题目链接：
http://poj.org/problem?id=1509
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Glass Beads
Description Once upon a time there was a famous actress. As you may expect, she played mostly Antique Comedies most of all. All the people loved her. But she was not interested in the crowds. Her big hobby were beads of any kind. Many bead makers were working for her and The necklace should be made of glass beads of different sizes connected to each other but without any thread running through the beads, so that means the beads can be disconnected at any point. The actress chose the succession of beads she wants to have and The description of the necklace is a string A = a1a2 ... am specifying sizes of the particular beads, where the last character am is considered to precede character a1 in circular fashion. The disjoint point i is said to be worse than the disjoint point j if and only if the string aiai+1 ... ana1 ... ai-1 is lexicografically smaller than the string ajaj+1 ... ana1 ... aj-1. String a1a2 ... an is lexicografically smaller than the string b1b2 ... Input The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly one line containing necklace description. Maximal length of each description is 10000 characters. Each bead is Output For each case, print exactly one line containing only one integer -- number of the bead which is the first at the worst possible disjoining, i.e.\ such i, that the string A[i] is lexicographically smallest among all the n possible disjoinings of a necklace. Sample Input 4 helloworld amandamanda dontcallmebfu aaabaaa Sample Output 10 11 6 5 Source |
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题目意思：
给一个字符串，求循环字符串中字典序最小的那个起始位置。
解题思路：
裸的最小表示。
注意当i,j改变后且i==j时，要i++或j++不然就是一样的了，因为是一个串。
代码：
//#include<CSpreadSheet.h>
#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define Maxn 210000
char save[Maxn];
int cal()
{
int n=strlen(save+1);
int i=1,j=2;
for(int i=n+1;i<=2*n;i++)
save[i]=save[i-n];
//save[2*n+1]=save[1];
while(i<=n&&j<=n)
{
int k=0;
while(save[i+k]==save[j+k]&&k<n)
k++;
if(k>=n)
break;
if(save[i+k]>save[j+k])
{
i=i+k+1;
if(i==j) //要手动改下
i++;
}
else
{
j=j+k+1;
if(i==j)
i++;
}
}
return min(i,j);
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int t;
scanf("%d",&t);
while(t--)
{
scanf("%s",save+1);
printf("%d\n",cal());
}
return 0;
}
[最小表示] poj 1509 Glass Beads,布布扣,bubuko.com