CF 486C Palindrome Transformation

题目给出一个字符串和光标所在位置，每次操作可以把光标向左，向右移动或者把当前字符串ASCII值 +- 1

那么问变成回文的最小代价

首先我们观察到，因为我们可以对字符串+或者-，所以显然清理左边和右边没有任何差别，代价都是字符串的距离（这题可以改成只能+1，这样就是一道稍微难一点的题了）

然后我们假如从k开始清理，最优解显然是先清理一边的，然后拐回头清理另外一边，所以有一段路走了两次，我们维护一个前半部分有字符串不对称的最小区间，然后进行计算就好

#include <bits/stdc++.h>
using namespace std;
constexpr int limit =  (200000 + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;

inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}
int n, m, k;
int a[limit];
int min_step(char c, char ch){//price from c to ch
    return min(abs(c - ch), 26 - abs(c - ch));
}
int min_step2(int idx, int idx2){
    return min(abs(idx - idx2), n - abs(idx - idx2));
}
struct node{
    int idx;
    int val;
    bool operator < (const node &b) const{
        return val > b.val;
    }
};
void solve(){
    cin>>n;
    cin>>k;
    string s;
    cin>>s;
    s = " " + s;
    int l,r;
    ll ans = INF;
    ll res = 0;
    //跟k没关系,我们只需要知道清除左半边和右半边的东西给的代价
    l = -1, r = -1;
    rep(i,1,(n + 1) / 2){
        ll tmp = min_step(s[i], s[n - i + 1]);
        res += tmp; //代价是一样的
        if(tmp){
            if(l == -1)l = i;
            r = i;
        }
    }
    if(r == -1){
        cout<<0<<endl;
        return;
    }
    if(k > (n + 1) / 2){
        k = n - k + 1; //对称
    }
    ll cost = min({min_step2(k, l), min_step2(k, r)});

    cost += abs(r - l);
    cout<<res + cost<<endl;
};
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
//    int kase;
//    cin>>kase;
//    while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
    return 0;
}