poj 2406 Power Strings(KMPnext性质)

【题目大意】：给出一个字符串，将其分解为若干子串的和，求可分解的最多子串的个数。Sample Input

abcd aaaa ababab .

Sample Output

1 4 3

【解题思路】：

利用next数组的性质：

•如果len%(len-next[len-1])==0，则字符串中必存在最小循环节，且循环次数即为 len/(len-next[len-1]) 其中len为所给字符串的长度！ 证明：•必要性：因为字符串中存在最小循环节(设长度为k)，next[len-1]=len-k,所以len%(len-next[len-1])==0;• 充分性：令k1=len-next[len-1],由于k1整除len,所以可以相应的把len划分为n片区域(n=len/(k1))，从小到大依次表示为•t1,t2...tn;由next数组的定义可知，t1=t2,t2=t3,...t(n-1)=tn,且相应的片区域即为最小，所以循环次数也为len/(len-next[len-1]);[/code]

【代码】：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>#include <cmath>#include <string>#include <cctype>#include <map>#include <iomanip>using namespace std;#define eps 1e-8#define pi acos(-1.0)#define inf 1<<30#define pb push_back#define lc(x) (x << 1)#define rc(x) (x << 1 | 1)#define lowbit(x) (x & (-x))#define ll long longchar s[1000010];int next[1000010],sn,temp;void solve_kmp_next(){next[0]=0;sn=strlen(s);for(int i = 1; i < sn; i++){temp=next[i-1];while(temp && s[temp]!=s[i])temp=next[temp-1];if(s[temp]==s[i])next[i]=temp+1;elsenext[i]=0;}return ;}int main(){while (~scanf("%s",s)){if (s[0]=='.') return 0;solve_kmp_next();if(sn%(sn-next[sn-1])==0)printf("%d\n",sn/(sn-next[sn-1]));elseprintf("1\n");}return 0;}

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