[ACM] POJ 3096 Surprising Strings （map使用）

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

ZGBG
X
EE
AAB
AABA
AABB
BCBABCC
*

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

解题思路：

推断一个字符串是不是surpring。

条件：该字符串的全部D-pairs,D是字符串中两个字母的距离, 假设全部的D-pairs都不同，那么该字符串是D-unique。

假设对全部的距离D，都满足D-unique，那么字符串就是surpring.

比方ZGBG

0-pairs : ZG GB BG 不同样，是0-unique

1-pairs : ZB GG 不同样，是1-unique

2-pairs: ZG 不同样。是2-unique

综上。ZGBG是surpring

每个D-pairs进行推断。假设在推断一个D-pairs过程中有同样的，那么该字符串肯定不是surpring。

用map<string,bool> 来推断是否字符串已经出现过。要注意其声明的位置，要在每一层D循环里面声明。

代码：

#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++)//距离d
        {
            bool dtap=1;//是D-unique
            map<string,bool>mp;//注意其声明的位置
            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])//该字符串出现过
                {
                    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;
}