uva 10716——Evil Straw Warts Live

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题意:给定一个字符串,然后判断最小经过若干次交换然后使这个串变成一个回文串(每次可以交换相邻两位)。


思路:贪心。如果一个串的奇数字母的个数为奇数个,那么一定是不可能的。以开头和结尾作为两头不断枚举最小交换次数直到字符串的个数为2.


code:

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <sstream>#include <string>#include <vector>#include <list>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef long double ld;const int INF=0x3fffffff;const int inf=-INF;const int N=1000000;const int M=2005;const int mod=1000000007;const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))#define cpy(x,a) memcpy(x,a,sizeof(a))#define ft(i,s,n) for (int i=s;i<=n;i++)#define frt(i,s,n) for (int i=s;i>=n;i--)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lrt  rt<<1#define rrt  rt<<1|1#define middle int m=(r+l)>>1#define lowbit(x) (x&-x)#define pii pair<int,int>#define mk make_pair#define IN freopen("in.txt","r",stdin);#define OUT freopen("out.txt","w",stdout);char s[150];int cnt[30];int main(){int T;scanf("%d",&T);while (T--){scanf("%s",s);cls(cnt,0);int l=0,r=strlen(s)-1,k=0,ans=0,m=-1;ft(i,l,r){if (cnt[s[i]-'a']) cnt[s[i]-'a']--,k--;else cnt[s[i]-'a']++,k++;}if (k>1) {puts("Impossible");continue;}if (k==1) ft(i,0,26) if(cnt[i]==1) m=i;while(l<r){if (s[l]==s[r]) {l++,r--;continue;}if (s[l]-'a'==m) {ft(i,l+1,r-1){if (s[i]==s[r]){frt(j,i,l+1) swap(s[j],s[j-1]),ans++;r--,l++;break;}}}else {frt(i,r-1,l+1)if (s[i]==s[l]){ft(j,i,r-1)swap(s[j],s[j+1]),ans++;r--,l++;break;}}}printf("%d\n",ans);}}


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