UVA - 10716 - Evil Straw Warts Live （简单模拟）

UVA - 10716 Evil Straw Warts Live Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu Submit Status Description Problem D: Evil Straw Warts Live 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" 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. 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 Output for Sample Input 3 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: Intermediate Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 4. Algorithm Design Root :: Prominent Problemsetters :: Gordon V. Cormack Submit Status

#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;
}