NOIP 模拟赛
思路：求 n , m 的 gcd，然后用 n , m 分别除以 gcd；若 n 或 m 为偶数，则输出 1/2.
特别的，当 n = m = 1 时，应输出 1/1
#include <algorithm> #include <iostream> #include <cstring> #include <cstdio> using namespace std; typedef long long LL; LL n, m; LL gcd(LL x, LL y) { return y == 0 ? x : gcd(y, x % y); } int main() { freopen("line.in","r",stdin); freopen("line.out","w",stdout); cin >> n >> m; if (n == m && m == 1) return printf("1/1\n"), 0; LL tmp = gcd(n, m); n /= tmp; m /= tmp; if (!(n & 1) || !(m & 1)) return printf("1/2\n"), 0; fclose(stdin); fclose(stdout); return 0; }
正解：题目要求的是比值，故当 m 与 n 不互质时，我们可以求出 m 和 n 的最大公约数 d，并将 m /= d, n /= d，并不影响结果。故我们现在假定 m 和 n 互质。若m 和 n 中有一个为偶数， 那么根据对称性， 答案就是 1/2。 如果 m 和 n 均为奇数，那么答案就是(n*m+1) / (2*m*n）。
#include<iostream> #include<cstring> #include<cstdlib> #include<cstdio> #include<cmath> using namespace std; typedef long long LL; LL n, m, d; LL gcd(LL x, LL y) { return x % y ? gcd(y, x % y) : y; } int main() { freopen("line.in","r",stdin); freopen("line.out","w",stdout); cin >> n >> m; d = gcd(n, m); n /= d, m /= d; if ((n + m) & 1) cout << "1/2" << endl; else cout << (n * m + 1) / 2 << "/" << n * m << endl; return 0; }
思路：线段树维护是否有浓雾。一开始都没有，建树时每个位置的值都赋为 1；若进行 1 或 2 操作，用 flag 标记，因为是区间直接赋值，所以两种操作只需要一个标记即可。查询写错了，导致100 -> 10，对题意理解稍有偏差。
#include <algorithm> #include <cstring> #include <cstdio> using namespace std; const int N = 100005; int n, m, k, x, y; struct nond { int ll, rr; int sum; int flag; } tree[N << 2]; inline void update(int now) { tree[now].sum = tree[now << 1].sum + tree[now << 1 | 1].sum; } inline void down(int now) { if (tree[now].flag == 1) { tree[now << 1].flag = 1; tree[now << 1 | 1].flag = 1; tree[now << 1].sum = tree[now << 1].rr - tree[now << 1].ll + 1; tree[now << 1 | 1].sum = tree[now << 1 | 1].rr - tree[now << 1 | 1].ll + 1; } else { tree[now << 1].flag = -1; tree[now << 1 | 1].flag = -1; tree[now << 1].sum = 0; tree[now << 1 | 1].sum = 0; } tree[now].flag = 0; return ; } inline void build(int now, int l, int r) { tree[now].ll = l; tree[now].rr = r; tree[now].flag = 0; tree[now].sum = 1; if (l == r) return ; int mid = (l + r) >> 1; build(now << 1, l, mid); build(now << 1 | 1, mid + 1, r); update(now); } inline void change(int now, int l, int r) { if (tree[now].ll == l && tree[now].rr == r) { tree[now].flag = -1; tree[now].sum = 0; return ; } if (tree[now].flag != 0) down(now); int mid = (tree[now].ll + tree[now].rr) >> 1; if (l <= mid && mid < r) change(now << 1, l, mid), change(now << 1 | 1, mid + 1, r); else if (r <= mid) change(now << 1, l, r); else change(now << 1 | 1, l, r); update(now); } inline void become(int now, int l, int r) { if (tree[now].ll == l && tree[now].rr == r) { tree[now].flag = 1; tree[now].sum = tree[now].rr - tree[now].ll + 1; return ; } if (tree[now].flag != 0) down(now); int mid = (tree[now].ll + tree[now].rr) >> 1; if (l <= mid && mid < r) become(now << 1, l, mid), become(now << 1 | 1, mid + 1, r); else if (r <= mid) become(now << 1, l, r); else become(now << 1 | 1, l, r); update(now); } inline int query(int now, int l, int r) { if (tree[now].ll == l && tree[now].rr == r) return tree[now].sum; if (tree[now].flag != 0) down(now); int mid = (tree[now].ll + tree[now].rr) >> 1; if (l <= mid && mid < r) return query(now << 1, l, mid) + query(now << 1 | 1, mid + 1, r); else if (r <= mid) return query(now << 1, l, r); else return query(now << 1 | 1, l, r); } int main() { freopen("explore.in","r",stdin); freopen("explore.out","w",stdout); scanf("%d%d", &n, &m); build(1, 1, n); for (int i = 1; i <= m; ++i) { scanf("%d%d", &k, &x); if (k == 1) { scanf("%d", &y); change(1, x, y); } else if (k == 2) { scanf("%d", &y); become(1, x, y); } else { if (query(1, x, x) == 0) printf("0\n"); else { if (query(1, 1, x) == x || query(1, x, n) == n - x + 1) printf("INF\n"); else for (int j = 1; j <= (n >> 1); ++j) { int tmp = query(1, x - j, x + j); if (tmp != j << 1 | 1) { printf("%d\n", tmp); break; } } } } } fclose(stdin); fclose(stdout); return 0; }
#include<iostream> #include<algorithm> #include<cstdio> #include<cstdlib> #define fortodo(i, f, t) for (i = f; i <= t; i++) using namespace std; int lsd[200001], rsd[200001], lsid[200001], rsid[200001], cov[200001], segsize; bool emp[200001]; int SEG_Build(int L, int R) { int Nid = ++segsize; lsd[Nid] = L; rsd[Nid] = R; emp[Nid] = true; cov[Nid] = 0; if (L == R) lsid[Nid] = rsid[Nid] = -1; else { lsid[Nid] = SEG_Build(L, (L + R) / 2); rsid[Nid] = SEG_Build((L + R) / 2 + 1, R); }; return Nid; }; bool SEG_Empty(int Nid) { if (cov[Nid] == 0) return true; if (cov[Nid] == 1) return false; return emp[Nid]; }; void SEG_Reemp(int Nid) { emp[Nid] = SEG_Empty(lsid[Nid]) && SEG_Empty(rsid[Nid]); }; void SEG_Inherit(int Nid) { if (cov[Nid] == -1) return; if (lsd[Nid] == rsd[Nid]) return; cov[lsid[Nid]] = cov[Nid]; cov[rsid[Nid]] = cov[Nid]; cov[Nid] = -1; SEG_Reemp(Nid); }; void SEG_Paint(int Nid, int L, int R, int Color) { SEG_Inherit(Nid); if ((L == lsd[Nid]) && (R == rsd[Nid])) { cov[Nid] = Color; return; }; int Div = (lsd[Nid] + rsd[Nid]) / 2; if (L > Div) SEG_Paint(rsid[Nid], L, R, Color); if (R <= Div) SEG_Paint(lsid[Nid], L, R, Color); if ((L <= Div) && (R > Div)) { SEG_Paint(lsid[Nid], L, Div, Color); SEG_Paint(rsid[Nid], Div + 1, R, Color); }; SEG_Reemp(Nid); }; bool SEG_Query(int Nid, int L, int R) { SEG_Inherit(Nid); if (SEG_Empty(Nid)) return true; if ((L == lsd[Nid]) && (R == rsd[Nid])) return SEG_Empty(Nid); int Div = (lsd[Nid] + rsd[Nid]) / 2; if (L > Div) return SEG_Query(rsid[Nid], L, R); if (R <= Div) return SEG_Query(lsid[Nid], L, R); return SEG_Query(lsid[Nid], L, Div) && SEG_Query(rsid[Nid], Div + 1, R); }; int S, Q; int i, j; int Opt, X, Y; void Answer(int P) { if (!SEG_Query(1, P, P)) { printf("0\n"); return; }; if ((SEG_Query(1, 1, P)) || (SEG_Query(1, P, S))) { printf("INF\n"); return; }; int L, R, M, Ans[2]; L = 2; R = P; while (L < R) { M = (L + R) / 2; if (SEG_Query(1, M, P)) R = M; else L = M + 1; }; Ans[0] = L; L = P; R = S - 1; while (L < R) { M = (L + R + 1) / 2; if (SEG_Query(1, P, M)) L = M; else R = M - 1; }; Ans[1] = L; printf("%d\n", Ans[1] - Ans[0] + 1); }; int main() { freopen("explore.in", "r", stdin); freopen("explore.out", "w", stdout); scanf("%d%d", &S, &Q); segsize = 0; SEG_Build(1, S); fortodo(i, 1, Q) { scanf("%d%d", &Opt, &X); if (Opt != 3) scanf("%d", &Y); if (Opt == 1) SEG_Paint(1, X, Y, 1); if (Opt == 2) SEG_Paint(1, X, Y, 0); if (Opt == 3) Answer(X); }; return 0; };
思路：
1.暴搜 写挂了，样例都过不了
2.树形DP 不会写 后来讲题时发现自己想法是错的
3.并查集 将直接连接两个有苹果的节点的边打上标记，记录边权和，但仍会存在多个苹果间接相连的情况，没有想到解决办法
然后我就交了唯一一个过了样例但是最不靠谱的 3
然后就暴零了 T^T。。。
#include <algorithm> #include <iostream> #include <cstring> #include <cstdio> using namespace std; const int M = 100005; int n, m, tot; int t[M], fc[M]; int fa[M], f[M]; struct nond { int u, v, w; int flag; }e[M]; int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); } bool flag = false; inline void dfs(int now) { //可以明显看到没写完 因为不会写了 qwq if (f[now] == 1) flag = 1; } long long ans; int main() { freopen("apple.in","r",stdin); freopen("apple.out","w",stdout); scanf("%d%d", &n, &m); for (int i = 0; i < n; ++i) fa[i] = i; for (int i = 1; i < n; ++i) { scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w); e[i].flag = 0; } for (int i = 1; i <= m; ++i) { scanf("%d", &t[i]); f[t[i]] = 1; } for (int i = 1; i < n; ++i) { if (f[e[i].u] == 1 && f[e[i].v] == 1) { e[i].flag = 1; ans += e[i].w; ++tot; continue; } int x = find(e[i].u), y = find(e[i].v); fa[x] = y; } if (tot == m - 1) cout << ans; else { for (int i = 0; i < n; ++i) ++fc[find(i)]; for (int i = 0; i < n; ++i) { if (fc[i] > 1) dfs(i); if (tot == m - 1) break; } cout << ans; } fclose(stdin); fclose(stdout); return 0; }
正解：
#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <string> #include <cstdio> #include <cmath> #define REP(i, n) for (int i = 0; i < (n); ++i) #define FOR(i, a, b) for (int i = (a); i <= (b); ++i) #define ROF(i, a, b) for (int i = (a); i >= (b); --i) #define FEC(p, u) for (edge *p = G.head[u]; p; p = p->nxt) using namespace std; typedef long long LL; LL inf = 1LL<<60; int n, rt; bool a[200000]; LL f[200000], g[200000]; struct edge { int b, len; edge *nxt; } e[300000], *le; struct graph { edge *head[200000]; void init() { le = e; REP(i, n) head[i] = NULL; } void add(int x, int y, int z) { le->b = y, le->len = z, le->nxt = head[x], head[x] = le++; } } G; int fa[200000]; LL pre[200000]; int q[200000]; void init() { int k, u, v, w; scanf("%d%d", &n, &k); G.init(); REP(i, n-1) { scanf("%d%d%d", &u, &v, &w); G.add(u, v, w); G.add(v, u, w); } while (k--) { scanf("%d", &u); a[u] = true; } rt = u; } void bfs() { int R = 0; q[0] = rt; REP(L, n) FEC(p, q[L]) if (p->b != fa[q[L]]) { fa[p->b] = q[L]; pre[p->b] = p->len; q[++R] = p->b; } } void work() { ROF(i, n - 1, 0) { int x = q[i]; if (a[x]) { FEC(p, x) if (p->b != fa[x]) g[x] += f[p->b]; f[x] = g[x]+pre[x]; } else { FEC(p, x) if (p->b != fa[x]) f[x] += f[p->b]; g[x] = inf; FEC(p, x) if (p->b != fa[x]) g[x] = min(g[x], f[x] - f[p->b] + g[p->b]); f[x] = min(f[x], g[x]+pre[x]); } } } int main() { freopen("apple.in", "r", stdin); freopen("apple.out", "w", stdout); init(); bfs(); work(); cout << f[rt] << endl; return 0; }