return x == par[x] ? x : find(par[x]); } 28 void merge(int u, int v) { 29 u = find(u), v = find(v); 30 if (u != v) { 31 if (sz[u] < sz[v]) 32 swap(u, v); 33 par[v] = u; 34 sz[u] += sz[v]; 35 st[++top] = v; 36 } 37 } 38 void del(int u) { 39 sz[par[u]] -= sz[u]; 40 par[u] = u; 41 } 42 43 struct SegmentTree { 44 vector<int> edges[MAXN << 2]; 45 bool solved[MAXN << 2]; 46 void insert(int ql, int qr, int v, int l=1, int r=qc, int o=1) { 47 if (ql > qr) 48 return; 49 if (ql == l && qr == r) { 50 edges[o].push_back(v); 51 //cerr << ql << ' ' << qr << ' ' << v << ' ' << l << ' ' << r << endl; 52 return; 53 } 54 int mid = (l + r) >> 1; 55 if (qr <= mid) 56 insert(ql, qr, v, l, mid, o << 1); 57 else if (mid < ql) 58 insert(ql, qr, v, mid + 1, r, o << 1 | 1); 59 else { 60 insert(ql, mid, v, l, mid, o << 1); 61 insert(mid + 1, qr, v, mid + 1, r, o << 1 | 1); 62 } 63 } 64 void solve(int v, int l=1, int r=qc, int o=1) { 65 if (solved[o]) 66 return; 67 int temp = top; 68 for (auto id : edges[o]) 69 merge(e[id].u, e[id].v); 70 if (l == r) { 71 if (ans[l] == -1 && find(q[l].first) == find(q[r].second)) { 72 ans[l] = v; 73 solved[o] = true; 74 } 75 } else { 76 int mid = (l + r) >> 1; 77 solve(v, l, mid, o << 1); 78 solve(v, mid + 1, r, o << 1 | 1); 79 solved[o] = solved[o << 1] & solved[o << 1 | 1]; 80 } 81 82 while (temp != top) { 83 del(st[top--]); 84 } 85 } 86 87 } segt; 88 89 int main() { 90 ios::sync_with_stdio(false); 91 int n, t; 92 cin >> n >> t; 93 for (int i = 1; i <= t; ++i) { 94 int opt, u, v; 95 int w; 96 cin >> opt; 97 cin >> u >> v; 98 if (u > v) 99 swap(u, v); 100 if (opt == 0) { 101 cin >> w; 102 e[++ec] = Edge(u, v, w, qc + 1, -1); 103 id[pii(u, v)] = ec; 104 } else if (opt == 1) { 105 e[id[pii(u, v)]].ed = qc; 106 } else if (opt == 2) { 107 q[++qc] = pii(u, v); 108 } 109 } 110 for (int i = 1; i <= ec; ++i) { 111 if (e[i].ed == -1) 112 e[i].ed = qc; 113 } 114 115 for (int i = 0; i < n; ++i) { 116 par[i] = i; 117 sz[i] = 1; 118 } 119 120 memset(ans, -1, sizeof(ans)); 121 //cerr << "NMSL\n"; 122 for (int danger = 0; danger <= MAXV; ++danger) { 123 for (int i = 1; i <= ec; ++i) { 124 if (e[i].w == danger) { 125 segt.insert(e[i].st, e[i].ed, i); 126 } 127 } 128 segt.solve(danger); 129 //cerr << "WDNMD\n"; 130 } 131 for (int i = 1; i <= qc; ++i) { 132 cout << ans[i] << endl; 133 } 134 }
C.
Attendance problems. Note that memory search.
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int MAXB = 40 + 10; 6 int n; 7 int64_t x; 8 char s[MAXB]; 9 int ans = MAXB; 10 map<int64_t, int> mp; 11 void dfs(int64_t cur, int step) { 12 if (cur == (1ll << n) - 1) { 13 ans = min(ans, step); 14 return; 15 } 16 if (1 << step >= n) { 17 return; 18 } 19 if (step >= ans) 20 return; 21 if (mp[cur] && mp[cur] <= step) { 22 return; 23 } 24 mp[cur] = step; 25 for (int i = 1; i < n; ++i) { 26 int64_t now = cur | (cur >> i); 27 if (now != cur) { 28 dfs(now, step + 1); 29 } 30 } 31 } 32 33 int main() { 34 ios::sync_with_stdio(false); 35 cin >> s; 36 n = strlen(s); 37 for (int i = 0; s[i]; ++i) { 38 x = x * 2 + s[i] - '0'; 39 } 40 if (s[0] == '0') { 41 cout << "-1\n"; 42 return 0; 43 } 44 dfs(x, 0); 45 cout << ans << endl; 46 }