T1、矩阵乘法

题目链接

给定一个 \(n \times n \ (n \leq 500)\) 的矩阵 \((a_{i,j} \leq 10 ^ 9)\)，\(q \ (q \leq 6 \times 10 ^ 5)\) 组询问，每次询问一个子矩阵的第 \(k\) 大元素 (保证存在)。

\(Sol\)：

整体二分主席树，注意常数因子带来的影响；
全场只有我一个常数怪 \(95\) 分。

时间复杂度 \(O(q \log_2^2 n)\)。

\(Source\)：

//#pragma GCC optimize(2)
#include <cstdio>
#include <cstring>
#include <algorithm>
inline int in() {int x = 0; char c = getchar(); bool f = 0;while (c < '0' || c > '9')f |= c == '-', c = getchar();while (c >= '0' && c <= '9')x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();return f ? -x : x;
}
template<typename T>inline void chk_min(T &_, T __) { _ = _ < __ ? _ : __; }
template<typename T>inline void chk_max(T &_, T __) { _ = _ > __ ? _ : __; }const int N = 505, Q = 60005;struct info {int id, x1, y1, x2, y2, k;
} b[N * N + Q];
int n, m, q, id[N * N + Q], tmp1[N * N + Q], tmp2[N * N + Q], pos[N * N], res[Q];
int nn, mp[N * N];struct persistable_segment_tree {int sum[N * N * 11], c[N * N * 11][2], rt[N * N], tot;void clear() {tot = 0;}int new_node() {++tot;sum[tot] = c[tot][0] = c[tot][1] = 0;return tot;}void modify(int pos, int tl, int tr, int pre, int &p) {p = new_node(), sum[p] = sum[pre];++sum[p];if (tl == tr)return ;int mid = (tl + tr) >> 1;if (mid >= pos) {c[p][1] = c[pre][1];modify(pos, tl, mid, c[pre][0], c[p][0]);} else {c[p][0] = c[pre][0];modify(pos, mid + 1, tr, c[pre][1], c[p][1]);}}int query(int l, int r, int tl, int tr, int pre, int p) {if (l <= tl && tr <= r)return sum[p] - sum[pre];int mid = (tl + tr) >> 1;if (mid < l)return query(l, r, mid + 1, tr, c[pre][1], c[p][1]);if (mid >= r)return query(l, r, tl, mid, c[pre][0], c[p][0]);return query(l, r, tl, mid, c[pre][0], c[p][0]) +query(l, r, mid + 1, tr, c[pre][1], c[p][1]);}
} T;void binary_search(int l, int r, int s, int t) {if (l > r || s > t)return ;if (l == r) {for (int i = s; i <= t; ++i)if (b[id[i]].id)res[b[id[i]].id] = mp[l];return ;}int mid = (l + r) >> 1, p1 = 0, p2 = 0, tot = 0;T.clear();for (int i = s; i <= t; ++i) {if (!b[id[i]].id) {if (b[id[i]].k <= mp[mid]) {T.modify(b[id[i]].y1, 1, n, T.rt[tot], T.rt[tot + 1]);tmp1[++p1] = id[i];pos[++tot] = b[id[i]].x1;} else {tmp2[++p2] = id[i];}} else {int x, y, w;x = std::lower_bound(pos + 1, pos + 1 + tot, b[id[i]].x1) - pos;y = std::upper_bound(pos + 1, pos + 1 + tot, b[id[i]].x2) - pos - 1;w = T.query(b[id[i]].y1, b[id[i]].y2, 1, n, T.rt[x - 1], T.rt[y]);if (w >= b[id[i]].k) {tmp1[++p1] = id[i];} else {b[id[i]].k -= w;tmp2[++p2] = id[i];}}}for (int i = 1; i <= p1; ++i)id[s + i - 1] = tmp1[i];for (int i = 1; i <= p2; ++i)id[s + p1 + i - 1] = tmp2[i];binary_search(l, mid, s, s + p1 - 1);binary_search(mid + 1, r, s + p1, t);
}int main() {//freopen("in", "r", stdin);n = in(), q = in();for (int i = 1; i <= n; ++i)for (int j = 1; j <= n; ++j) {b[++m] = (info){0, i, j, 0, 0, in()};mp[++nn] = b[m].k;}for (int i = 1; i <= q; ++i)b[++m] = (info){i, in(), in(), in(), in(), in()};for (int i = 1; i <= m; ++i)id[i] = i;std::sort(mp + 1, mp + 1 + nn);nn = std::unique(mp + 1, mp + 1 + nn) - mp - 1;binary_search(0, nn, 1, m);for (int i = 1; i <= q; ++i)printf("%d\n", res[i]);return 0;
}

T2、Tree

题目链接

给定一张 \(n \ (n \leq 100)\) 个点 \(m \ (m \leq 2000)\) 条边的无相连通图，有边权 \(C_i \ (0 \leq C_i \leq 100)\)，求最小标准差生成树。

\(Sol\)：

可以将答案看做一个关于平均数的函数：
\[ f(x) = \sqrt{ \frac{\sum (C_i - x) ^ 2}{n - 1} } \]
枚举 \(x\)，将边权设为 \(C_i - x\)，求最小生成树，再按最小生成树的真实平均数更新答案。
具体证明我不会，大概是对于每个 \(x\) 得到的真实平均值一定是与 \(x\) 相邻的。

时间复杂度 \(O(\)能跑多少跑多少\()\)。

\(Source\)：

#include <cstdio>
#include <algorithm>
#include <cmath>
typedef double db;
int in() {int x = 0; char c = getchar(); bool f = 0;while (c < '0' || c > '9')f |= c == '-', c = getchar();while (c >= '0' && c <= '9')x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();return f ? -x : x;
}
template<typename T>inline void chk_min(T &_, T __) { _ = _ < __ ? _ : __; }
template<typename T>inline void chk_max(T &_, T __) { _ = _ > __ ? _ : __; }const int N = 105, M = 2005;
const db eps = 1e-5;db res, now;
int n, m;
int fa[N];
struct edge {int u, v, w;inline bool operator < (const edge &b) const {return (this->w - now) * (this->w - now) < (b.w - now) * (b.w - now);}
} e[M];int get_fa(int u) {return fa[u] == u ? u : fa[u] = get_fa(fa[u]);
}void calc() {res = 1 << 30;for (now = 100; now > 0; now -= 0.25) {std::sort(e + 1, e + 1 + m);int now_edge = 0, _sum = 0, sum = 0;for (int i = 1; i <= n; ++i)fa[i] = i;for (int i = 1; i <= m && now_edge < n - 1; ++i) {int fx = get_fa(e[i].u), fy = get_fa(e[i].v);if (fx == fy)continue;fa[fx] = fy;++now_edge;_sum += e[i].w * e[i].w, sum += e[i].w;}db tmp = sqrt(((db)(n - 1) * _sum - sum * sum) / (n - 1) / (n - 1));if (res - tmp > eps)res = tmp;}
}int main() {//freopen("in", "r", stdin);n = in(), m = in();if (n == 1) {puts("0.0000");return 0;}for (int i = 1; i <= m; ++i)e[i] = (edge){in(), in(), in()};calc();printf("%.4lf\n", res);return 0;
}

T3、Points and Segments

题目链接

数轴上给定 \(n \ (n \leq 10 ^ 5)\) 条线段，每条线段为红色或蓝色，给出一种颜色方案，使得数轴上所有点被不同颜色覆盖的次数差绝对值不超过 \(1\)。

\(Sol\)：

可以把原数轴上的点看作线段，点被颜色覆盖可以看作线段被覆盖；
题目中\([l,r]\) 线段就可以当作新图中 \([l,r + 1]\) 中所有线段被覆盖。

问题转化为所有线段被覆盖的绝对值不超过 \(1\)；
\(l\) 向 \(r + 1\) 连无向边，\(l\) 到 \(r + 1\) 即为染成蓝色，否则染成红色；
这样如果该图有欧拉回路 (所有点都是偶点)，则两色覆盖次数都相等；
若有奇点，那么奇点个数一定是偶数 (一条线段有两个端点)，则从左到右第 \(2i\) 和 \(2i + 1\) 个奇点连边；
该图一定有欧拉回路，所有被覆盖的次数都相等；
因为新加的边之间一定不会相交，所以把他们删掉后每个点只会少覆盖一次。

时间复杂度 \(O(n)\)。

\(Source\)：

#include <cstdio>
#include <cstring>
#include <algorithm>
inline int in() {int x = 0; char c = getchar(); bool f = 0;while (c < '0' || c > '9')f |= c == '-', c = getchar();while (c >= '0' && c <= '9')x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();return f ? -x : x;
}
template<typename T>inline void chk_min(T &_, T __) { _ = _ < __ ? _ : __; }
template<typename T>inline void chk_max(T &_, T __) { _ = _ > __ ? _ : __; }const int N = 1e5 + 5;struct node {int l, r;
} a[N];
struct edge {int next, to;
} e[N << 2];
int ecnt = 1, head[N << 1];
int n, nn, mp[N << 1], res[N], deg[N << 1];
bool vis[N << 1], used[N << 1];inline void jb(const int u, const int v) {e[++ecnt] = (edge){head[u], v}, head[u] = ecnt;e[++ecnt] = (edge){head[v], u}, head[v] = ecnt;++deg[u], ++deg[v];
}void prep() {for (int i = 1; i <= n; ++i) {a[i] = (node){in(), in()};mp[++nn] = a[i].l, mp[++nn] = a[i].r;}std::sort(mp + 1, mp + 1 + nn);nn = std::unique(mp + 1, mp + 1 + nn) - mp - 1;for (int i = 1; i <= n; ++i) {a[i].l = std::lower_bound(mp + 1, mp + 1 + nn, a[i].l) - mp;a[i].r = std::lower_bound(mp + 1, mp + 1 + nn, a[i].r) - mp;jb(a[i].l, a[i].r);}int last = 0;for (int i = 1; i <= nn; ++i)if (deg[i] & 1) {if (!last)last  = i;elsejb(last, i), last = 0;}
}void dfs(const int u) {used[u] = 1;for (int i = head[u]; i; i = e[i].next) {if (!vis[i >> 1]) {vis[i >> 1] = 1;dfs(e[i].to);if (u > e[i].to)res[i >> 1] = 1;elseres[i >> 1] = 0;}}
}void work() {for (int i = 1; i <= nn; ++i)if (!used[i])dfs(i);
}int main() {//freopen("in", "r", stdin);n = in();prep();work();for (int i = 1; i <= n; ++i)printf("%d ", res[i]);puts("");return 0;
}