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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#189966#4432. Jungle Trailberarchegas#AC ✓1117ms297924kbC++204.6kb2023-09-28 03:09:372023-09-28 03:09:37

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你现在查看的是最新测评结果

  • [2023-09-28 03:09:37]
  • 评测
  • 测评结果:AC
  • 用时:1117ms
  • 内存:297924kb
  • [2023-09-28 03:09:37]
  • 提交

answer

#include <bits/stdc++.h>
#include <iostream>
// #define int long long
#define ld long double
#define endl "\n"
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define per(i, a, b) for(int i = b-1; i>=a ; i--)
#define trav(a, x) for(auto& a : x)
#define allin(a , x) for(auto a : x)
#define Unique(v) sort(all(v)),v.erase(unique(all(v)),v.end());
#define sz(v) ((int)v.size())
using namespace std;
typedef vector<int> vi;
#define y1 abacaba
//#define left oooooopss
#define db(x) cerr << #x <<" == "<<x << endl;
#define db2(x,y) cerr<<#x <<" == "<<x<<", "<<#y<<" == "<<y<<endl;
#define db3(x,y,z) cerr << #x<<" == "<<x<<", "<<#y<<" == "<<y<<", "<<#z<<" == "<<z<<endl;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<ll> vl;
// std::mt19937_64 rng64((int) std::chrono::steady_clock::now().time_since_epoch().count());
std::mt19937 rng(
  
//  (int) std::chrono::steady_clock::now().time_since_epoch().count()
   1239010
);
ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down
//inline ll mod(ll n, ll m ){ ll ret = n%m; if(ret < 0) ret += m; return ret; }
ll gcd(ll a, ll b){return (b == 0LL ? a : gcd(b, a%b));}
ll exp(ll b,ll e,ll m){
    b%=m;
    ll ans = 1;
    for (; e; b = b * b % m, e /= 2)
        if (e & 1) ans = ans * b % m;
    return ans;
}
// debug:
template<class T>void print_vector(vector<T> &v){
    rep(i,0,sz(v))cout<<v[i]<<" \n"[i==sz(v)-1];
}
void dbg_out() {cerr << endl; }
template<typename Head, typename ... Tail> void dbg_out(Head H,Tail... T){
    cerr << ' ' << H;
    dbg_out(T...);
}
#ifdef LOCAL
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__), cerr << endl
#else
#define dbg(...) 42
#endif
//

const int N = 2050;
string s[N];
bool can[N][N][2][2];
bool vis[N][N][2][2];

struct nx{
    int i , j , l , r;
} nxt[N][N][2][2];

int n, m;
bool A[N] , B[N];
bool dp(int i ,int j , int l , int r){
    if(i >= n || j >= m)
        return 0;
    if(s[i][j] == '#')
        return 0;
    char p = s[i][j];
    bool flipped = l ^ r;
    if((p == '@' && !flipped) || (p == 'O' && flipped)){
       return can[i][j][l][r] = false;  
    }
    if(i == n - 1&& j == m - 1){
        can[i][j][l][r] = 1 , nxt[i][j][l][r] = {-1,-1,-1,-1};
        return true;
    }
    if(vis[i][j][l][r])
        return can[i][j][l][r];
    vis[i][j][l][r] = true;
    bool a = dp(i+1,j,0,r);
    bool b = dp(i,j+1,l,0);
    bool c = dp(i+1,j,1,r);
    bool d = dp(i,j+1,l,1);
    if(a)
        can[i][j][l][r] = 1 , nxt[i][j][l][r] = {i+1,j,0,r};
    if(b)
        can[i][j][l][r] = 1 , nxt[i][j][l][r] = {i,j+1,l,0};
    if(c)
        can[i][j][l][r] = 1 , nxt[i][j][l][r] = {i+1,j,1,r};
    if(d)
        can[i][j][l][r] = 1 , nxt[i][j][l][r] = {i,j+1,l,1};
    

    return can[i][j][l][r];
}
string ans;
void recover(int i , int j , int l , int r){
    if(i == -1 )
        return;
    if(l)
        A[i] = true;
    if(r)
        B[j] = true;
    auto [ii, jj , ll, rr] = nxt[i][j][l][r];
    if(jj > j)
        ans.push_back('P');
    else if(ii > i)
        ans.push_back('D');
    recover(ii, jj , ll,  rr);
}

void solve(){
    cin >> n >> m;
    ans.clear();
    for(int i = 0 ; i < n; i ++){
        cin >> s[i];
        A[i] = 0;
        for(int j = 0 ; j < m ; j ++)
            for(int l = 0 ; l < 2; l ++)
                for(int r = 0 ; r < 2 ; r ++)
                    can[i][j][l][r] = vis[i][j][l][r] = B[j] = 0;
    }
    bool pt = dp(0,0,0,0);
    bool pt2 = dp(0,0,1,0);
    bool pt3 = dp(0,0,0,1);
    bool pt4 = dp(0,0,1,1);
    if(pt || pt2 || pt3 || pt4){
        cout << "TAK" << endl;
        if(pt)
            recover(0,0,0,0);
        else if(pt2)
            recover(0,0,1,0);
        else if(pt3)
            recover(0,0,0,1);
        else if(pt4)
            recover(0,0,1,1);
        for(int i = 0 ; i < n; i ++)   
            cout << (A[i] ? 'T' : 'N');
        cout << endl;
        for(int i = 0 ; i < m ; i ++)
            cout << (B[i] ? 'T' : 'N');
        cout << endl;
        cout << ans << endl;
    }
    else{
        cout << "NIE" << endl;
    }
}

int32_t main(){
    fastio;
    int t;
    cin >> t;
    while(t--){
        solve();
    }

    // math -> gcd it all
    // Did you swap N,M? N=1?
}

详细

Test #1:

score: 100
Accepted
time: 1117ms
memory: 297924kb

input:

486
4 5
..#..
@@O@@
##@#O
..@.@
2 2
OO
OO
2 2
@@
@@
2 2
@@
#@
2 2
@O
#@
2 2
@@
OO
2 2
O#
@O
2 2
@#
#@
2 2
@.
.@
2 2
@#
.O
2 2
OO
.O
10 10
@O@O#O@@@#
OO#@#@@#OO
#@#@#O##O@
OO##@@O#@O
O##@@#@O#@
OO@OO@@@O@
@O#@#@O#@O
@OOOOO@##.
O@OOO##O@@
OO@@OOOO#@
10 10
@@#OOO#O@@
#@@OO@@.O@
#.O@@O#@@O
OO@@#O@#O@
.#...

output:

TAK
NNTN
NTNTT
PDPPPDD
TAK
NN
NN
PD
TAK
TT
NN
DP
TAK
TT
NN
PD
TAK
TN
NT
PD
TAK
TN
NN
PD
TAK
NT
NT
DP
NIE
TAK
TN
NT
PD
TAK
TT
NT
DP
TAK
NT
NT
DP
NIE
TAK
TTTTNNTTTN
NNTNNNNNNT
PDDPPDDDPPPDPPPDDD
TAK
NTTNNNTTTT
NTNTNTTTNT
DDPPDPPDPDDDPDPPPD
TAK
NTTTNNNNNT
NNTTTTNNTT
DPPDDPPPPDPPDDPDDD
TAK
NTTTNTNTNT
NT...

result:

ok ac (486 test cases)

Extra Test:

score: 0
Extra Test Passed