Section 1.1

省选准备/assets/day1/poj1236.cpp

@@ -0,0 +1,53 @@
+#include <cstdio>
+inline int min(int a, int b) { return a < b ? a : b; }
+const int maxn = 110, maxm = 10100;
+int head[maxn], next[maxm], to[maxm], ecnt, f[maxn], g[maxn];
+inline void addEdge(int f, int t)
+{
+    ecnt++;
+    next[ecnt] = head[f];
+    head[f] = ecnt;
+    to[ecnt] = t;
+}
+int dfn[maxn], low[maxn], stk[maxn], scc[maxn], scccnt, top, idx;
+void tarjan(int x)
+{
+    dfn[x] = low[x] = ++idx;
+    stk[top++] = x;
+    for (int i = head[x]; i; i = next[i])
+        if (!dfn[to[i]])
+            tarjan(to[i]), low[x] = min(low[x], low[to[i]]);
+        else if (!scc[to[i]])
+            low[x] = min(low[x], dfn[to[i]]);
+    if (dfn[x] == low[x])
+    {
+        scccnt++;
+        do
+            scc[stk[--top]] = scccnt;
+        while (stk[top] != x);
+    }
+}
+int main()
+{
+    int n;
+    scanf("%d", &n);
+    for (int i = 1, x; i <= n; i++)
+        for (scanf("%d", &x); x; scanf("%d", &x))
+            addEdge(i, x);
+    for (int i = 1; i <= n; i++)
+        if (!dfn[i]) tarjan(i);
+    for (int i = 1; i <= n; i++)
+        for (int j = head[i]; j; j = next[j])
+            if (scc[i] != scc[to[j]])
+                f[scc[i]]++, g[scc[to[j]]]++;
+    int ans1 = 0, ans2 = 0;
+    if (scccnt == 1)
+        printf("1\n0");
+    else
+    {
+        for (int i = 1; i <= scccnt; i++)
+            ans1 += f[i] == 0, ans2 += g[i] == 0;
+        printf("%d\n%d", ans2, ans1 > ans2 ? ans1 : ans2);
+    }
+    return 0;
+}

省选准备/assets/day1/poj2186.cpp

@@ -0,0 +1,65 @@
+#include <cstdio>
+inline int min(int a, int b) { return a < b ? a : b; }
+int head[10010], next[50010], to[50010], ecnt;
+int dfn[10010], low[10010], stk[10010], scc[10010], top, idx, scccnt;
+bool instk[10010];
+int deg[10010];
+inline void addEdge(int f, int t)
+{
+    ecnt++;
+    next[ecnt] = head[f];
+    head[f] = ecnt;
+    to[ecnt] = t;
+}
+void tarjan(int x)
+{
+    dfn[x] = low[x] = ++idx;
+    instk[stk[top++] = x] = true;
+    for (int cur = head[x]; cur; cur = next[cur])
+        if (!dfn[to[cur]])
+            tarjan(to[cur]), low[x] = min(low[x], low[to[cur]]);
+        else if (instk[to[cur]])
+            low[x] = min(low[x], dfn[to[cur]]);
+    if (dfn[x] == low[x])
+    {
+        scccnt++;
+        do
+        {
+            top--;
+            scc[stk[top]] = scccnt;
+            instk[stk[top]] = false;
+        } while (stk[top] != x);
+    }
+}
+int main()
+{
+    int n, m;
+    scanf("%d%d", &n, &m);
+    for (int i = 0, x, y; i < m; i++)
+    {
+        scanf("%d%d", &x, &y);
+        addEdge(x, y);
+    }
+    for (int i = 1; i <= n; i++)
+        if (!dfn[i])
+            tarjan(i);
+    for (int i = 1; i <= n; i++)
+        for (int cur = head[i]; cur; cur = next[cur])
+            if (scc[i] != scc[to[cur]])
+                deg[scc[i]]++;
+    int zcnt = 0, id = 0;
+    for (int i = 1; i <= scccnt; i++)
+        if (deg[i] == 0)
+            zcnt++, id = i;
+    if (zcnt != 1)
+        putchar('0');
+    else
+    {
+        int ans = 0;
+        for (int i = 1; i <= n; i++)
+            if (scc[i] == id)
+                ans++;
+        printf("%d", ans);
+    }
+    return 0;
+}

省选准备/assets/day1/poj3180.cpp

@@ -0,0 +1,46 @@
+#include <cstdio>
+inline int min(int a, int b) { return a < b ? a : b; }
+const int maxn = 100010;
+int head[maxn], next[maxn << 1], to[maxn << 1], ecnt, n, m;
+int dfn[maxn], scc[maxn], cnt[maxn], scccnt, stk[maxn], low[maxn], idx, top;
+inline void addEdge(int f, int t)
+{
+    ecnt++;
+    next[ecnt] = head[f];
+    head[f] = ecnt;
+    to[ecnt] = t;
+}
+void tarjan(int x)
+{
+    dfn[x] = low[x] = ++idx;
+    stk[top++] = x;
+    for (int i = head[x]; i; i = next[i])
+        if (!dfn[to[i]])
+            tarjan(to[i]), low[x] = min(low[x], low[to[i]]);
+        else if (!scc[to[i]])
+            low[x] = min(low[x], dfn[to[i]]);
+    if (dfn[x] == low[x])
+    {
+        scccnt++;
+        do
+            scc[stk[--top]] = scccnt;
+        while (stk[top] != x);
+    }
+}
+int main()
+{
+    scanf("%d%d", &n, &m);
+    for (int i = 0, x, y; i < m; i++)
+    {
+        scanf("%d%d", &x, &y);
+        addEdge(x, y);
+    }
+    for (int i = 1; i <= n; i++)
+        if (!dfn[i]) tarjan(i);
+    int ans = 0;
+    for (int i = 1; i <= n; i++) cnt[scc[i]]++;
+    for (int i = 1; i <= scccnt; i++)
+        if (cnt[i] > 1) ans++;
+    printf("%d", ans);
+    return 0;
+}

省选准备/assets/day1/tarjan.cpp

@@ -0,0 +1,14 @@
+void tarjan(int u)
+{
+    dfn[u] = low[u] = ++idx;
+    st[top++] = u;
+    for (Edge cur : G[u])
+        if (!dfn[cur.to])
+            tarjan(cur.to),
+            low[u] = min(low[u], low[cur.to]);
+        else if (!scc[cur.to])
+            low[u] = min(low[u], dfn[cur.to]);
+    if (dfn[u] == low[u] && ++cnt)
+        do scc[st[--top]] = cnt;
+        while (st[top] != u);
+}

省选准备/省选基础算法.tex

@@ -0,0 +1,69 @@
+\documentclass[]{cpp}
+\title{省选基础算法}
+\author{雷宇辰}
+\begin{document}
+	\setcounter{page}{0}
+	\maketitle
+	\newpage
+	\tableofcontents
+	\newpage
+    \setcounter{page}{1}
+	\section{day1 图论}
+	\subsection{有向图强连通分量的 Tarjan 算法}
+	\paragraph{定义}
+	在\textbf{有向图$G$}中，如果两个顶点$u,v$间存在一条路径$u$到$v$的路径且也存在一条$v$到$u$的路径，则称这两个顶点$u,v$是\textbf{强连通的(strongly connected)}。如果有向图$G$的每两个顶点都强连通，称$G$是一个\textbf{强连通图}。有向非强连通图的 极大强连通子图，称为\textbf{强连通分量(strongly connected components)}。若将有向图中的强连通分量都缩为一个点，则原图会形成一个 DAG（有向无环图）。
+	\subparagraph{极大强连通子图}
+	G 是一个极大强连通子图当且仅当 G 是一个强连通子图且不存在另一个强连通子图 G’使得 G 是 G’的真子集。
+	\paragraph{Tarjan 算法}
+	定义$dfn(u)$为结点$u$搜索的次序编号，给出函数$low(u)$使得\\
+	$low(u) = min$
+	\\$\{$\\
+	\verb|    |$dfn(u),$\\
+	\verb|    |$low(v),$  \quad $(u,v)$为树枝边，$u$为$v$的父结点\\
+	\verb|    |$dfn(v)\;$ \quad $(u,v)$为后向边或指向栈中结点的横叉边
+	\\$\}$\\
+	当结点$u$的搜索过程结束后，若$dfn(u)=low(u)$，则以$u$为根的搜索子树上所有还在栈中的结点是一个强连通分量。
+	\subparagraph{代码}$\\$
+	\codeinput[tarjan - SCC]{assets/day1/tarjan.cpp}
+	\paragraph{练习题}
+	\subparagraph{
+	\href{http://poj.org/problem?id=2186}{POJ2186}/
+	\href{http://www.lydsy.com/JudgeOnline/problem.php?id=1051}{BZOJ1051}
+	 - Popular Cows} 双倍的快乐
+	\codeinput[Popular Cows]{assets/day1/poj2186.cpp}
+	\subparagraph{\href{http://poj.org/problem?id=3180}{POJ3180} - The Cow Prom}
+	\verb|The N (2 <= N <= 10,000) cows are so excited.|
+	\codeinput[The Cow Prom]{assets/day1/poj3180.cpp}
+	\subparagraph{\href{http://poj.org/problem?id=3180}{POJ1236} - Network of Schools}
+	强连通分量缩点求出度为0的和入度为0的分量个数
+	\codeinput[Network of Schools]{assets/day1/poj1236.cpp}
+    \newpage
+	\section{day2 }
+    \newpage
+	\section{day3 }
+    \newpage
+	\section{day4 }
+    \newpage
+	\section{day5 }
+    \newpage
+	\section{day6 }
+    \newpage
+	\section{day7 }
+    \newpage
+	\section{day8 }
+    \newpage
+	\section{day9 }
+    \newpage
+	\section{day10 }
+    \newpage
+	\section{day11 }
+    \newpage
+	\section{day12 }
+    \newpage
+	\section{day13 }
+    \newpage
+	\section{day14 }
+    \newpage
+	\section{day15 }
+    \newpage
+\end{document}