poj 1854 Evil Straw Warts Live 变成回文要几次

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 1799		Accepted: 523

Description

A palindrome is a string of symbols that is equal to itself when reversed. Given an input string, not necessarily a palindrome, compute the number of swaps necessary to transform the string into a palindrome. By swap we mean reversing the order of two adjacent symbols. For example, the string "mamad" may be transformed into the palindrome "madam" with 3 swaps:
swap "ad" to yield "mamda"
swap "md" to yield "madma"
swap "ma" to yield "madam"

Input

The first line of input gives n, the number of test cases. For each test case, one line of input follows, containing a string of up to 8000 lowercase letters.

Output

Output consists of one line per test case. This line will contain the number of swaps, or "Impossible" if it is not possible to transform the input to a palindrome.

Sample Input

3
mamad
asflkj
aabb

Sample Output

3
Impossible
2

题意：一个字符串变成回文最少要几次

题解：
1、用回文的性质判断是否可以变成回文：个数为奇数的字母出现的次数小于等于1
2、从字符串的两端出发（l，r）,在字符串两端同时找能使s[l]==s[r]的位置，
3、取使左右两端相等步数最少的那一端，并从内到外交换位置（若从外到内交换会疯狂WA）

#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<vector>
#include<stack>
#include<math.h>
#define mod 998244353
#define ll long long
#define MAX 0x3f3f3f3f
using namespace std;
int vis[26];
string s;

int check(string ss)
{
    int cnt=0;
    memset(vis,0,sizeof(vis));
    for(int i=0;i<ss.length();i++)
        vis[ss[i]-'a']++;
    for(int i=0;i<26;i++)
    {
        if(vis[i]%2==1)
            cnt++;
    }
    if(cnt>1)
        return 1;
    else
        return 0;
}
int main()
{
    
    int n,ans;
    cin>>n;
    while(n--)
    {
        ans=0;
        cin>>s;
        if(check(s))
            cout<<"Impossible"<<endl;
        else
        {
            int len=s.length(),l,r,j,mid,x=MAX,y=MAX;

            for(int i=0;i<len/2;i++)
            {
                l=i;
                r=len-1-i;
                if(s[l]!=s[r])
                {
                    for(j=l+1;s[j]!=s[r];j++);
                    x=j;
                    for(j=r-1;s[j]!=s[l];j--);
                    y=j;
                    if(x-l>r-y)
                    {
                        ans=ans+r-y;
                        for(int j=y;j<r;j++)
                            s[j]=s[j+1];
                    }
                    else
                    {
                        ans=ans+x-l;
                        for(int j=x;j>l;j--)
                            s[j]=s[j-1];
                    }
                }
            }
            cout<<ans<<endl; 

        }
    }
    return 0;
}