POJ Evil Straw Warts Live (贪心）

Evil Straw Warts Live

Time Limit: 2000MS
Memory Limit: 65536KTotal Submissions: 850
Accepted: 239

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

3mamadasflkjaabb

Sample Output

3Impossible2

题目大意：给一个单词，为了让它变成一个回文串，允许相邻的两个数交换。求最少的交换次数，使得单词变成一个回文串。

解题思路：首先判断一下字母出现的奇数次的个数是否大于1， 大于1不满足。然后从左右开始遍历判断tmp[left] 和 tmp[right]是否相同，相同跳过， 不相同的话找到从左边开始第一个使得tmp[left + l] == tmp[right], 和从右边开始第一个使得tmp[left] == tmp[right + r]。比较l 和 r 的大小然后考虑该移动哪一边。

但是在这里我只用了temp[left]==temp[right-r]，将所有的r求和即可。

 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; char arr[10000]; int b[28]; int solve() {     int ans=0,len=strlen(arr),high=len-1;     for(int i=0;i<(len+1)/2-1;i++){        if(b[arr[i]-'a']>1){            for(int j=high;j>i;j--){                if(arr[i]==arr[j]){                    for(int k=j+1;k<=high;k++){                        arr[k-1]=arr[k];                        ans++;                    }                    arr[high]=arr[i];                    high--;                    break;                }            }            b[arr[i]-'a']-=2;        }        else{            swap(arr[i],arr[i+1]);            ans++;            --i;        }     }     return ans; } int main() {     int n;     scanf("%d",&n);     while(n--){        memset(b,0,sizeof(b));        scanf("%s",arr);        for(int i=0;arr[i];++i){            ++b[arr[i]-'a'];        }        int cnt=0;        for(int i=0;i<27;i++){            if(b[i]%2) cnt++;        }        if(cnt>=2){            printf("Impossible\n");            continue;        }        else{            printf("%d\n",solve());        }     }     return 0; }