2025-11-13~15 hetao1733837的刷题记录

2025-11-13~15 hetao1733837的刷题记录

11-13

[JOISC 2014]Water Bottle

原题链接1：[P14422 [JOISC 2014] 水桶 / Water Bottle]([P14422 JOISC 2014] 水桶 / Water Bottle - 洛谷)

原题链接2：#2876. 「JOISC 2014 Day2」水壶

洛谷题名怎么和LOJ不一样啊/(ㄒoㄒ)/~~

分析

呃，不知从哪冒出了瓶颈最短路这个词，被mhh认为是正确的，因此引发了瓶颈生成树（整张图），因为由定义“无向图$G$的瓶颈生成树是这样的一个生成树，它的最大的边权值在$G$的所有生成树中最小。”而瓶颈生成树又是最小生成树，所以转化为原图求最小生成树，两点之间距离转化为树上路径，可以直接建$Kruskal$重构树（最小生成树多维护一下最大两点距离而已）跑倍增，斜二倍增甚至树剖都可以。但是问题在于$O(n^{2})$建图是无法接受的，因为原图是个矩阵……线段树优化建图，？可行但是吃多了。看一眼$tag$，惊奇发现$bfs$，容我稍考……多源$bfs$即可？好吧，我基础是一坨……

那么，就可以写代码了？我真是吃多了……

正解

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int HW = 2005, P = 400005;
char c[HW][HW];
pair<int, int> vis[HW][HW];
struct node{int a, b;
}inp[P];
int h, w, p, q;
int dx[] = {1, 0, -1, 0},dy[] = {0, -1, 0, 1};
int fa[P * 2]; 
int getfa(int x){return x == fa[x] ? x : fa[x] = getfa(fa[x]);
}
struct node1{int u, v, w;
}e[HW * HW];
bool cmp(node1 o1, node1 o2){return o1.w < o2.w;
}
vector<int> g[P * 2]; 
int we[P * 2], de[P * 2]; 
pair<int, int> f[P * 2][21]; 
void dfs(int u){for (int i = 1; i <= 20; i++){f[u][i].first = f[f[u][i - 1].first][i - 1].first;f[u][i].second = max(f[u][i - 1].second, f[f[u][i - 1].first][i - 1].second);}for (auto v : g[u]){f[v][0].first = u;f[v][0].second = we[v];de[v] = de[u] + 1;dfs(v);}return ;
}
int ans;
void lca(int x, int y){ans = 0;if (de[x] < de[y])swap(x, y);for (int i = 20; i >= 0; i--){if (de[f[x][i].first] >= de[y]){ans = max(ans, f[x][i].second);x = f[x][i].first;}} if (x == y){return ;}else{for (int i = 20; i >= 0; i--){if (f[x][i].first != f[y][i].first){ans = max(ans, f[x][i].second);ans = max(ans, f[y][i].second);x = f[x][i].first;y = f[y][i].first;}}ans = max(ans, we[x]);ans = max(ans, we[y]);ans = max(ans, we[f[x][0].first]);}return ;
}
signed main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);//inputcin >> h >> w >> p >> q;for (int i = 1; i <= h; i++){string s;cin >> s;for (int j = 0; j < w; j++){c[i][j + 1] = s[j];}}for (int i = 1; i <= h; i++){for (int j = 1; j <= w; j++){vis[i][j] = {0, 0};}}queue<pair<int, int>> qu;for (int i = 1; i <= p; i++){cin >> inp[i].a >> inp[i].b;vis[inp[i].a][inp[i].b] = {0, i};qu.push({inp[i].a, inp[i].b});}//bfsint top = 0;while (!qu.empty()){int x = qu.front().first;int y = qu.front().second;qu.pop();for (int i = 0; i < 4; i++){int tx = x + dx[i], ty = y + dy[i];if (tx >= 1 && tx <= h && ty >= 1 && ty <= w && c[tx][ty] == '.'){if (!vis[tx][ty].second){vis[tx][ty] = {vis[x][y].first + 1, vis[x][y].second};qu.push({tx, ty});}else if (vis[x][y].second != vis[tx][ty].second){e[++top] = {vis[x][y].second, vis[tx][ty].second, vis[x][y].first + vis[tx][ty].first};}}}}//Kruskalfor (int i = 1; i < p; i++)e[++top] = {i, i + 1, 0x3f3f3f3f3f3f3f3f};sort(e + 1, e + top + 1, cmp);int cur = p;for (int i = 1; i <= p * 2; i++)fa[i] = i;for (int i = 1; i <= top; i++){int x = e[i].u, y = e[i].v, z = e[i].w;int fx = getfa(x), fy = getfa(y);if (fx == fy)continue;cur++;g[cur].push_back(fx);g[cur].push_back(fy);fa[fx] = fa[fy] = cur; we[cur] = z;}for (int i = 1; i <= cur; i++){for (int j = 0; j <= 20; j++){f[i][j] = {0, 0};}}de[cur] = 1;f[cur][0] = {cur, we[cur]};dfs(cur);//solvewhile (q--){int x, y;cin >> x >> y;if (x == y){cout << 0 << '\n';continue;}ans = 0;lca(x, y);if (ans == 0x3f3f3f3f3f3f3f3f)cout << -1 << '\n';elsecout << ans << '\n';}return 0;
}

吃饱了……

[JOISC 2014] Making Friends is Fun

原题链接1：[JOISC 2014] 有趣的交朋友 / Making Friends is Fun

原题链接2：「JOISC 2014 Day2」交朋友

分析

利用人类智慧一下就模拟出来了，哈哈哈哈哈哈哈哈哈……计算机，靠你了……靠不住/(ㄒoㄒ)/~~

像这样一个点指向两外两个点，有点像树啊……加上边像基环树？我在胡言乱语……问一下……$tag$里给出了并查集，给我一些恍惚的启发……难道就是这样的“有向三角”？Oh，我有了一些思路，额外记录每个点的入度和出度，然后对于没有入度的点，对其出度进行排列组合，呃，好像还要做一些去重，这些是$tag$里的$bfs$吗？我需要询问。

呃，花花的思路大概是打标记打标记，然后啊吧啊吧啊吧啊，我也并非深刻理解。

但是，aoao的很具启发性，是一种很数学的做法，类似排列组合，虽然并未理解精髓。

正解

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 100005;
int n, m, a, b;
vector<int> e[N];
int fa[N], sz[N];
int getfa(int x){return x == fa[x] ? fa[x] : fa[x] = getfa(fa[x]);
}
void merge(int x, int y){x = getfa(x);y = getfa(y);if (x != y){if (sz[x] > sz[y])swap(x, y);fa[x] = y;sz[y] += sz[x];}
}
void dfs(int u){for (auto v : e[u]){if (getfa(u) == getfa(v))continue;merge(u, v);dfs(v);}
}
signed main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin >> n >> m;for (int i = 1; i <= m; i++){cin >> a >> b;e[a].push_back(b);}for (int i = 1; i <= n; i++){fa[i] = i;sz[i] = 1;}for (int i = 1; i <= n; i++){int tmp = 0;for (auto v : e[i]){if (!tmp){tmp = v;continue;}merge(tmp, v);tmp = v;}}for (int i = 1; i <= n; i++)if (sz[getfa(i)] >= 2)dfs(i);int ans = 0;for (int i = 1; i <= n; i++){if (getfa(i) == i)ans += sz[i] * (sz[i] - 1);for (auto v : e[i]){if (getfa(v) == getfa(i)){continue;}ans++;}}cout << ans;
}

花花思路也很巧妙！对于两个点，如果连了双向边，那么就可以直接合并它们的儿子，对于一些还没有合并儿子的，直接合并然后递归儿子即可，那么打一下标记即可。代码我会(will)写吗？

11-14

[CSP-S 2023] 结构体

原题链接：[CSP-S 2023] 结构体

分析

吃饭前写了$op1$和$op2$两个，交了一发全WA了。

${\color{Black}{16:20}}$ 不会后两个，感觉要额外记录，初觉不对。决定找AI把前两个修一下，瞅一眼题解。

${\color{Black}{16:35}}$ 吃不动了，决定he题解。发现有一篇思路十分接近。

wcnm，为啥题解是错的！！！

™不写了。

[CSP-S 2025] 道路修复 / road

原题链接：[CSP-S 2025] 道路修复 / road

分析

考场上在仅剩的40分钟里想到了最小生成树，结果，忘了板子，以为赛季结束了/(ㄒoㄒ)/~~

正解

#include <bits/stdc++.h>
using namespace std;
const int N = 2000005, M = 10000005;
int n, m, k;
int fa[N];
struct node{int u, v, w;
}e[M], g[N];
int c[N];
bool vis[N];
bool cmp(node x, node y){return x.w < y.w;
}
int getfa(int x){return fa[x] == x ? x : fa[x] = getfa(fa[x]);
}
int main(){cin >> n >> m >> k;for (int i = 1; i <= m; i++){cin >> e[i].u >> e[i].v >> e[i].w;} sort(e + 1, e + m + 1, cmp);for (int i = 1; i <= n; i++)fa[i] = i;int cnt = 0;for (int i = 1; i <= m; i++){int fu = getfa(e[i].u), fv = getfa(e[i].v);if (fu == fv) continue;fa[fu] = fv;g[++cnt] = e[i];if (cnt == n - 1) break;}cnt = m;for (int i = 1; i <= k; i++){cin >> c[i];for (int j = 1; j <= n; j++){cnt++;e[cnt].u = j;e[cnt].v = n + i;cin >> e[cnt].w;}}sort(e + 1, e + cnt + 1, cmp);long long ans = 0x3f3f3f3f3f3f3f3f;for (int mask = 0; mask < (1 << k); mask++){long long cur = 0;int num = 0;for (int j = 0; j < k; j++){if (mask >> j & 1){num++;cur += c[j + 1];}}for (int i = 1; i <= n + k; i++)fa[i] = i;int need = n + num - 1;for (int i = 1; i <= cnt; i++){int u = e[i].u, v = e[i].v;if (v > n) {int id = v - n;if (!(mask >> (id - 1) & 1)) continue;}int fu = getfa(u), fv = getfa(v);if (fu == fv) continue;fa[fu] = fv;cur += e[i].w;need--;if (need == 0) break;}if (need == 0)ans = min(ans, cur);}cout << ans;return 0;
}

两遍$Kruskal$，行吧……我是sugar这启示我们，$\color{Black}{多重限制，时间复杂度允许，就多跑几层，分层的思想还是很常用的。}$

好吧，要去听烫烫选科讲座了，我不明白，竞赛班不是选物化生吗？反正我是这么打算的。毕竟文科从开学以来没咋听过(●'◡'●)

OSU!

原题链接：OSU!

分析

维护$x1$表示$x$的期望，$x2$表示$x^2$的期望。

$x1_i=(x1_{i-1}+1)\times p_i$

$x2_i=(x2_{i-1}+2\times x1_{i - 1} + 1)\times p_i$

$ans_i=ans_{i-1}+(3 \times x2_{i - 1}+ 3 \times x1_{i-1} + 1) \times p_i$

正解

#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n;
double x1[N], x2[N], ans[N], p[N];
int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin >> n;for (int i = 1; i <= n; i++){cin >> p[i];}for (int i = 1; i <= n; i++){x1[i] = (x1[i - 1] + 1) * p[i];x2[i] = (x2[i - 1] + 2 * x1[i - 1] + 1) * p[i];ans[i] = ans[i - 1] + (3 * x2[i - 1] + 3 * x1[i - 1] + 1) * p[i];}cout << fixed << setprecision(1) << ans[n];
}

[JXOI2018] 游戏

原题链接：[JXOI2018]游戏

分析

设有$k$个关键点，把序列分成$k+1$段。对于每个非关键点，排在关键点后面概率为$P=\frac{1}{k+1}$，期望$E=(n-k)P=\frac{n-k}{k+1}$，最后一个关键点期望$E_n=\frac{k(n+1)}{k+1}$，最终答案$\frac{k}{k+1}(n+1)!$。

正解

#include <bits/stdc++.h>
#define mod 1000000007
#define int long long
using namespace std;
const int N = 10000005;
int l, r;
bool vis[N];
signed main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin >> l >> r;int k = 0;for (int i = l; i <= r; i++){if (!vis[i]){k++;for (int j = i * 2; j <= r; j += i){vis[j] = 1;}}}int ans = k;for (int i = 1; i <= r - l + 2; i++){if (i != k + 1)ans = ans * i % mod;}cout << ans;
}

11.15

[JOISC 2014] 邮戳收集 / Collecting Stamps

原题链接1：Collecting Stamps

原题链接2：邮戳拉力赛

分析

$08点33分$ 上下行给我一种二分图的感觉，两站台之间打卡点，也就是说要蛇形走位？$unk$，那么转化为二分图的……（我还没想好）好困……woc，洛谷比赛开了，AeeE5x说T1是$mex$，不想打了/(ㄒoㄒ)/~~

$\color{Black}{11点51分}$ 这比赛是人？不打了，呱呱为啥过了T1，为啥我大样例RE了/ll

还是看《星轨》吧……

ber，我LOJ怎么亖了/ll

cao了，模拟赛10分。

好吧，我要吃了这题。

$\color{Black}{2025.11.16}$ LOJ活了/ll

并非二分图，但是与匹配联系紧密，考虑$DP$。设$f_{i,j}$表示考虑前$i$站，走了$j$次回头路的最小时间花费。

那么转移有：

$f_{i,j}=f_{i-1,j}+u_i+v_i+2jT$。

$f_{i,j}=f_{i-1,j}+e_i+d_i+2jT$，其中$j>0$。

$f_{i,j}=f_{i-1,j+1}+u_i+e_i+2jT$。

$f_{i,j}=f_{i-1,j-1}+d_i+v_i+2jT$，其中$j>0$。

$f_{i,j}=f_{i,j-1}+v_i+d_i$，其中$j>0$。

$f_{i,j}=f_{i,j+1}+u_i+e_i$，其中$n-1\ge j\ge 0$

答案即为$ans=f[n][0]+(n+1)\times T$

$\color{Black}{16:51}$ 连交四发，全部WA掉，开he题解。

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 3010;
int n, T, dp[N][N];
signed main() {ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin >> n >> T;memset(dp, 0x3f, sizeof(dp));dp[0][0] = 0;for (int i = 1; i <= n; i++) {int u, v, e, d;cin >> u >> v >> d >> e;for (int j = 0; j <= n; j++)dp[i - 1][j] += 2 * T * j;for (int j = 0; j <= n; j++)dp[i][j] = min(dp[i][j], dp[i - 1][j] + u + v);for (int j = 1; j <= n; j++)dp[i][j] = min(dp[i][j], dp[i - 1][j] + d + e);for (int j = 1; j <= n; j++)dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + v + d);for (int j = 0; j < n; j++)dp[i][j] = min(dp[i][j], dp[i - 1][j + 1] + u + e);for (int j = 1; j <= n; j++)dp[i][j] = min(dp[i][j], dp[i][j - 1] + v + d);for (int j = n - 1; j >= 0; j--)dp[i][j] = min(dp[i][j], dp[i][j + 1] + u + e);}cout << dp[n][0] + (n + 1) * T;return 0;
}

你说他shit吗？从码量上，并非，从思维上应该降绿？我的问题。

【模板】AC 自动机

原题链接：【模板】AC 自动机

正解

#include <bits/stdc++.h>
using namespace std;
const int N = 200005, V = 26, M = 2000005;
int reads(char *s){int t = 0;char c = getchar();while (c < 'a' || c > 'z')c = getchar();while (c >= 'a' && c <='z'){s[++t] = c - 'a';c = getchar();}return t;
}
int n, ncnt, child[N][V], fail[N], m, ans[N], res[N], indeg[N];
char s[M];
int rt;
vector<int> vec[N];
void init(){ncnt = rt = 1;
}
void ins(char *s, int n, int id){int cur = rt;for (int i = 1; i <= n; i++){if (!child[cur][s[i]])child[cur][s[i]] = ++ncnt;cur = child[cur][s[i]];}vec[cur].emplace_back(id);
}
queue<int> q;
void build(){for (int i = 0; i < V; i++){if (child[rt][i]){fail[child[rt][i]] = rt;q.push(child[rt][i]);} else{child[rt][i] = rt;  }}while (!q.empty()){int u = q.front();q.pop();for (int i = 0; i < V; i++){if (child[u][i]){fail[child[u][i]] = child[fail[u]][i];q.push(child[u][i]);} else {child[u][i] = child[fail[u]][i];}}}
}
void query(char *s, int n){int cur = rt;for (int i = 1; i <= n; i++){cur = child[cur][s[i]];ans[cur]++;}
}
void solve(){for (int i = 1; i <= ncnt; i++){if (fail[i] != i) {  indeg[fail[i]]++;}}for (int i = 1; i <= ncnt; i++){if (indeg[i] == 0 && i != rt) {q.push(i);}}while (!q.empty()){int u = q.front();q.pop();if (fail[u] != u) {ans[fail[u]] += ans[u];indeg[fail[u]]--;if (indeg[fail[u]] == 0 && fail[u] != rt){q.push(fail[u]);}}}for (int i = 1; i <= ncnt; i++){for (int p : vec[i]){res[p] = ans[i];}}
}
int main(){init();cin >> n;for (int i = 1; i <= n; i++){m = reads(s);ins(s, m, i);}build();m = reads(s);query(s, m);solve();for (int i = 1; i <= n; i++)cout << res[i] << '\n';return 0;
}

不知道PYD1咋写的，十分扭曲，套了$bfs$序，我没抄全，加了个拓扑排序优化才过，我需要另一个干练一点的AC自动机版本。

AC 自动机（简单版）

原题链接：AC 自动机（简单版）

正解

#include <bits/stdc++.h>
using namespace std;
const int N = 1000005, V = 26, M = 2000005;
int reads(char *s){int t = 0;char c = getchar();while (c < 'a' || c > 'z')c = getchar();while (c >= 'a' && c <='z'){s[++t] = c - 'a';c = getchar();}return t;
}
int n, ncnt, child[N][V], fail[N], m, ans[N], res[N], indeg[N];
char s[M];
int rt;
vector<int> vec[N];
void init(){ncnt = rt = 1;
}
void ins(char *s, int n, int id){int cur = rt;for (int i = 1; i <= n; i++){if (!child[cur][s[i]])child[cur][s[i]] = ++ncnt;cur = child[cur][s[i]];}vec[cur].emplace_back(id);
}
queue<int> q;
void build(){for (int i = 0; i < V; i++){if (child[rt][i]){fail[child[rt][i]] = rt;q.push(child[rt][i]);} else{child[rt][i] = rt;  }}while (!q.empty()){int u = q.front();q.pop();for (int i = 0; i < V; i++){if (child[u][i]){fail[child[u][i]] = child[fail[u]][i];q.push(child[u][i]);} else {child[u][i] = child[fail[u]][i];}}}
}
void query(char *s, int n){int cur = rt;for (int i = 1; i <= n; i++){cur = child[cur][s[i]];ans[cur]++;}
}
void solve(){for (int i = 1; i <= ncnt; i++){if (fail[i] != i) {  indeg[fail[i]]++;}}for (int i = 1; i <= ncnt; i++){if (indeg[i] == 0 && i != rt) {q.push(i);}}while (!q.empty()){int u = q.front();q.pop();if (fail[u] != u) {ans[fail[u]] += ans[u];indeg[fail[u]]--;if (indeg[fail[u]] == 0 && fail[u] != rt){q.push(fail[u]);}}}for (int i = 1; i <= ncnt; i++){for (int p : vec[i]){res[p] = ans[i];}}
}
int main(){init();cin >> n;for (int i = 1; i <= n; i++){m = reads(s);ins(s, m, i);}build();m = reads(s);query(s, m);solve();int cnt = 0;for (int i = 1; i <= n; i++)if (res[i])cnt++;cout << cnt;return 0;
}

经过对PYD1代码稍加的修改，得到神秘代码通过此题。