結果
問題 | No.273 回文分解 |
ユーザー |
|
提出日時 | 2023-02-20 17:00:58 |
言語 | Go (1.23.4) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 5,835 bytes |
コンパイル時間 | 14,382 ms |
コンパイル使用メモリ | 229,172 KB |
実行使用メモリ | 5,376 KB |
最終ジャッジ日時 | 2024-07-21 09:01:04 |
合計ジャッジ時間 | 15,714 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 32 |
ソースコード
package mainimport ("bufio""fmt""os")const INF int = 1 << 60func main() {// https://yukicoder.me/problems/no/273in := bufio.NewReader(os.Stdin)out := bufio.NewWriter(os.Stdout)defer out.Flush()var s stringfmt.Fscan(in, &s)n := len(s)dp1 := make([]int, n+3)for i := range dp1 {dp1[i] = -INF}dp2 := make([]int, n+1)for i := range dp2 {dp2[i] = -INF}tree := NewPalindromicTree("")dp2[0] = 1for i := 0; i < n; i++ {tree.AddChar(s[i])res := tree.UpdateDp(func(pos, start int) {if i+1-start == n {dp1[pos] = dp2[start]} else {dp1[pos] = max(dp2[start], i+1-start)}}, func(pos, pre int) {if tree.Nodes[pos].Len == n {dp1[pos] = max(dp1[pos], dp1[pre])} else {dp1[pos] = max(dp1[pos], max(dp1[pre], tree.Nodes[pos].Len))}})for _, p := range res {dp2[i+1] = max(dp2[i+1], dp1[p])}}fmt.Fprintln(out, dp2[n])}func max(a, b int) int {if a > b {return a}return b}type PalindromicTree struct {Chars []byteNodes []*Nodeptr int // 新添加一个字母后所形成的最长回文后缀表示的节点}type Node struct {Len int // 结点代表的回文串的长度Indexes []int // 以哪些索引结尾的最长回文后缀是当前回文串Fail int // 指向当前回文串的最长回文后缀的位置next map[byte]int // 当前回文串前后都加上字符c形成的回文串deltaLink int // 以当前回文串结尾的长度不同的回文串的位置(ertre -> e)?}func NewPalindromicTree(s string) *PalindromicTree {res := &PalindromicTree{}res.Nodes = append(res.Nodes, res.newNode(0, -1)) // 長さ -1 (奇数)res.Nodes = append(res.Nodes, res.newNode(0, 0)) // 長さ 0 (偶数)res.AddString(s)return res}// !添加一个字符,返回以这个字符为后缀的最长回文串的位置pos.func (pt *PalindromicTree) AddChar(x byte) int {pos := len(pt.Chars)pt.Chars = append(pt.Chars, x)// 如果在这个回文串前后都加上字符c形成的回文串原串的后缀,那么就添加儿子,否则就沿着fail指针往上找cur := pt.findPrevPalindrome(pt.ptr)_, hasKey := pt.Nodes[cur].next[x]if !hasKey {pt.Nodes[cur].next[x] = len(pt.Nodes)}pt.ptr = pt.Nodes[cur].next[x]if !hasKey {pt.Nodes = append(pt.Nodes, pt.newNode(-1, pt.Nodes[cur].Len+2))if pt.Nodes[len(pt.Nodes)-1].Len == 1 {pt.Nodes[len(pt.Nodes)-1].Fail = 1} else {pt.Nodes[len(pt.Nodes)-1].Fail = pt.Nodes[pt.findPrevPalindrome(pt.Nodes[cur].Fail)].next[x]}if pt.diff(pt.ptr) == pt.diff(pt.Nodes[len(pt.Nodes)-1].Fail) {pt.Nodes[len(pt.Nodes)-1].deltaLink = pt.Nodes[pt.Nodes[len(pt.Nodes)-1].Fail].deltaLink} else {pt.Nodes[len(pt.Nodes)-1].deltaLink = pt.Nodes[len(pt.Nodes)-1].Fail}}pt.Nodes[pt.ptr].Indexes = append(pt.Nodes[pt.ptr].Indexes, pos)return pt.ptr}func (pt *PalindromicTree) AddString(s string) {if len(s) == 0 {return}for i := 0; i < len(s); i++ {pt.AddChar(s[i])}}// 在每次调用AddChar(x)之后使用,更新dp.// - init(pos, start): 初始化顶点pos的dp值,对应回文串s[start:].// - apply(pos, prePos): 用prePos(fail指针指向的位置)更新pos.// 返回值: 本次更新的回文串的顶点.func (pt *PalindromicTree) UpdateDp(init func(int, int), apply func(int, int)) (update []int) {i := len(pt.Chars) - 1id := pt.ptrfor pt.Nodes[id].Len > 0 {init(id, i+1-pt.Nodes[pt.Nodes[id].deltaLink].Len-pt.diff(id))if pt.Nodes[id].deltaLink != pt.Nodes[id].Fail {apply(id, pt.Nodes[id].Fail)}update = append(update, id)id = pt.Nodes[id].deltaLink}return}// 求出每个顶点代表的回文串出现的次数.func (pt *PalindromicTree) GetFrequency() []int {res := make([]int, pt.Size())for i := pt.Size() - 1; i > 0; i-- {res[i] += len(pt.Nodes[i].Indexes)res[pt.Nodes[i].Fail] += res[i] // 长回文包含短回文}return res}// 以当前字符结尾的回文串个数.func (pt *PalindromicTree) CountPalindromes() int {res := 0for i := 1; i < pt.Size(); i++ {res += len(pt.Nodes[i].Indexes)}return res}// 输出每个顶点代表的回文串.func (pt *PalindromicTree) GetPalindrome(pos int) []string {if pos == 0 {return []string{"-1"}}if pos == 1 {return []string{"0"}}res := []byte{}pt.outputDfs(0, pos, &res)pt.outputDfs(1, pos, &res)start := len(res) - 1if pt.Nodes[pos].Len&1 == 1 {start--}for i := start; i >= 0; i-- {res = append(res, res[i])}return []string{string(res)}}// 回文树中的顶点个数.(包含两个奇偶虚拟顶点)func (pt *PalindromicTree) Size() int {return len(pt.Nodes)}// 返回pos位置的回文串顶点.func (pt *PalindromicTree) GetNode(pos int) *Node {return pt.Nodes[pos]}// 当前位置的回文串长度减去当前回文串的最长后缀回文串的长度.func (pt *PalindromicTree) diff(pos int) int {if pt.Nodes[pos].Fail <= 0 {return -1}return pt.Nodes[pos].Len - pt.Nodes[pt.Nodes[pos].Fail].Len}func (pt *PalindromicTree) newNode(suf, pLen int) *Node {return &Node{next: make(map[byte]int),Fail: suf,Len: pLen,deltaLink: -1,}}// 沿着失配指针找到第一个满足 x+s+x 是原串回文后缀的位置.func (pt *PalindromicTree) findPrevPalindrome(cur int) int {pos := len(pt.Chars) - 1for {rev := pos - 1 - pt.Nodes[cur].Lenif rev >= 0 && pt.Chars[rev] == pt.Chars[len(pt.Chars)-1] {break}cur = pt.Nodes[cur].Fail}return cur}func (pt *PalindromicTree) outputDfs(v, id int, res *[]byte) bool {if v == id {return true}for key, next := range pt.Nodes[v].next {if pt.outputDfs(next, id, res) {*res = append(*res, key)return true}}return false}