HDU 3746 Cyclic Nacklace (KMP:補齊迴圈節)
Cyclic NacklaceTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 16284 Accepted Submission(s): 6747 Problem Description CC always becomes very depressed at the end of this month, he has checked his credit card yesterday, without any surprise, there are only 99.9 yuan left. he is too distressed and thinking about how to tide over the last days. Being inspired by the entrepreneurial spirit of "HDU CakeMan", he wants to sell some little things to make money. Of course, this is not an easy task.As Christmas is around the corner, Boys are busy in choosing christmas presents to send to their girlfriends. It is believed that chain bracelet is a good choice. However, Things are not always so simple, as is known to everyone, girl's fond of the colorful decoration to make bracelet appears vivid and lively, meanwhile they want to display their mature side as college students. after CC understands the girls demands, he intends to sell the chain bracelet called CharmBracelet. The CharmBracelet is made up with colorful pearls to show girls' lively, and the most important thing is that it must be connected by a cyclic chain which means the color of pearls are cyclic connected from the left to right. And the cyclic count must be more than one. If you connect the leftmost pearl and the rightmost pearl of such chain, you can make a CharmBracelet. Just like the pictrue below, this CharmBracelet's cycle is 9 and its cyclic count is 2:
Now CC has brought in some ordinary bracelet chains, he wants to buy minimum number of pearls to make CharmBracelets so that he can save more money. but when remaking the bracelet, he can only add color pearls to the left end and right end of the chain, that is to say, adding to the middle is forbidden. CC is satisfied with his ideas and ask you for help.
Input The first line of the input is a single integer T ( 0 < T <= 100 ) which means the number of test cases.
Output For each case, you are required to output the minimum count of pearls added to make a CharmBracelet.
Sample Input 3 aaa abca abcde
Sample Output 0 2 5
Author possessor WC
Source HDU 3rd “Vegetable-Birds Cup” Programming Open Contest
Recommend lcy
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演算法分析：
題意：
給你一個串,要你在串頭或尾新增最少的字元,使得該串至少有2個迴圈節,問你最少需要加幾個字元.
分析：
我們首先推出一個重要的性質len-next[i]為此字串s[1...i]的最小迴圈節(i為字串的結尾)的長度，另外如果len%(len-next[i])==0,此字串的最小週期就為len/(len-next[i]);
分析：poj 1961 Period (KMP+最小迴圈節)
所以我們可以排除一種情況，如果len/(len-next[i])==0，答案為0；
如果len/(len-next[i])不為0呢，還是那一個重要性質：
重要的性質len-next[i]為此字串s[1...i]的最小迴圈節(i為字串的結尾)的長度
所以直接：len-nxt[len]-len%(len-nxt[len]
還有一種特殊情況需要判斷就是最後next[i]==0時，直接增加全部字元，舉幾個例子就可以推出。
大佬部落格：https://blog.csdn.net/u013480600/article/details/22954037
程式碼實現：
#include<cstdio>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<functional>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<numeric>
#include<cctype>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<map>
using namespace std;
const int N = 1000002;
int nxt[N];
char T[N];
int tlen;
void getNext()
{
int j, k;
j = 0; k = -1;
nxt[0] = -1;
while(j < tlen)
if(k == -1 || T[j] == T[k])
{
nxt[++j] = ++k;
if (T[j] != T[k]) //се╩╞
nxt[j] = k;
}
else
k = nxt[k];
}
int main()
{
int t;
cin>>t;
while(t--)
{
scanf("%s",&T);
tlen= strlen(T);
getNext();
if(nxt[tlen]==0)
{
printf("%d\n",tlen) ;
continue;
}
if(tlen%(tlen-nxt[tlen])==0)
printf("0\n");
else
{
printf("%d\n",tlen-nxt[tlen]-tlen%(tlen-nxt[tlen]));
}
}
return 0;
}