poj 1854 Evil Straw Warts Live

Evil Straw Warts Live

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 728 Accepted: 198

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

题目的大概意思是让你判断输入的字符串是否能变为回文数，若不能，则输出Impossible，若能，则输出从该字符串变为回文串所需要的最少的步数，（一次只能交换相邻的两个数）。

如何判断该串能否变为回文串：题中已经说明，输入的字符串中只有小写的字母，所以统计各个字母出现的频率，若出现两个或以上的字母频率为奇数，则该字符串，不可能变为回文串。

怎么找出最少的步数：这里用到了贪心的思想。具体看代码。。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define maxn 8010int n,ans;char arr[maxn];int b[27];void solve();int main(){    scanf("%d",&n);    while(n--){        memset(b,0,sizeof(b));        scanf("%s",arr);        int cnt = 0;        ans = 0;        for(int i = 0 ; arr[i];i++){            b[arr[i] - 'a'] ++;        }        for(int i = 0; i < 27;i++){            if(b[i] % 2 == 1){                cnt ++;            }        }         if(cnt >= 2){            printf("Impossible\n");            continue;        }        solve();        printf("%d\n",ans);    }    return 0;}void solve(){    int len = strlen(arr);    int high = len - 1;    for(int i = 0; i < (len+1)/ 2 - 1;i++){        if(b[arr[i] - 'a'] % 2 == 0 || b[arr[i] - 'a'] > 1){            for(int j = high; j > i; j--){                if(arr[j] == arr[i]){                    char tmp = arr[j];                    for(int k = j + 1; k <= high; k++){                        arr[k-1] = arr[k];                        ans++;                    }                    arr[high] = tmp;                    high --;                    break;                }            }            b[arr[i] - 'a'] -= 2;        }        else{            swap(arr[i],arr[i+1]);            i--;            ans ++;        }    }}