本来想做数论的……但是别的dalao都在做制胡窜
所以……
Chapter I KMP
KMP 最关键的不是这个半暴力的单模匹配
而是这个nxt数组 经常出一些奇怪的题 尤其是循环节可以直接由T-nxt[T]得到……神啊
总之记住nxt就是最长公共前后缀中前缀的尾指针
T1 poj3461 Oulipo
Time cost: 10min
纯纯的板子 真没啥可说的
就是每次清空nxt
Code:
1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 const int N = 1000005; 5 #define rep(i,a,n) for(int i = a;i <= n;i++) 6 #define ms(a,b) memset(a,b,sizeof a) 7 8 char s[N],t[N]; 9 int S,T; 10 int nxt[N]; 11 int ans; 12 void clr(){ms(nxt,0),ans = 0;} 13 void gnxt() { 14 int tmp = 0; 15 rep(i,2,T) { 16 while(tmp && t[tmp+1] != t[i]) tmp = nxt[tmp]; 17 if(t[tmp+1] == t[i]) nxt[i] = ++tmp; 18 } 19 } 20 void KMP() { 21 int tmp = 0; 22 rep(i,1,S) { 23 while(tmp && t[tmp+1] != s[i]) tmp = nxt[tmp]; 24 if(t[tmp+1] == s[i]) { 25 tmp++; 26 if(tmp == T) ans++; 27 } 28 } 29 } 30 31 32 int main() { 33 int q; 34 scanf("%d",&q); 35 while(q--) { 36 clr(); 37 scanf("%s",t+1),scanf("%s",s+1); 38 S = strlen(s+1),T = strlen(t+1); 39 gnxt(); 40 KMP(); 41 printf("%d ",ans); 42 } 43 }
T2 poj 2406 Power strings
Time cost:55min
之前曾经遇到过……最后暴力滚粗
今天算是靠着打表理解了
由于nxt定义的时候刨去了串本身是自己的最长公共前后缀
在最后一次求nxt的时候 我们考虑被nxt[T]抛弃的部分（记为len）
如果这一段能整除T，那么考虑第二段len长的串
它与第一段len长的串是相等的
同理可以推到T/len倍
同时 nxt取最大值 故T-nxt[T]取最小
所以如果T是T-nxt[T]的整数倍 那么T-nxt[T]就是最小循环节
Code:
1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 const int N = 1000005; 5 #define rep(i,a,n) for(int i = a;i <= n;i++) 6 #define ms(a,b) memset(a,b,sizeof a) 7 8 char s[N],t[N]; 9 int S,T; 10 int nxt[N]; 11 int ans; 12 void clr(){ms(nxt,0),ans = 0;} 13 void gnxt() { 14 int tmp = 0; 15 rep(i,2,T) { 16 while(tmp && t[tmp+1] != t[i]) tmp = nxt[tmp]; 17 if(t[tmp+1] == t[i]) nxt[i] = ++tmp; 18 } 19 } 20 void KMP() { 21 int tmp = 0; 22 rep(i,1,S) { 23 while(tmp && t[tmp+1] != s[i]) tmp = nxt[tmp]; 24 if(t[tmp+1] == s[i]) { 25 tmp++; 26 if(tmp == T) ans++; 27 } 28 } 29 } 30 31 32 int main() { 33 while(1) { 34 clr(); 35 scanf("%s",t+1); 36 T = strlen(t+1); 37 if(T == 1 && t[1] == '.') return 0; 38 gnxt(); 39 if(T % (T - nxt[T]) == 0) printf("%d ",T/(T-nxt[T])); 40 else puts("1"); 41 } 42 }
T3 CF526D Om Nom and Necklace
Time cost:45min
有了前一道题做铺垫 还是比较容易想到的
交叉没法做 我们视AB为循环节
由于k给定 所以通过i和k可以求出一个循环节长度范围
如果这个长度是由nxt求出的最短循环节的整数倍就OK
实现的时候是 除len然后判断合法区间是否存在 这样就可以自动减掉后面剩的一段A
但是要注意 A或B为空的情况就是分成k段或者k+1段纯循环节（没有后缀） 可以特判
（实际A为空已经处理过了）
Code:
#include<cstdio> #include<cstring> using namespace std; const int N = 1000005; #define rep(i,a,n) for(int i = a;i <= n;i++) #define ms(a,b) memset(a,b,sizeof a) char s[N],t[N]; int S,T; int nxt[N]; int ans; void clr(){ms(nxt,0),ans = 0;} void gnxt() { int tmp = 0; rep(i,2,T) { while(tmp && t[tmp+1] != t[i]) tmp = nxt[tmp]; if(t[tmp+1] == t[i]) nxt[i] = ++tmp; } } void KMP() { int tmp = 0; rep(i,1,S) { while(tmp && t[tmp+1] != s[i]) tmp = nxt[tmp]; if(t[tmp+1] == s[i]) { tmp++; if(tmp == T) ans++; } } } int n,k; int main() { scanf("%d%d",&n,&k); scanf("%s",t+1); T = n; gnxt(); rep(i,1,T) { int len = i - nxt[i]; if(i % (k+1) == 0 && (i / (k+1)) % len == 0) putchar('1');//lenb==0 else { //circular subsequence range int upp = i / k,down = i / (k + 1) + 1; upp = upp / len; down = (down + len - 1) / len; if(upp >= down) putchar('1'); else putchar('0'); } } puts(""); return 0; }
Chapter II Trie
字典树 比较重要的地方就是空间
一定要考虑好空间问题不要超限
其他的话还是比较好理解的 当做一棵26叉树
T1 poj3630 Phone list
Time cost: 25min
经典的Trie树板子题
先把所有串压进去然后再每一个串查
Code:
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #define rep(i,a,n) for(int i = a;i <= n;i++) 7 #define per(i,n,a) for(int i = n;i >= a;i--) 8 #define inf 2147483647 9 using namespace std; 10 typedef long long ll; 11 inline int read() { 12 int as = 0,fu = 1; 13 char c = getchar(); 14 while(c < '0' || c > '9') { 15 if(c == '-') fu = -1; 16 c = getchar(); 17 } 18 while(c >= '0' && c <= '9') { 19 as = as * 10 + c - '0'; 20 c = getchar(); 21 } 22 return as * fu; 23 } 24 const int N = 10005; 25 const int M = 12; 26 #define ms(a) memset(a,0,sizeof(a)); 27 //head 28 char s[N][M]; 29 struct trie { 30 int nxt[N * M][12],val[N * M]; 31 int cnt; 32 inline void clear() { 33 ms(nxt); 34 ms(val); 35 cnt = 0; 36 } 37 inline void build(char p[]) { 38 int now = 0; 39 rep(i,0,strlen(p) - 1) { 40 if(!nxt[now][p[i] - '0']) 41 nxt[now][p[i] - '0'] = ++cnt; 42 now = nxt[now][p[i] - '0']; 43 } 44 val[now] = 1; 45 } 46 inline bool query(char p[]) { 47 int now = 0; 48 rep(i,0,strlen(p) - 1) { 49 if(val[now]) return 0; 50 now = nxt[now][p[i] - '0']; 51 } 52 return 1; 53 } 54 } T; 55 56 int main() { 57 int q = read(); 58 while(q--) { 59 bool flag = 1; 60 ms(s); 61 T.clear(); 62 int n = read(); 63 rep(i,1,n) { 64 scanf("%s",s[i]); 65 T.build(s[i]); 66 } 67 rep(i,1,n) { 68 flag = T.query(s[i]); 69 if(flag == 0) break; 70 } 71 flag ? printf("YES ") : printf("NO ") ; 72 } 73 return 0; 74 }
T2 poj2945 Find the clones
Time cost:25min
题目要求输出被复制x次的有多少种
依然先把所有串建好树
那么就在query的时候开一个res表示val[now]的有多少个
注意每次找到的时候要把val清空
Code:
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #define rep(i,a,n) for(int i = a;i <= n;i++) 7 #define per(i,n,a) for(int i = n;i >= a;i--) 8 #define inf 2147483647 9 using namespace std; 10 typedef long long ll; 11 inline int read() { 12 int as = 0,fu = 1; 13 char c = getchar(); 14 while(c < '0' || c > '9') { 15 if(c == '-') fu = -1; 16 c = getchar(); 17 } 18 while(c >= '0' && c <= '9') { 19 as = as * 10 + c - '0'; 20 c = getchar(); 21 } 22 return as * fu; 23 } 24 const int N = 20005; 25 const int M = 25; 26 #define ms(a) memset(a,0,sizeof(a)); 27 //head 28 char s[N][M]; 29 struct trie { 30 int nxt[N * M][5],val[N * M]; 31 int res[N]; 32 int cnt; 33 inline void clear() { 34 ms(nxt); 35 ms(val); 36 ms(res); 37 cnt = 0; 38 } 39 inline int cng(char c) { 40 if(c == 'A') return 0; 41 if(c == 'G') return 1; 42 if(c == 'T') return 2; 43 if(c == 'C') return 3; 44 else return 4; 45 } 46 inline void build(char p[]) { 47 int now = 0; 48 rep(i,0,strlen(p) - 1) { 49 int c = cng(p[i]); 50 if(!nxt[now][c]) 51 nxt[now][c] = ++cnt; 52 now = nxt[now][c]; 53 } 54 val[now]++; 55 } 56 inline void query(char p[]) { 57 int now = 0; 58 rep(i,0,strlen(p) - 1) { 59 int c = cng(p[i]); 60 now = nxt[now][c]; 61 } 62 res[val[now]]++; 63 val[now] = 0; 64 } 65 } T; 66 67 int main() { 68 while(1) { 69 int n = read(); 70 int m = read(); 71 if(!n && !m) break; 72 T.clear(); 73 ms(s); 74 rep(i,1,n) { 75 scanf("%s",s[i]); 76 T.build(s[i]); 77 } 78 rep(i,1,n) T.query(s[i]); 79 rep(i,1,n) { 80 printf("%d ",T.res[i]); 81 } 82 } 83 return 0; 84 }
T3 poj1816 Wild words
Time cost:65min
好像这题有个名字叫模糊查询……
?的话就额外开一个节点 查的时候单独查
*的话就暴搜除掉的位数……
注意S* 也能匹配S
所以搜*的时候要在搜结束节点之前
然后就是空间……这道题nxt数组开1e5*30*30都不行 每个节点存答案的时候必须用vector
Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 using namespace std; 6 typedef long long ll; 7 #define ms(a,b) memset(a,b,sizeof a) 8 #define rep(i,a,n) for(int i = a;i <= n;i++) 9 #define per(i,n,a) for(int i = n;i >= a;i--) 10 const int N = 100001; 11 const int M = 28; 12 //HEAD 13 int m,n; 14 vector<int> mem[N*M]; 15 int ans[N],top; 16 int nxt[N][M],cnt; 17 void clr() { 18 cnt = 0; 19 ms(nxt,0); 20 } 21 int cng(char op) { 22 if(op == '*') return 26; 23 if(op == '?') return 27; 24 return op - 'a'; 25 } 26 27 char p[35]; 28 int len; 29 void ins(int idx) { 30 int cur = 0; 31 rep(i,0,len - 1) { 32 if(!nxt[cur][cng(p[i])]) 33 nxt[cur][cng(p[i])] = ++cnt; 34 cur = nxt[cur][cng(p[i])]; 35 } 36 mem[cur].push_back(idx); 37 } 38 39 void query(int cur,int pos) { 40 if(nxt[cur][26]) 41 rep(i,pos,len) query(nxt[cur][26],i); 42 43 if(pos == len) { 44 int sze = mem[cur].size(); 45 rep(i,0,sze - 1) 46 ans[++top] = mem[cur][i]; 47 return; 48 } 49 if(nxt[cur][cng(p[pos])]) 50 query(nxt[cur][cng(p[pos])],pos+1); 51 if(nxt[cur][27]) 52 query(nxt[cur][27],pos+1); 53 } 54 55 int main() { 56 scanf("%d%d",&n,&m); 57 clr(); 58 rep(i,1,n) { 59 scanf("%s",p); 60 len = strlen(p); 61 ins(i-1); 62 } 63 while(m--) { 64 scanf("%s",p); 65 top = 0; 66 len = strlen(p); 67 query(0,0); 68 if(!top) { 69 puts("Not match"); 70 continue; 71 } 72 sort(ans+1,ans+1+top); 73 top = unique(ans+1,ans+1+top) - ans - 1; 74 rep(i,1,top) printf("%d ",ans[i]); 75 puts(""); 76 } 77 return 0; 78 }
T4 CF633C Spy Syndrome 2
Time cost:105min
发现做字符串时间格外长……还是不熟练
一开始想着先反过来再匹配……
事实证明考虑前一位是否匹配的时候如果是正的还是要遍历一遍
所以直接就倒着做就行
O(n^2)可以做出类似链式前向星的东西 然后从最后一位输出就可以了
注意标记初值要设成-1
然后递归输出的时候注意顺序不要反了就行
Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 #include<queue> 6 #include<vector> 7 #define ms(a,b) memset(a,b,sizeof a) 8 #define rep(i,a,n) for(int i = a;i <= n;i++) 9 #define per(i,n,a) for(int i = n;i >= a;i--) 10 #define inf 2147483647 11 using namespace std; 12 typedef long long ll; 13 ll read() { 14 ll as = 0,fu = 1; 15 char c = getchar(); 16 while(c < '0' || c > '9') { 17 if(c == '-') fu = -1; 18 c = getchar(); 19 } 20 while(c >= '0' && c <= '9') { 21 as = as * 10 + c - '0'; 22 c = getchar(); 23 } 24 return as * fu; 25 } 26 const int N = 100005; 27 //head 28 int n,m; 29 char S[10005]; 30 char p[N][1005]; 31 int cng(char op){return op >= 'a' ? op - 'a' : op - 'A';} 32 33 int nxt[1000005][26],val[1000005],tot; 34 void ins(char *s,int id) { 35 int cur = 0; 36 int len = strlen(s); 37 rep(i,0,len-1) { 38 int c = cng(s[i]); 39 if(!nxt[cur][c]) nxt[cur][c] = ++tot; 40 cur = nxt[cur][c]; 41 } 42 val[cur] = id; 43 } 44 45 int pre[N],ans[N]; 46 void query(int x) { 47 int cur = 0; 48 per(i,x,0) { 49 int c = cng(S[i]); 50 cur = nxt[cur][c]; 51 if(!cur) return; 52 if(val[cur]) { 53 if(ans[i-1] || !i) { 54 pre[x] = i-1; 55 ans[x] = val[cur]; 56 return; 57 } 58 } 59 } 60 } 61 62 void output(int x) { 63 if(~pre[x]) output(pre[x]); 64 printf("%s ",p[ans[x]]); 65 } 66 67 int main() { 68 n = read(); 69 scanf("%s",S); 70 m = read(); 71 rep(i,1,m) { 72 scanf("%s",p[i]); 73 ins(p[i],i); 74 } 75 ms(pre,-1); 76 rep(i,0,n-1) query(i); 77 output(n-1),puts(""); 78 return 0; 79 }
Chapter III AC自动机
我太菜了……板子调了好久……
总体来讲 AC自动机与KMP相同的是 fail很重要
不同的是 这个匹配也很重要
都可以拿来出题
总体来讲 沿着fail的路线也可以搞事情（poj4052），沿着trie树向上走也可以做标记
总之 只有节点维护标记 节点之间的联系只有trie边和fail 考虑清楚就行了
T1 poj 4052 Hrinity
Time cost:95min
这么水一个题我做这么长时间……太菜了
题意大概是给出一堆模式串（还有缩写）在文本串里找
其中被包含的子串不计数
不能在文本串里去掉因为可能剩下的还能匹配小串
所以就第一次先把所有匹配了的串都打上标记 然后再扫一遍把被普通节点覆盖的终止节点去掉
注意：
1.不要把自己去了
2.转换的时候用strcpy比较方便 但别用反了……
3.去掉小串之前要先把访问数组清空
Code:
1 Source Code 2 3 Problem: 4052 User: dvb 4 Memory: 36408K Time: 594MS 5 Language: G++ Result: Accepted 6 Source Code 7 #include<cstdio> 8 #include<cstring> 9 #include<algorithm> 10 #include<cmath> 11 #include<queue> 12 #include<vector> 13 #define itn int 14 #define inf 2147483647 15 #define eps 1e-9 16 #define ms(a,b) memset(a,b,sizeof a) 17 #define rep(i,a,n) for(int i = a;i <= n;i++) 18 #define per(i,n,a) for(int i = n;i >= a;i--) 19 using namespace std; 20 typedef long long ll; 21 ll read() { 22 ll as = 0,fu = 1; 23 char c = getchar(); 24 while(c < '0' || c > '9') { 25 if(c == '-') fu = -1; 26 c = getchar(); 27 } 28 while(c >= '0' && c <= '9') { 29 as = as * 10 + c - '0'; 30 c = getchar(); 31 } 32 return as * fu; 33 } 34 //head 35 int n; 36 char p[2505][1105],S[5100005]; 37 char P[5100005]; 38 39 void cng(char s[]) { 40 int len = strlen(s); 41 int cur = 0; 42 rep(i,0,len-1) { 43 if(s[i] == '[') {i++; 44 int num = 0; 45 char op; 46 while(s[i] != ']') { 47 if(s[i] >= '0' && s[i] <= '9') num = num * 10 + s[i] - '0'; 48 else op = s[i]; 49 i++; 50 } 51 rep(i,1,num) P[cur++] = op; 52 } 53 else P[cur++] = s[i]; 54 } 55 P[cur] = ' '; 56 strcpy(s,P); 57 } 58 59 60 struct node { 61 int val,fail; 62 int nxt[26]; 63 void init() { 64 ms(nxt,0),val = fail = 0; 65 } 66 }t[2750005]; 67 int tot = 1; 68 69 void ins(char *s,int id) { 70 int len = strlen(s); 71 int cur = 1; 72 rep(i,0,len-1) { 73 int c = s[i] - 'A'; 74 if(!t[cur].nxt[c]) 75 t[cur].nxt[c] = ++tot,t[tot].init(); 76 cur = t[cur].nxt[c]; 77 } 78 t[cur].val = id; 79 } 80 81 82 void getfail() { 83 rep(i,0,25) t[0].nxt[i] = 1; 84 queue<int> q; 85 q.push(1); 86 while(!q.empty()) { 87 int u = q.front(),v,w; 88 q.pop(); 89 rep(i,0,25) { 90 v = t[u].fail; 91 while(!t[v].nxt[i]) v = t[v].fail; 92 v = t[v].nxt[i]; 93 w = t[u].nxt[i]; 94 if(w) t[w].fail = v,q.push(w); 95 else t[u].nxt[i] = v; 96 } 97 } 98 } 99 100 101 bool vis[2750005]; 102 bool flag[2505]; 103 void query() { 104 int len = strlen(S); 105 int cur = 1; 106 rep(i,0,len-1) { 107 cur = t[cur].nxt[S[i]-'A']; 108 if(t[cur].val) { 109 flag[t[cur].val] = 1; 110 continue; 111 } 112 int tmp = cur; 113 while(tmp && !vis[tmp]) { 114 vis[tmp] = 1; 115 flag[t[tmp].val] = 1; 116 tmp = t[tmp].fail; 117 } 118 } 119 } 120 121 void del(char *s,int id) { 122 int len = strlen(s); 123 int cur = 1; 124 rep(i,0,len-1) { 125 cur = t[cur].nxt[s[i]-'A']; 126 int tmp = cur; 127 while(tmp && !vis[tmp]) { 128 if(t[tmp].val && t[tmp].val ^ id) { 129 vis[tmp] = 1; 130 flag[t[tmp].val] = 0; 131 } 132 tmp = t[tmp].fail; 133 } 134 } 135 } 136 137 void clr() { 138 rep(i,1,tot) t[i].init(); 139 tot = 1; 140 ms(flag,0); 141 } 142 143 int main() { 144 int T = read(); 145 while(T--) { 146 clr(); 147 148 n = read(); 149 rep(i,1,n) { 150 scanf("%s",p[i]); 151 cng(p[i]); 152 ins(p[i],i); 153 } 154 155 getfail(); 156 157 scanf("%s",S); 158 cng(S); 159 ms(vis,0),query(); 160 161 ms(vis,0); 162 rep(i,1,n) 163 if(flag[i]) del(p[i],i); 164 165 int ans = 0; 166 rep(i,1,n) ans += flag[i]; 167 printf("%d ",ans); 168 169 } 170 return 0; 171 }
Chapter IV manacher
回文串...看起来比较不常见 但是出了就得会
听说后面有个回文自动机? 看到了再说吧...
T1 poj3974 Palindrome
Time cost: 35min
基础的回文串题 调了一下之前的板子 换了种写法
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 #include<queue> 6 #define ms(a,b) memset(a,b,sizeof a) 7 #define rep(i,a,n) for(int i = a;i <= n;i++) 8 #define per(i,n,a) for(int i = n;i >= a;i--) 9 #define inf 2147483647 10 using namespace std; 11 typedef long long ll; 12 ll read() { 13 ll as = 0,fu = 1; 14 char c = getchar(); 15 while(c < '0' || c > '9') { 16 if(c == '-') fu = -1; 17 c = getchar(); 18 } 19 while(c >= '0' && c <= '9') { 20 as = as * 10 + c - '0'; 21 c = getchar(); 22 } 23 return as * fu; 24 } 25 const int N = 1000005; 26 //head 27 char s[N<<1],t[N]; 28 int R[N<<1]; 29 int n,m,T,maxx; 30 void manacher() { 31 int id = 0,pos = 0,r = 0; 32 rep(i,1,n) { 33 if(pos > i) r = min(R[id * 2 - i],pos - i); 34 else r = 1; 35 while(s[i - r] == s[i + r]) r++; 36 if(r + i > pos) { 37 pos = r + i; 38 id = i; 39 } 40 R[i] = r; 41 } 42 } 43 44 void clr() { 45 ms(R,0); 46 maxx = 0; 47 } 48 int main() { 49 while(scanf("%s",t+1),t[1] != 'E') { 50 clr(); 51 m = strlen(t+1); 52 s[n = 0] = 0; 53 rep(i,1,m) { 54 s[++n] = '#'; 55 s[++n] = t[i]; 56 } 57 s[++n] = '#'; 58 s[n+1] = 1; 59 manacher(); 60 rep(i,1,n) maxx = max(maxx,R[i]); 61 printf("Case %d: %d ",++T,maxx - 1); 62 } 63 return 0; 64 }
T2 luogu P1659 [国家集训队]拉拉队排练
Time cost:45min
发现自己还是不太明白原理....
这道题做法就是找出所有长为奇数的回文串
发现长的回文串一定包含较短的
所以求一遍最长长度的后缀和就是真正的个数
数据范围较大 使用快速幂
Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 #include<queue> 6 #define ms(a,b) memset(a,b,sizeof a) 7 #define rep(i,a,n) for(int i = a;i <= n;i++) 8 #define per(i,n,a) for(int i = n;i >= a;i--) 9 #define inf 2147483647 10 using namespace std; 11 typedef long long ll; 12 ll read() { 13 ll as = 0,fu = 1; 14 char c = getchar(); 15 while(c < '0' || c > '9') { 16 if(c == '-') fu = -1; 17 c = getchar(); 18 } 19 while(c >= '0' && c <= '9') { 20 as = as * 10 + c - '0'; 21 c = getchar(); 22 } 23 return as * fu; 24 } 25 const int N = 1000005; 26 const ll mod = 19930726ll; 27 //head 28 char s[N]; 29 int r[N]; 30 ll n,k,ans = 1; 31 ll num[N]; 32 void manacher() { 33 int id = 0,pos = 0,x = 0; 34 rep(i,1,n) { 35 if(pos > i) x = min(r[(id<<1)-i] , pos - i); 36 else x = 1; 37 while(s[i+x] == s[i-x]) x++; 38 if(x + i > pos) pos = x + i,id = i; 39 r[i] = x; 40 num[r[i]-1<<1|1]++; 41 } 42 } 43 ll ksm(ll a,ll b) { 44 ll tmp = 1; 45 while(b) { 46 if(b & 1) tmp = tmp * a % mod; 47 a = a * a % mod; 48 b >>= 1; 49 } 50 return tmp; 51 } 52 53 54 int main() { 55 n = read(),k = read(); 56 scanf("%s",s+1); 57 s[0] = '&'; 58 manacher(); 59 per(i,n+1>>1,1) { 60 ll l = i-1<<1|1; 61 num[l] += num[l+2]; 62 if(k >= num[l]) { 63 ans = (ans * ksm(l,(ll)num[l])) % mod; 64 k -= num[l]; 65 } else { 66 ans = (ans * ksm(l,k)) % mod; 67 printf("%lld ",ans); 68 k = 0; 69 break; 70 } 71 } 72 if(k) puts("-1"); 73 return 0; 74 }
T3 luogu P4555 [国家集训队]最长双回文串
Time cost:85min
难度明显上来了
不能直接暴力 考虑枚举中间的连接点
然后思路就是找以当前点为左右端点的最远另一端点（记为L，R）
注意真正的回文串长度是所求回文半径长度-1
所以就在manacher里面找到最长回文串的时候求出其左右端点的答案
注意这个答案还可以被L（R）更新
所以我们找到所有的下标为‘#’（插入的特殊字符）的位置互相更新（具体见代码main）
复杂度O(n)
Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 #include<queue> 6 #define ms(a,b) memset(a,b,sizeof a) 7 #define rep(i,a,n) for(int i = a;i <= n;i++) 8 #define per(i,n,a) for(int i = n;i >= a;i--) 9 #define inf 2147483647 10 using namespace std; 11 typedef long long ll; 12 ll read() { 13 ll as = 0,fu = 1; 14 char c = getchar(); 15 while(c < '0' || c > '9') { 16 if(c == '-') fu = -1; 17 c = getchar(); 18 } 19 while(c >= '0' && c <= '9') { 20 as = as * 10 + c - '0'; 21 c = getchar(); 22 } 23 return as * fu; 24 } 25 const int N = 200005; 26 //head 27 int n,maxx; 28 char s[N],t[N]; 29 int r[N],L[N],R[N]; 30 void manacher() { 31 int pos = 0,maxr = 0,x = 0; 32 rep(i,1,n) { 33 if(maxr > i) x = min(r[(pos<<1)-i],maxr-i); 34 else x = 1; 35 while(s[i + x] == s[i - x]) x++; 36 if(i + x > maxr) maxr = i+x,pos = i; 37 r[i] = x; 38 L[i+r[i]-1] = max(L[i+r[i]-1],r[i]-1); 39 R[i-r[i]+1] = max(R[i-r[i]+1],r[i]-1); 40 } 41 } 42 43 int main() { 44 scanf("%s",t+1); 45 s[++n] = 0; 46 int l = strlen(t+1); 47 rep(i,1,l) { 48 s[++n] = '#'; 49 s[++n] = t[i]; 50 } 51 s[++n] = '#'; 52 s[++n] = '%'; 53 manacher(); 54 rep(i,1,n>>1) R[i<<1] = max(R[i<<1],R[i-1<<1]-2); 55 per(i,n>>1,1) L[i<<1] = max(L[i<<1],L[i+1<<1]-2); 56 rep(i,1,n>>1) if(R[i<<1] && L[i<<1]) maxx = max(maxx,L[i<<1] + R[i<<1]); 57 printf("%d ",maxx); 58 return 0; 59 }
To be continued...
https://www.luogu.org/problemnew/show/P1659