UVA - 10716
Description
Problem D: Evil Straw Warts LiveA 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
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 100 lowercase letters. Output consists of one line per test case. Sample Input3 mamad asflkj aabb Output for Sample Input3 Impossible 2 Gordon V. Cormack Source Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 1. Algorithm Design :: General Problem Solving Techniques :: Exercises: Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 4. Algorithm Design Root :: Prominent Problemsetters :: Gordon V. Cormack |
给出一个字符串，看这个字符串能否组成回文串，如果能则输出需要交换字母的次数，否则输出Impossible
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
int T;
int num[30];
char str[105];
int main() {
scanf("%d", &T);
while(T--) {
memset(num, 0, sizeof(num));
scanf("%s", str);
int len = strlen(str), cnt = 0;
for(int i = 0; i < len; i++) num[str[i] - 'a'] ++;
for(int i = 0; i < 26; i++)
if(num[i] & 1) cnt++;
if(cnt >= 2) {
printf("Impossible\n");
continue;
}
int ans = 0, max = len - 1;
for(int i = 0; i < len / 2; i++) {
int j;
for(j = max; j > i; j--) {
if(str[j] == str[i]) break;
}
if(i == j) {
swap(str[i + 1], str[i]);
i--;
ans++;
continue;
}
ans += max - j;
for(int k = j + 1; k <= max; k++) str[k - 1] = str[k];
max--;
}
printf("%d\n", ans);
}
return 0;
}