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\n",ans);
	return 0;
}