NOIP2016题目整合
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今天终于拿到官方数据,兴致勃勃地全 A 了。
Day 1 T1 toy
处理一下正负号加加减减取模乱搞就好了。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <algorithm> using namespace std; int read() { int x = 0, f = 1; char c = getchar(); while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); } while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); } return x * f; } #define maxn 100010 int n, q; struct Toy { int tp; char S[15]; } ts[maxn]; int main() { freopen("toy.in", "r", stdin); freopen("toy.out", "w", stdout); n = read(); q = read(); for(int i = 0; i < n; i++) { int t = read(); scanf("%s", ts[i].S); if(!t) ts[i].tp = 1; else ts[i].tp = -1; } int p = 0; while(q--) { int a = read(), b = read(); if(a) a = 1; else a = -1; p += a * ts[p].tp * b; p = (p % n + n) % n; } printf("%s\n", ts[p].S); return 0; }
Day 1 T2 running
受到“链”和“S 恒为 1”和“T 恒为 1”的特殊点的启发,我们发现可以将每条链 [Si, Ti] 分成 [Si, Ci] 和 [Ci, Ti] 两条,然后对于一个全部可观测到的链 [Ci, Ti] 将所有 Wi 减去深度是一个定值,对于 [Si, Ci] 的部分所有 Wi 加上深度是一个定值。于是树链剖分再打打标记统计就好了。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <algorithm> using namespace std; int read() { int x = 0, f = 1; char c = getchar(); while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); } while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); } return x * f; } #define maxn 300010 #define maxm 600010 #define maxs 12000010 int n, q, m, head[maxn], next[maxm], to[maxm], W[maxn]; struct Player { int s, t, c; Player() {} Player(int _, int __): s(_), t(__) {} } ps[maxn]; void AddEdge(int a, int b) { to[++m] = b; next[m] = head[a]; head[a] = m; swap(a, b); to[++m] = b; next[m] = head[a]; head[a] = m; return ; } int fa[maxn], dep[maxn], siz[maxn], son[maxn], top[maxn], pos[maxn], pid[maxn], clo; void build(int u) { siz[u] = 1; for(int e = head[u]; e; e = next[e]) if(to[e] != fa[u]) { fa[to[e]] = u; dep[to[e]] = dep[u] + 1; build(to[e]); siz[u] += siz[to[e]]; if(siz[son[u]] < siz[to[e]]) son[u] = to[e]; } return ; } void gett(int u, int tp) { top[u] = tp; pid[++clo] = u; pos[u] = clo; if(son[u]) gett(son[u], tp); for(int e = head[u]; e; e = next[e]) if(to[e] != fa[u] && to[e] != son[u]) gett(to[e], to[e]); return ; } int lca(int a, int b) { int f1 = top[a], f2 = top[b]; while(f1 != f2) { if(dep[f1] < dep[f2]) swap(f1, f2), swap(a, b); a = fa[f1]; f1 = top[a]; } return dep[a] < dep[b] ? a : b; } struct Info { int c, fir[maxn], aft[maxs], val[maxs]; void clear() { c = 0; memset(fir, 0, sizeof(fir)); return ; } void AddInfo(int x, int v) { val[++c] = v; aft[c] = fir[x]; fir[x] = c; return ; } } add, del; int tot[maxn<<1], ans[maxn]; void process(int x, int t, int a, int v) { int f = top[a]; while(f != top[t]) { add.AddInfo(pos[f], v); del.AddInfo(pos[a], v); a = fa[f]; f = top[a]; } add.AddInfo(pos[t], v); del.AddInfo(pos[a], v); return ; } int main() { freopen("running.in", "r", stdin); freopen("running.out", "w", stdout); n = read(); q = read(); for(int i = 1; i < n; i++) { int a = read(), b = read(); AddEdge(a, b); } for(int i = 1; i <= n; i++) W[i] = read(); for(int i = 1; i <= q; i++) { int s = read(), t = read(); ps[i] = Player(s, t); } build(1); gett(1, 1); for(int i = 1; i <= n; i++) W[i] -= dep[i]; add.clear(); del.clear(); memset(tot, 0, sizeof(tot)); for(int i = 1; i <= q; i++) { ps[i].c = lca(ps[i].s, ps[i].t); process(i, ps[i].c, ps[i].t, dep[ps[i].s] - dep[ps[i].c] - dep[ps[i].c]); } for(int i = 1; i <= n; i++) { for(int e = add.fir[i]; e; e = add.aft[e]) { int v = add.val[e] + n; tot[v]++; } int u = pid[i]; ans[u] += tot[W[u]+n]; for(int e = del.fir[i]; e; e = del.aft[e]) { int v = del.val[e] + n; tot[v]--; } } for(int i = 1; i <= n; i++) W[i] += (dep[i] << 1); add.clear(); del.clear(); memset(tot, 0, sizeof(tot)); for(int i = 1; i <= q; i++) process(i, ps[i].c, ps[i].s, dep[ps[i].s]); for(int i = 1; i <= n; i++) { for(int e = add.fir[i]; e; e = add.aft[e]) { int v = add.val[e]; tot[v]++; } int u = pid[i]; ans[u] += tot[W[u]]; for(int e = del.fir[i]; e; e = del.aft[e]) { int v = del.val[e]; tot[v]--; } } for(int i = 1; i <= q; i++) if(W[ps[i].c] == dep[ps[i].s]) ans[ps[i].c]--; for(int i = 1; i <= n; i++) printf("%d%c", ans[i], i < n ? ‘ ‘ : ‘\n‘); return 0; }
Day 1 T3 classroom
这题我考场上想到正解但是因为邻接矩阵连边时忘记取 min 就 sb 了。。。。。
设 f[0][i][j] 表示前 i 条请求使用了 j 条,最后一条没取的最小期望距离;f[1][i][j] 表示前 i 条请求使用了 j 条,最后一条取了的最小期望距离。然后因为每次走哪条边是独立的,转移时直接乘上概率累加上就好了。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <algorithm> using namespace std; int read() { int x = 0, f = 1; char c = getchar(); while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); } while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); } return x * f; } #define maxn 2010 #define maxv 310 #define oo 1000000000 int n, m, v, e, c[maxn], d[maxn], D[maxn][maxn]; double p[maxn], f[2][maxn][maxn]; void up(double& a, double b) { a = min(a, b); return ; } int main() { freopen("classroom.in", "r", stdin); freopen("classroom.out", "w", stdout); n = read(); m = read(); v = read(); e = read(); for(int i = 1; i <= n; i++) c[i] = read(); for(int i = 1; i <= n; i++) d[i] = read(); for(int i = 1; i <= n; i++) scanf("%lf", &p[i]); for(int i = 1; i <= v; i++) { D[i][i] = 0; for(int j = i + 1; j <= v; j++) D[i][j] = D[j][i] = oo; } for(int i = 1; i <= e; i++) { int a = read(), b = read(), c = read(); D[a][b] = min(D[a][b], c); D[b][a] = min(D[b][a], c); } for(int k = 1; k <= v; k++) for(int i = 1; i <= v; i++) for(int j = 1; j <= v; j++) D[i][j] = min(D[i][j], D[i][k] + D[k][j]); for(int i = 0; i <= n; i++) for(int j = 0; j <= m; j++) f[0][i][j] = f[1][i][j] = 1e9; f[0][0][0] = 0.0; for(int i = 0; i <= n; i++) for(int j = 0; j <= min(i + 1, m); j++) { double P = p[i], np = p[i+1], p1 = 1.0 - P, np1 = 1.0 - np; up(f[0][i+1][j], f[0][i][j] + D[c[i]][c[i+1]]); up(f[1][i+1][j+1], f[0][i][j] + np * D[c[i]][d[i+1]] + np1 * D[c[i]][c[i+1]]); up(f[0][i+1][j], f[1][i][j] + P * D[d[i]][c[i+1]] + p1 * D[c[i]][c[i+1]]); up(f[1][i+1][j+1], f[1][i][j] + P * np * D[d[i]][d[i+1]] + P * np1 * D[d[i]][c[i+1]] + p1 * np * D[c[i]][d[i+1]] + p1 * np1 * D[c[i]][c[i+1]]); } double ans = 1e9; for(int i = 0; i <= m; i++) up(ans, min(f[0][n][i], f[1][n][i])); printf("%.2lf\n", ans); return 0; }
Day 2 T1 problem
用递推法求组合数,实时模 k。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <algorithm> using namespace std; int read() { int x = 0, f = 1; char c = getchar(); while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); } while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); } return x * f; } #define maxn 2010 int C[2][maxn], f[maxn][maxn]; int main() { freopen("problem.in", "r", stdin); freopen("problem.out", "w", stdout); int size = 2000; bool cur = 0; int T = read(), k = read(); for(int i = 0; i <= size; i++, cur ^= 1) { C[cur][0] = 1; C[cur][i] = 1; for(int j = 1; j < i; j++) C[cur][j] = (C[cur^1][j-1] + C[cur^1][j]) % k; for(int j = 0; j <= i; j++) { f[i][j] = (!C[cur][j]); int t = 0; if(j) f[i][j] += f[i][j-1], t++; if(j <= i - 1) f[i][j] += f[i-1][j], t++; if(t == 2 && i && j) f[i][j] -= f[i-1][j-1]; } } while(T--) { int n = read(), m = read(); printf("%d\n", f[n][min(n,m)]); } return 0; }
Day 2 T2 earthworm
受到 q = 0 即蚯蚓长度没有增加的数据启发,我们发现每次坎出的两条蚯蚓的长度一定是单调降的,于是可以 O(n) 做了。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <algorithm> using namespace std; int read() { int x = 0, f = 1; char c = getchar(); while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); } while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); } return x * f; } #define maxn 100010 #define maxm 7100010 #define LL long long int n, m, q, u, v, t, A[maxn], B[maxm], C[maxm], lb, rb, lc, rc, ans[maxm], cnt; bool cmp(int a, int b) { return a > b; } void process(int tmp, int i, int len) { int nb = (LL)tmp * u / v, nc = tmp - nb; B[++rb] = nb - len - q; C[++rc] = nc - len - q; if(i % t == 0) ans[++cnt] = tmp; return ; } int main() { freopen("earthworm.in", "r", stdin); freopen("earthworm.out", "w", stdout); n = read(); m = read(); q = read(); u = read(); v = read(); t = read(); for(int i = 1; i <= n; i++) A[i] = read(); sort(A + 1, A + n + 1, cmp); // for(int i = 1; i <= n; i++) printf("%d%c", A[i], i < n ? ‘ ‘ : ‘\n‘); lb = 1; rb = 0; lc = 1; rc = 0; int pa = 1, len = 0; for(int i = 1; i <= m; i++, len += q) { int a, b, c; a = (pa <= n) ? A[pa] + len : -1; b = (lb <= rb) ? B[lb] + len : -1; c = (lc <= rc) ? C[lc] + len : -1; if(a >= b && a >= c) { pa++; process(a, i, len); } else if(b >= a && b >= c) { lb++; process(b, i, len); } else { lc++; process(c, i, len); } } for(int i = 1; i <= cnt; i++) printf("%d%c", ans[i], i < cnt ? ‘ ‘ : ‘\n‘); if(!cnt) putchar(‘\n‘); cnt = 0; for(int i = 1; i <= n + m; i++) { int a, b, c; a = (pa <= n) ? A[pa] + len : -1; b = (lb <= rb) ? B[lb] + len : -1; c = (lc <= rc) ? C[lc] + len : -1; if(a >= b && a >= c) { pa++; if(i % t == 0) ans[++cnt] = a; } else if(b >= a && b >= c) { lb++; if(i % t == 0) ans[++cnt] = b; } else { lc++; if(i % t == 0) ans[++cnt] = c; } } for(int i = 1; i <= cnt; i++) printf("%d%c", ans[i], i < cnt ? ‘ ‘ : ‘\n‘); if(!cnt) putchar(‘\n‘); return 0; }
Day 2 T3 angrybirds
状压 dp,设 f[S] 表示干掉集合 S 的猪需要的最少抛物线条数,转移时找到第一个没有被干掉的猪,再枚举另一头猪,两点确定一条过 (0, 0) 的抛物线,然后转移到当前集合与抛物线经过猪 的集合的并集。注意到每条抛物线经过猪的集合是可以预处理的。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <algorithm> #include <cmath> using namespace std; int read() { int x = 0, f = 1; char c = getchar(); while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); } while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); } return x * f; } #define maxn 23 #define maxs 362154 const double eps = 1e-6; struct Point { double x, y; Point() {} Point(double _, double __): x(_), y(__) {} bool operator < (const Point& t) const { return x != t.x ? x < t.x : y < t.y; } } ps[maxn]; int f[maxs], ls[maxn][maxn]; bool on(double a, double b, Point p) { return fabs(a * p.x * p.x + b * p.x - p.y) <= eps; } void up(int& a, int b) { if(a < 0) a = b; else a = min(a, b); return ; } int main() { freopen("angrybirds.in", "r", stdin); freopen("angrybirds.out", "w", stdout); int T = read(); while(T--) { int n = read(); read(); for(int i = 0; i < n; i++) scanf("%lf%lf", &ps[i].x, &ps[i].y); sort(ps, ps + n); memset(f, -1, sizeof(f)); memset(ls, 0, sizeof(ls)); f[0] = 0; for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) if(i != j) { double x1 = ps[i].x, y1 = ps[i].y, x2 = ps[j].x, y2 = ps[j].y; double b = (x1 * x1 * y2 - x2 * x2 * y1) / (x1 * x1 * x2 - x2 * x2 * x1); double a = (y1 - b * x1) / (x1 * x1); if(a >= 0.0) continue; int S = 0; for(int k = 0; k < n; k++) if(((S >> k & 1) ^ 1) && on(a, b, ps[k])) S |= (1 << k); ls[i][j] = S; } int all = (1 << n) - 1; for(int S = 0; S <= all; S++) if(f[S] >= 0) for(int j = 0; j < n; j++) if((S >> j & 1) ^ 1) { int tS = S | (1 << j); up(f[tS], f[S] + 1); for(int i = j + 1; i < n; i++) if((tS >> i & 1) ^ 1) up(f[tS | ls[i][j]], f[S] + 1); break; } printf("%d\n", f[all]); } return 0; }
这次 NOIP 为什么都是第二题最难 TAT
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