Opencup XII, Grand Prix of Tokyo F.Robots

Posted 2023-04-23 tiany7

F.Robots

来复健，但是没料到是GP场，就不该来，应该去vp

这道题的题意是给两个数组a和b，对于每个b，会有最近的依然active的a来配对，问最小的距离是多少，并且给出一种删b的方案

样例给了：

3
1 2 3
11 12 13

答案是30，我们先删掉13，然后删掉12，最后删掉11， 为什么不从前往后呢？

因为我们没法保证所有的a[i] <> b[i]，会出现一种情况，当前的b抢走了前面的a，或者前面的b抢走了当前的a，为了保证一致性，我们可以分类讨论

我们固定a[i] < b[i] 为唯一的ordering，那么在我们新iterate到的节点，我们看看这个节点的a[i]会不会被前面抢走，如果会的话，先把前面爆了，然后再塞入当前的b，这样就保证了一致性，如果不会被爆，那么我们就把第i个节点爆了，从而保证无后效性。

那么怎么维护呢？我们可以用一个栈来维护。

#include <bits/stdc++.h>
#include <ext/rope>
using namespace std;
constexpr int limit =  (4e5  + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f
#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 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;
int a[limit];
int b[limit];
bitset<limit>vis;
struct node
    int i, aa, bb;
;
list<node>rp;
void solve()
    cin>>n;
    rep(i,1,n)cin>>a[i];
    rep(i,1,n)cin>>b[i];
    ll ans = 0;
    vector<int>res;
    rep(i,1, n)
        if(a[i] < b[i])
            rp.push_back(int(i), a[i], b[i]);
        else if(a[i] == b[i])
            res.push_back(i);
        else
            if(rp.size() and llabs(rp.back().bb - rp.back().aa) < llabs(rp.back().bb - a[i]))
                while(rp.size() and llabs(rp.back().bb - rp.back().aa) < llabs(rp.back().bb - a[i]))
                    res.push_back(rp.back().i);
                    rp.pop_back();
                
                rp.push_back(int(i), a[i], b[i]);
            else
                res.push_back(i);
            
        
    
    for(auto &i : rp| views::reverse)
        res.push_back(i.i);
    
    for(auto &i : res)
        ans += llabs(a[i] - b[i]);
    
    cout<<ans<<endl;
    for(auto &i : res )cout<<i<<" ";
    cout<<endl;
;
int32_t main() 
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
//    int kase;
//    cin>>kase;
//    while (kase--)
        invoke(solve);
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s";
    return 0;

天才选手zerol的主页：https://zerol.me/ | WeepingDemon的个人主页：https://weepingdemon.gitee.io/blog/

2015-2016 XVI Open Cup, Grand Prix of Bashkortostan, SKB Kontur Cup Stage 2

地址

Rank	Solved	A	B	C	D	E	F	G	H	I	J	K	L	M
72/213	8/13	O	.	O	O	.	O	.	O	O	O	.	O	.

O: 当场通过

?: 赛后通过

.: 尚未通过

A Abstract Picture

solved by chelly

---

chelly‘s solution

B Battle Mage

unsolved

---

C Constant Ratio

solved by chelly

---

chelly‘s solution

D dir -C

solved by chelly

---

chelly‘s solution

E Extreme Permutations

unsolved

---

F Find the Length

solved by chelly

---

chelly‘s solution

G Game ``Minesweeper‘‘

unsolved

---

H Hierarchy

solved by chelly

---

chelly‘s solution

I Interactive Casino

solved by chelly

---

chelly‘s solution

J Judgement

solved by chelly

---

chelly‘s solution

K Krotek

unsolved

---

L Liesbeth and the String

solved by ch&chelly

---

chelly‘s solution

M Matrix, The

unsolved

---

Dirt Replay