partial Section 2.4

OnlineJudges/poj/3974.cpp

@@ -0,0 +1,30 @@
+#include <cstdio>
+#include <cstring>
+inline int min(int a, int b) { return a < b ? a : b; }
+inline int max(int a, int b) { return a > b ? a : b; }
+char buf[1000100], str[2000200];
+int R[2000200], T;
+int main()
+{
+    while (scanf("%s", buf), buf[0] != 'E')
+    {
+        int m = int(strlen(buf)), n = 0;
+        str[n++] = '!';
+        str[n++] = '#';
+        for (int i = 0; i < m; i++)
+            str[n++] = buf[i], str[n++] = '#';
+        str[n++] = '#';
+        str[n++] = '?';
+        int p = 0, mx = 0, ans = 0;
+        for (int i = 1; i < n; i++)
+        {
+            R[i] = mx > i ? min(R[2 * p - i], mx - i) : 1;
+            while (str[i + R[i]] == str[i - R[i]]) R[i]++;
+            if (R[i] + i > mx)
+                mx = i + R[p = i];
+            ans = max(ans, R[i]);
+        }
+        printf("Case %d: %d\n", ++T, ans - 1);
+    }
+    return 0;
+}

省选准备/assets/day2/hdu2222.cpp

@@ -0,0 +1,78 @@
+#include <cstdio>
+#include <cstring>
+struct node
+{
+    node *trans[26], *fail;
+    int cnt;
+} nodes[500010], virt;
+node *que[500010];
+int tot;
+node *root;
+node *new_node() { return &nodes[tot++]; }
+void insert(char *str)
+{
+    node *cur = root;
+    for (; *str; str++)
+    {
+        if (cur->trans[*str - 'a'] == 0)
+            cur->trans[*str - 'a'] = new_node();
+        cur = cur->trans[*str - 'a'];
+    }
+    cur->cnt++;
+}
+void buildFail()
+{
+    int h = 0, t = 0;
+    root->fail = &virt;
+    que[t++] = root;
+    while (h < t)
+    {
+        node *cur = que[h++];
+        for (int i = 0; i < 26; i++)
+        {
+            node *f = cur->fail;
+            while (f->trans[i] == 0) f = f->fail;
+            f = f->trans[i];
+            if (cur->trans[i])
+                (que[t++] = cur->trans[i])->fail = f;
+            else
+                cur->trans[i] = f;
+        }
+    }
+}
+char buf[1000010];
+int vis[500010];
+int main()
+{
+    memset(vis, -1, sizeof(vis));
+    int T, n;
+    scanf("%d", &T);
+    while (T--)
+    {
+        memset(nodes, 0, sizeof(nodes));
+        tot = 0, root = new_node();
+        for (int i = 0; i < 26; i++) virt.trans[i] = root;
+        scanf("%d", &n);
+        for (int i = 0; i < n; i++)
+        {
+            scanf("%s", buf);
+            insert(buf);
+        }
+        buildFail();
+        scanf("%s", buf);
+        node *cur = root, *tmp;
+        int ans = 0;
+        for (char *ch = buf; *ch; ch++)
+        {
+            tmp = cur = cur->trans[*ch - 'a'];
+            while (tmp != &virt && vis[tmp - nodes] != T)
+            {
+                vis[tmp - nodes] = T;
+                ans += tmp->cnt;
+                tmp = tmp->fail;
+            }
+        }
+        printf("%d\n", ans);
+    }
+    return 0;
+}

省选准备/assets/day2/poj3974.cpp

@@ -0,0 +1,30 @@
+#include <cstdio>
+#include <cstring>
+inline int min(int a, int b) { return a < b ? a : b; }
+inline int max(int a, int b) { return a > b ? a : b; }
+char buf[1000100], str[2000200];
+int R[2000200], T;
+int main()
+{
+    while (scanf("%s", buf), buf[0] != 'E')
+    {
+        int m = int(strlen(buf)), n = 0;
+        str[n++] = '!';
+        str[n++] = '#';
+        for (int i = 0; i < m; i++)
+            str[n++] = buf[i], str[n++] = '#';
+        str[n++] = '#';
+        str[n++] = '?';
+        int p = 0, mx = 0, ans = 0;
+        for (int i = 1; i < n; i++)
+        {
+            R[i] = mx > i ? min(R[2 * p - i], mx - i) : 1;
+            while (str[i + R[i]] == str[i - R[i]]) R[i]++;
+            if (R[i] + i > mx)
+                mx = i + R[p = i];
+            ans = max(ans, R[i]);
+        }
+        printf("Case %d: %d\n", ++T, ans - 1);
+    }
+    return 0;
+}

省选准备/省选基础算法.tex

@@ -168,5 +168,11 @@ $n$个基因片段，每个长度为$m$，输出$n$行表示重复出现$i$次$(
 \paragraph{练习题}
 \subparagraph{\href{http://acm.hdu.edu.cn/showproblem.php?pid=2222}{HDU2222} - Keywords Search}
 AC 自动机模板题，注意统计答案时，每个节点只能统计一次不要重复统计。
-\codeinput[Keywords Search]{assets/day2/poj3630.cpp}
+\codeinput[Keywords Search]{assets/day2/hdu2222.cpp}
+\subsection{Manacher}
+求出字符串每一位的回文半径，算法流程奥妙重重，不易让常人理解，然后我就抄了份板子改了改，然后就比讲义上的标程快了$20\%$。
+\paragraph{练习题}
+\subparagraph{\href{http://poj.org/problem?id=3974}{POJ3974} - Palindrome}
+Manacher 模板题
+\codeinput[Palindrome]{assets/day2/poj3974.cpp}
 \end{document}