51nod 1277 字符串中的最大值

题目链接

51nod 1277 字符串中的最大值

题解

对于单串，考虑多串的fail树，发现next数组的关系形成树形结构
建出next树，对于每一个前缀，他出现的次数就是他子树的大小

代码

#include<cstdio> 
#include<cstring> 
#include<algorithm>  


inline int read() { 
	int x = 0,f = 1; 
	char c = getchar(); 
	while(c < '0' || c > '9')c = getchar(); 
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar(); 
	return x * f; 
} 

const int maxn = 100007;  
struct node  {
	int v,next; 
} edge[maxn << 1]; 
int num = 0,head[maxn]; 
inline void add_edge(int u,int v) { 
	edge[++ num].v = v; edge[num].next = head[u];head[u] = num; 
} 
long long siz[maxn]; 
char s[maxn]; 
int next[maxn]; 
int len = 0; 
void getnext() { 
	next[0] = -1; 
	for(int j = 0,i = 2;i <= len;++ i) {
		for(;s[i] != s[j + 1] && j > 0;j = next[j]);   
		if(s[i] == s[j + 1]) { next[i] = j + 1,++ j; add_edge(j,i); }  
	}  
}  
long long ans = 0; 
void dfs(int x) { 
	siz[x] = 1; 
	for(int i = head[x];i;i = edge[i].next) { 
		dfs(edge[i].v); 
		siz[x] += siz[edge[i].v]; 
	} 
	ans = std::max(ans, (long long) x * siz[x]); 
} 
int main() { 
	scanf("%s",s + 1); 
	len = strlen(s + 1); 
	getnext();   
	for(int i = 1;i <= len;++ i) if(!next[i]) dfs(i); 
	printf("%lld
",ans); 
	return 0; 
}