[ACM] POJ 3096 Surprising Strings （map使用）

標籤：

Surprising Strings

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5783		Accepted: 3792

Description

The D-pairs of a string of letters are the ordered pairs of letters that are distance D from each other. A string is D-unique if all of its D-pairs are different. A string is surprising if it is D-unique for every possible distance D.

Consider the string ZGBG. Its 0-pairs are ZG, GB, and BG. Since these three pairs are all different, ZGBG is 0-unique. Similarly, the 1-pairs of ZGBG are ZB and GG, and since these two pairs are different, ZGBG is 1-unique. Finally, the only 2-pair of ZGBG is ZG, so ZGBG is 2-unique. Thus ZGBG is surprising. (Note that the fact that ZG is both a 0-pair and a 2-pair of ZGBG is irrelevant, because 0 and 2 are different distances.)

Acknowledgement: This problem is inspired by the "Puzzling Adventures" column in the December 2003 issue of Scientific American.

Input

The input consists of one or more nonempty strings of at most 79 uppercase letters, each string on a line by itself, followed by a line containing only an asterisk that signals the end of the input.

Output

For each string of letters, output whether or not it is surprising using the exact output format shown below.

Sample Input

ZGBGXEEAABAABAAABBBCBABCC*

Sample Output

ZGBG is surprising.X is surprising.EE is surprising.AAB is surprising.AABA is surprising.AABB is NOT surprising.BCBABCC is NOT surprising.

Source

Mid-Central USA 2006

解題思路：

推斷一個字串是不是surpring。

條件：該字串的全部D-pairs,D是字串中兩個字母的距離, 假設全部的D-pairs都不同，那麼該字串是D-unique。

假設對全部的距離D，都滿足D-unique，那麼字串就是surpring.

比方ZGBG

0-pairs :  ZG  GB BG  不同樣，是0-unique

1-pairs :  ZB  GG        不同樣，是1-unique

2-pairs:   ZG               不同樣。是2-unique

綜上。ZGBG是surpring

每個D-pairs進行推斷。假設在推斷一個D-pairs過程中有同樣的，那麼該字串肯定不是surpring。

用map<string,bool> 來推斷是否字串已經出現過。要注意其聲明的位置，要在每一層D迴圈裡面聲明。

代碼：

#include <iostream>#include <stdio.h>#include <map>#include <string.h>using namespace std;char str[80];int main(){    while(scanf("%s",str)!=EOF&&str[0]!='*')    {        int len=strlen(str);        if(len<=2)        {            cout<<str<<" is surprising."<<endl;            continue;        }        bool ok=1;//是surpring        for(int d=0;d<=len-2;d++)//距離d        {            bool dtap=1;//是D-unique            map<string,bool>mp;//注意其聲明的位置            for(int s=0;s<=len-2&&s+d+1<len;s++)            {                string temp="";                temp+=str[s];                temp+=str[s+d+1];                if(mp[temp])//該字串出現過                {                    dtap=0;                    break;                }                else                    mp[temp]=1;            }            if(!dtap)            {                ok=0;                break;            }        }        if(ok)            cout<<str<<" is surprising."<<endl;        else            cout<<str<<" is NOT surprising."<<endl;    }    return 0;}

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[ACM] POJ 3096 Surprising Strings （map使用）