CodeForces Gym 101190简要题解
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NEERC天天被搬还是做了好了。
Abbreviation
模拟。
#define FIO "abbreviation"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 500005;int n, p[MAXN];string t[MAXN];char s[MAXN];inline bool In(int x, int l, int r){ return x >= l && x <= r;}inline int Find(int x){ if (!In(s[x], 'A', 'Z') || !In(s[x + 1], 'a', 'z')) return 0; for (x ++; In(s[x], 'a', 'z'); x ++); if (In(s[x], 'A', 'Z')) return 0; return x;}inline int Chk(int x){ int y, z; if (!(y = Find(x)) || s[y] != ' ' || !(z = Find(y + 1))) return 0; t[x] += s[x], t[x] += s[y + 1]; while (true) { if (s[z] != ' ' || !(y = Find(z + 1))) return z - 1; t[x] += s[z + 1]; z = y; }}int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif for (char c = getchar(); ~c; c = getchar()) s[++ n] = c; for (int i = 1, j; i <= n; i ++) if (!In(s[i - 1], 'A', 'Z') && !In(s[i - 1], 'a', 'z')) if (j = Chk(i)) p[i] = 1, p[j] = 2, i = j; for (int i = 1; i <= n; i ++) { if (p[i] == 1) printf("%s (", t[i].c_str()); putchar(s[i]); if (p[i] == 2) putchar(')'); }}Binary Code
暴力是枚举两个点的两种状态，看是否一个是另一个的前缀，跑2-SAT。
考虑用trie树优化建边，分别是前缀和和后缀和(表示选没选)。
#define FIO "binary"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 1000005;const int MAXM = 4000005;int n, tot, len, cnt, num, scc[MAXM], par[MAXN], pos[MAXN], st[MAXN], ed[MAXN], tst[MAXN], ted[MAXN], dfn[MAXM], nxt[MAXN][2];vector <int> adj[MAXM], vec[MAXN], G[MAXM];char s[MAXN], ch[MAXN];bool res[MAXN];inline void Add(int p){ int cur = 0; for (int i = 1; i <= len; cur = nxt[cur][s[i ++] - '0']) if (!nxt[cur][s[i] - '0']) par[nxt[cur][s[i] - '0'] = ++ cnt] = cur; pos[p] = cur; vec[cur].pb(p);}inline void Build(){ for (int i = 1; i <= cnt; i ++) if (!vec[i].empty()) { for (int j = 0; j < vec[i].size(); j ++) if (!j) adj[++ tot].pb(vec[i][j] ^ 1); else adj[vec[i][j]].pb(tot), adj[tot + 1].pb(tot), adj[++ tot].pb(vec[i][j] ^ 1); tst[i] = tot; for (int j = 0; j < vec[i].size(); j ++) if (!j) adj[vec[i][j]].pb(++ tot); else adj[tot].pb(vec[i][j] ^ 1), adj[tot].pb(tot + 1), adj[vec[i][j]].pb(++ tot); ted[i] = tot; } for (int i = 2; i <= (n << 1 | 1); i ++) for (int j = par[pos[i]]; j; j = par[j]) if (!vec[j].empty()) adj[i].pb(tst[j]), adj[ted[j]].pb(i ^ 1);}int sta[MAXM], low[MAXM], tim, tp;inline void Dfs(int x){ dfn[x] = low[x] = ++ tim; sta[++ tp] = x; for (auto y : adj[x]) if (!dfn[y]) Dfs(y), low[x] = min(low[x], low[y]); else if (!scc[y]) low[x] = min(low[x], dfn[y]); if (dfn[x] == low[x]) { int cur = 0; num ++; while (cur ^ x) scc[cur = sta[tp --]] = num; }}int vis[MAXM], opp[MAXM], deg[MAXM], q[MAXM], ql, qr;inline void Print(){ for (int i = 1; i <= n; i ++) opp[scc[i << 1]] = scc[i << 1 | 1], opp[scc[i << 1 | 1]] = scc[i << 1]; for (int x = 2; x <= tot; x ++) for (auto y : adj[x]) if (scc[x] ^ scc[y]) G[scc[y]].pb(scc[x]), deg[scc[x]] ++; for (int i = 1; i <= num; i ++) if (!deg[i]) q[++ qr] = i; while (ql ^ qr) { int x = q[++ ql]; if (!vis[x]) vis[x] = 1, vis[opp[x]] = -1; for (auto y : G[x]) if (!-- deg[y]) q[++ qr] = y; } for (int i = 2; i <= (n << 1 | 1); i ++) if (~vis[scc[i]]) res[i >> 1] = i & 1;}int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif n = Read(), tot = n << 1 | 1; for (int i = 1, t = 1; i <= n; i ++) { scanf("%s", s + 1), len = strlen(s + 1); st[i] = t; for (int j = 1; j <= len; j ++) ch[t ++] = s[j]; ed[i] = t; for (int j = 1; j <= len; j ++) if (s[j] == '?') { s[j] = '0', Add(i << 1); s[j] = '1', Add(i << 1 | 1); break; } else if (j == len) Add(i << 1), Add(i << 1 | 1); } Build(); for (int i = 2; i <= tot; i ++) if (!dfn[i]) Dfs(i); for (int i = 1; i <= n; i ++) if (scc[i << 1] == scc[i << 1 | 1]) return puts("NO"), 0; puts("YES"); Print(); for (int i = 1; i <= n; putchar(10), i ++) for (int j = st[i]; j < ed[i]; j ++) if (ch[j] == '?') putchar(res[i] + '0'); else putchar(ch[j]); return 0;}Cactus Construction
先把所有环和桥边分开，然后随便构造一下就好了。
详见代码。
#define FIO "cactus"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 50005;struct Node{ int opt, a, b, c; Node(int _opt = 0, int _a = 0, int _b = 0, int _c = 0) { opt = _opt, a = _a, b = _b, c = _c; }};int n, m, tim, cnt, par[MAXN], a[MAXN], dfn[MAXN];vector <int> adj[MAXN], hav[MAXN], cir[MAXN];vector <Node> ans;bool f[MAXN];inline void Dfs(int x){ dfn[x] = ++ tim; for (auto y : adj[x]) if (y ^ par[x]) if (!dfn[y]) par[y] = x, Dfs(y); else if (dfn[y] < dfn[x]) { hav[y].pb(++ cnt); cir[cnt].pb(y); for (int i = x; i ^ y; i = par[i]) cir[cnt].pb(i); }}inline void Solve(int x){ for (auto c : hav[x]) { for (auto y : cir[c]) if (y ^ x) f[y] = 1; for (auto y : cir[c]) if (y ^ x) Solve(y); for (int i = 1; i < cir[c].size(); i ++) { int p = cir[c][i - 1], q = cir[c][i]; ans.pb(Node(2, p, 1, 3)); ans.pb(Node(1, p, q)); ans.pb(Node(3, p, 1, 3)); ans.pb(Node(2, p, 3, i == 1 ? 4 : 2)); } int y = cir[c].back(); ans.pb(Node(3, x, 1, 4)); ans.pb(Node(2, x, 1, 2)); ans.pb(Node(2, x, 4, 1)); } for (auto y : adj[x]) if (!f[y]) { f[y] = 1, Solve(y); ans.pb(Node(2, y, 1, 3)); ans.pb(Node(1, x, y)); ans.pb(Node(3, x, 1, 3)); ans.pb(Node(2, x, 3, 2)); }}int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif n = Read(), m = Read(); while (m --) { int x = Read(); for (int i = 1; i <= x; i ++) a[i] = Read(); for (int i = 1; i < x; i ++) adj[a[i]].pb(a[i + 1]), adj[a[i + 1]].pb(a[i]); } Dfs(1); f[1] = 1; Solve(1); printf("%d\n", ans.size()); for (auto e : ans) if (e.opt == 1) printf("j %d %d\n", e.a, e.b); else if (e.opt == 2) printf("r %d %d %d\n", e.a, e.b, e.c); else if (e.opt == 3) printf("c %d %d %d\n", e.a, e.b, e.c);}Delight for a Cat
考虑费用流，一个点如果选择睡觉那就会影响之后的
#define FIO "delight"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 1005;const int MAXM = 6005;const int INF = 0x3f3f3f3f;namespace Flow{ struct edge { int p, v, w, c; } e[MAXM]; int V, S, T, hed[MAXN], e_cnt = 1; LL ans, dis[MAXN]; bool vis[MAXN]; inline void Init(int n) { V = n, e_cnt = 1; for (int i = 0; i < V; i ++) hed[i] = 0; } inline void Add(int x, int y, int w, int c) { e[++ e_cnt] = {y, hed[x], w, c}; hed[x] = e_cnt; } inline void Addedge(int x, int y, int w, int c) { Add(x, y, w, c); Add(y, x, 0, -c); } inline bool Spfa() { for (int i = 0; i < V; i ++) dis[i] = 1LL << 60, vis[i] = 0; queue <int> q; dis[S] = 0; q.push(S); while (!q.empty()) { int x = q.front(); q.pop(); vis[x] = 0; for (int i = hed[x]; i; i = e[i].v) if (e[i].w && dis[e[i].p] > dis[x] + e[i].c) { dis[e[i].p] = dis[x] + e[i].c; if (!vis[e[i].p]) vis[e[i].p] = 1, q.push(e[i].p); } } return dis[T] < 1LL << 60; } inline int Dfs(int x, int f) { vis[x] = 1; if (x == T) return f; int ret = 0, d; for (int i = hed[x]; i; i = e[i].v) if (e[i].w && dis[e[i].p] == dis[x] + e[i].c && !vis[e[i].p]) { d = Dfs(e[i].p, min(f - ret, e[i].w)); e[i].w -= d; e[i ^ 1].w += d; ret += d; if (ret == f) return ret; } return ret; } inline int Solve() { int ret = 0; while (Spfa()) { vis[T] = 1; while (vis[T]) { for (int i = 0; i < V; i ++) vis[i] = 0; int d = Dfs(S, INF); ret += d; ans += d * dis[T]; } } return ret; }}int n, k, p, q, a[MAXN];int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif using namespace Flow; n = Read(), k = Read(), p = k - Read(), q = Read(); Flow::Init(n + 3), Flow::T = n + 2; for (int i = 1; i <= n; i ++) a[i] -= Read(), ans += a[i]; for (int i = 1; i <= n; i ++) a[i] += Read(); for (int i = 1; i <= n; i ++) Addedge(i, i + k <= n ? i + k : T, 1, -a[i]); for (int i = 1; i <= n; i ++) Addedge(i, i == n ? T : i + 1, p - q, 0); for (int i = 1; i <= k; i ++) Addedge(n + 1, i, INF, 0); Addedge(S, n + 1, p, 0); Solve(); printf("%lld\n", -ans); for (int i = 1; i <= n; i ++) putchar(e[i << 1].w ? 'S' : 'E'); putchar(10); return 0;}Expect to Wait
随便推一推发现是个一次函数的形式，离线之后打上差分标记。
#define FIO "expect"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 100005;int n, m, t[MAXN], c[MAXN], l[MAXN];LL p, q, a[MAXN], b[MAXN], ans[MAXN];pii e[MAXN];char ch[2];int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif n = Read(), m = Read(); for (int i = 1, x, y; i <= n; i ++) { scanf("%s", ch); if (ch[0] == '+') t[i] = Read(), c[i] = Read(); else t[i] = Read(), c[i] = -Read(); } for (int i = 1; i <= m; i ++) e[i].xx = Read(), e[i].yy = i; sort(e + 1, e + m + 1); for (int i = 1; i <= m; i ++) l[i] = e[i].xx; for (int i = 1, s = 0, d; i <= n; i ++) { s += c[i], d = t[i + 1] - t[i]; if (s < 0) { int p = lower_bound(l + 1, l + m + 1, -s) - l; if (i ^ n) a[0] -= d, b[0] -= 1LL * d * s, a[p] += d, b[p] += 1LL * d * s; else for (int j = 1; j < p; j ++) ans[e[j].yy] = -1; } } for (int i = 0; i <= m; i ++) { p += a[i], q += b[i]; if (~ans[e[i].yy]) ans[e[i].yy] = p * e[i].xx + q; } for (int i = 1; i <= m; i ++) if (ans[i] == -1) puts("INFINITY"); else printf("%lld\n", ans[i]); return 0;}Foreign Postcards
暴力想法是枚举一对
#define FIO "foreign"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 1000005;long double ans;int n, a[MAXN];LL p, q, c, d;char s[MAXN];int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif scanf("%s", s + 1); n = strlen(s + 1); for (int i = 1; i <= n; i ++) a[i] = s[i] == 'W'; for (int i = 1; i <= n; i ++) c += a[i], d += !a[i], p += c, q += d; for (int i = 0; i < n; i ++) { double r = i ? 1.0 / (1LL * (n - i) * (n - i + 1)) : 1.0 / n; ans += r * (a[i + 1] ? q : p); p -= a[i + 1] * (n - i); q -= (!a[i + 1]) * (n - i); } return printf("%.10Lf\n", ans), 0;}Game on Graph
考虑记
#define FIO "game"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 100005;vector <int> adj[MAXN], rev[MAXN];bool a[MAXN][2], b[MAXN][2];int n, m, ql, qr, d[MAXN];pii q[MAXN << 1];inline void Solve(bool f[MAXN][2]){ ql = qr = 0; for (int i = 1; i <= n; i ++) for (int j = 0; j < 2; j ++) if (f[i][j]) q[++ qr] = mp(i, j); for (int i = 1; i <= n; i ++) d[i] = adj[i].size(); while (ql ^ qr) { int x = q[++ ql].xx, p = q[ql].yy; if (p) { for (auto y : rev[x]) if (!f[y][0] && !-- d[y]) f[y][0] = 1, q[++ qr] = mp(y, 0); } else for (auto y : rev[x]) if (!f[y][1]) f[y][1] = 1, q[++ qr] = mp(y, 1); }}int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif n = Read(), m = Read(); for (int i = 1, x, y; i <= m; i ++) x = Read(), y = Read(), adj[x].pb(y), rev[y].pb(x); for (int i = 1; i <= n; i ++) if (adj[i].empty()) a[i][0] = a[i][1] = b[i][0] = 1; Solve(a); Solve(b); for (int i = 1; i <= n; i ++) if (b[i][0]) putchar('L'); else if (!a[i][0]) putchar('D'); else putchar('W'); putchar(10); for (int i = 1; i <= n; i ++) if (b[i][1]) putchar('W'); else if (a[i][1]) putchar('L'); else putchar('D'); putchar(10); return 0;}Hard Refactoring
模拟。
#define FIO "hard"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 100005;int a[MAXN], n = 1 << 15;vector <pii> ans;char s[MAXN];int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif while (fgets(s, MAXN, stdin)) { int l, r; if (sscanf(s, "x >= %d && x <= %d", &l, &r) ^ 2) if (sscanf(s, "x >= %d", &l) ^ 1) sscanf(s, "x <= %d", &r), l = -n; else r = n - 1; if (l > r) continue; a[l + n] ++, a[r + n + 1] --; } for (int i = 1; i < MAXN; i ++) a[i] += a[i - 1]; for (int i = 0; i < MAXN; i ++) if (a[i]) { int p = i; while (a[p + 1]) p ++; ans.pb(mp(i - n, p - n)); i = p; } if (ans.empty()) return puts("false"), 0; if (ans.size() == 1 && ans[0].xx == -n && ans[0].yy == n - 1) return puts("true"), 0; for (int i = 0; i < ans.size(); i ++) { pii x = ans[i]; if (x.xx == -n) printf("x <= %d", x.yy); else if (x.yy == n - 1) printf("x >= %d", x.xx); else printf("x >= %d && x <= %d", x.xx, x.yy); if (i < ans.size() - 1) puts(" ||"); } putchar(10); return 0;}Indiana Jones and the Uniform Cave
挺厉害的交互题。
考虑用
如果一个点走到了不在栈中的点，那么我们先向上跳(
在栈中的也是走两遍就知道深度了。
#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}int n, m, L = 0, R = 1, C = 2, adj[25][25];inline int Next(){ char s[15]; scanf("%s", s); if (s[0] == 'l') return L; if (s[0] == 'r') return R; if (s[0] == 'c') return C; exit(0);}inline int Print(int p, int t, int v){ printf("%d %s %d\n", p, t ? "right" : "left", v); fflush(stdout); return Next();}inline void Dfs(int x, int dep){ for (int i = 0; i < m; i ++) { int val = Print(1, R, 1), cnt = 0; if (val == C) Dfs(++ n, adj[x][i] = dep + 1), val = L; if (val == L) { for (; Print(0, L, 0) == L; ); for (; Print(0, R, 0) == R; cnt ++); for (; Print(0, L, 0) == L; ); for (int i = 1; i <= cnt; i ++) Print(0, R, 0); adj[x][i] = dep - cnt; } else { for (; Print(0, L, 0) == R; cnt ++); for (int i = 0; i <= cnt; i ++) Print(0, R, 0); for (int i = 1; i <= cnt; i ++) Print(0, R, 0); adj[x][i] = dep - cnt; } } if (x ^ 1) { int pos = -1, val; for (int i = 0; i < m; i ++) if (!~pos || adj[x][i] < adj[x][pos]) pos = i; val = Print((pos + 1) % m, L, (pos + 1) % m); for (; val == L; val = Print(0, L, 0)); for (; Print(0, R, 0) == R; ); }}int main(){ m = Read(); Next(); Dfs(++ n, 1); return 0;}Jenga Boom
暴力。
#define FIO "jenga"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 5005;int n, m, h, w, l[MAXN], r[MAXN], c[MAXN], s[MAXN];bool f[MAXN][MAXN << 1];inline bool chk(){ LL cnt[2] = {0, 0}, sum[2] = {0, 0}; for (int i = h, cur = i & 1; i > 1; i --, cur ^= 1) { cnt[cur] += c[i], sum[cur] += s[i]; if ((sum[cur ^ 1] + cnt[cur] * n) <= (cnt[0] + cnt[1]) * (l[i - 1] * 2 - 2)) return 1; if ((sum[cur ^ 1] + cnt[cur] * n) >= (cnt[0] + cnt[1]) * (r[i - 1] * 2)) return 1; } return 0;}int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif n = Read(), w = Read(), h = Read(), m = Read(); for (int i = 1; i <= h; i ++) l[i] = 1, r[i] = n, c[i] = n, s[i] = n * n; for (int i = 1; i <= m; i ++) { int x = Read(), y = Read(); f[x][y] = 1; s[x] -= y * 2 - 1; if (-- c[x]) { while (f[x][l[x]]) l[x] ++; while (f[x][r[x]]) r[x] --; } else if (x == h) h --; else return printf("yes\n%d\n", i), 0; if (chk()) return printf("yes\n%d\n", i), 0; } return puts("no"), 0;}Kids Designing Kids
三个图异或为
#define FIO "kids"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}set <pii> A, B, C;pii a, b, c;char ch[1005];inline void Init(set <pii> &S, pii &s){ int n = Read(), m = Read(), f = 0; for (int i = 1; i <= n; i ++) { scanf("%s", ch + 1); for (int j = 1; j <= m; j ++) if (ch[j] == '*') { if (!f) f = 1, s = mp(-i, -j); S.insert(mp(i + s.xx, j + s.yy)); } }}inline bool Chk(set <pii> &A, pii &a, set <pii> &B, pii &b, set <pii> &C, pii &c){ set <pii> S; for (auto x : A) S.insert(x); for (auto x : B) if (S.find(x) != S.end()) S.erase(x); else S.insert(x); if (S.size() != C.size()) return 0; pii s = *S.begin(); for (auto x : C) if (S.find(mp(x.xx + s.xx, x.yy + s.yy)) == S.end()) return 0; c = mp(c.xx + s.xx, c.yy + s.yy); return 1;}int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif Init(A, a); Init(B, b); Init(C, c); if (Chk(A, a, B, b, C, c) || Chk(A, a, C, c, B, b) || Chk(B, b, C, c, A, a)) return printf("YES\n%d %d\n", b.yy - a.yy, b.xx - a.xx), 0; return puts("NO"), 0;}List of Primes
注意到最后质数不会很大，预处理每个质数的贡献就可以开始爆搜了。
#define FIO "list"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 2340;LL f[MAXN][MAXN], g[MAXN][MAXN], l, r;int n, cnt, now, p[MAXN], v[MAXN], use[MAXN];bool chk[MAXN];char ch[MAXN];inline void Init(){ for (int i = 2; i < MAXN; i ++) { if (!chk[i]) p[++ n] = i; for (int j = 1; i * p[j] < MAXN; j ++) { chk[i * p[j]] = 1; if (i % p[j] == 0) break; } }}inline void Print(LL l, LL r){ int tot = 0, cur; ch[++ tot] = '['; for (int i = 1, t; i <= cnt; i ++) { t = p[use[i]], cur = tot + v[use[i]]; while (t) ch[cur --] = t % 10 + '0', t /= 10; tot += v[use[i]]; ch[++ tot] = ',', ch[++ tot] = ' '; } ch[-- tot] = ']', ch[++ tot] = ',', ch[++ tot] = ' '; for (LL i = l; i <= min(r, (LL)tot); i ++) putchar(ch[i]);}inline void Dfs(int sum, LL l, LL r){ int i; for (i = use[cnt] + 1; i <= n && p[i] < sum; i ++) { LL t = f[sum - p[i]][i + 1] * (now + v[i] + 2) + g[sum - p[i]][i + 1]; if (t < l) l -= t, r -= t; else { use[++ cnt] = i, now += v[i] + 2; Dfs(sum - p[i], l, r); cnt --, now -= v[i] + 2; l = 1, r -= t; if (r <= 0) return ; } } if (p[i] == sum) use[++ cnt] = i, Print(l, r), cnt --; return ;}int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif cin >> l >> r; Init(); for (int i = 1; i <= n; i ++) v[i] = p[i] >= 1000 ? 4 : (p[i] >= 100 ? 3 : (p[i] >= 10 ? 2 : 1)), f[p[i]][i] = 1, g[p[i]][i] = v[i] + 4; for (int i = 2; i < MAXN; i ++) for (int j = n; j; j --) { f[i][j] += f[i][j + 1]; g[i][j] += g[i][j + 1]; if (i >= p[j]) f[i][j] += f[i - p[j]][j + 1], g[i][j] += g[i - p[j]][j + 1] + f[i - p[j]][j + 1] * (v[j] + 2); } for (int i = 2; i < MAXN; i ++) { if (g[i][1] < l) l -= g[i][1], r -= g[i][1]; else { Dfs(i, l, r); l = 1, r -= g[i][1]; if (r <= 0) return 0; } }}Mole Tunnels
暴力是费用流。
完全二叉树深度是
#define FIO "mole"#include <bits/stdc++.h>#define xx first#define yy second#define mp make_pair#define pb push_back#define mset(x, y) memset(x, y, sizeof x)#define mcpy(x, y) memcpy(x, y, sizeof x)using namespace std;typedef long long LL;typedef pair <int, int> pii;inline int Read(){ int x = 0, f = 1, c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; return x * f;}const int MAXN = 300005;int n, m, c[MAXN], f[MAXN], g[MAXN], h[MAXN], p[MAXN], ans;inline void Upd(int x){ f[x] = 1 << 30, p[x] = 0; if (c[x]) f[x] = 0, p[x] = x; if ((x << 1) <= n && f[x] > f[x << 1] + (h[x << 1] ? -1 : 1)) f[x] = f[x << 1] + (h[x << 1] ? -1 : 1), p[x] = p[x << 1]; if ((x << 1 | 1) <= n && f[x] > f[x << 1 | 1] + (h[x << 1 | 1] ? -1 : 1)) f[x] = f[x << 1 | 1] + (h[x << 1 | 1] ? -1 : 1), p[x] = p[x << 1 | 1];}int main(){#ifdef wxh010910 freopen("data.in", "r", stdin);#else freopen(FIO".in", "r", stdin); freopen(FIO".out", "w", stdout);#endif n = Read(), m = Read(); for (int i = 1; i <= n; i ++) c[i] = Read(); for (int i = n; i; i --) Upd(i); while (m --) { int x = Read(), y, l, cur = 0, ret = 1 << 30; for (int i = x; i; cur += g[i] ? -1 : 1, i >>= 1) if (f[i] + cur < ret) ret = f[i] + cur, l = i; c[y = p[l]] --; for (int i = x; i ^ l; i >>= 1) if (g[i]) g[i] --; else h[i] ++; for (int i = y; i ^ l; i >>= 1) if (h[i]) h[i] --; else g[i] ++; for (int i = x; i; i >>= 1) Upd(i); for (int i = y; i; i >>= 1) Upd(i); printf("%d\n", ans += ret); }}阅读全文
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