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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
#325615#4833. Tetra-puzzletuanlinh1230 1ms3912kbC++205.4kb2024-02-11 17:56:022024-02-11 17:56:02

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你现在查看的是最新测评结果

  • [2024-02-11 17:56:02]
  • 评测
  • 测评结果:0
  • 用时:1ms
  • 内存:3912kb
  • [2024-02-11 17:56:02]
  • 提交

answer

#include<bits/stdc++.h>
#define ll int
#define pll pair<ll, ll>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ld long double
using namespace std;
inline ll IDX(ll i, ll j) {return i*5+j;}
inline bool BIT(ll mask, ll i, ll j) {return (mask&1<<IDX(i, j))!=0;}

struct piece
{
    ll val=0, h=0, w=0;
    piece(){}
    piece(char s)
    {
        vector <vector <bool>> a(5, vector <bool> (5, 0));
        if (s=='I') a[0][0]=a[1][0]=a[2][0]=a[3][0]=1, h=4, w=1;
        if (s=='L') a[0][0]=a[1][0]=a[2][0]=a[2][1]=1, h=3, w=2;
        if (s=='O') a[0][0]=a[0][1]=a[1][0]=a[1][1]=1, h=w=2;
        if (s=='T') a[0][0]=a[0][1]=a[0][2]=a[1][1]=1, h=2, w=3;
        if (s=='Z') a[0][0]=a[0][1]=a[1][1]=a[1][2]=1, h=2, w=3;
        for (ll i=0; i<5; i++)
            for (ll j=0; j<5; j++)
                if (a[i][j]) val|=1<<IDX(i, j);
    }

    inline void fdiag()
    {
        ll res=0;
        for (ll i=0; i<5; i++)
            for (ll j=0; j<5; j++)
                if (BIT(val, i, j))
                    res|=1<<IDX(j, i);
        val=res, swap(h, w);
    }
    inline void fhori()
    {
        ll res=0;
        for (ll i=0; i<5; i++)
            for (ll j=0; j<5; j++)
                if (BIT(val, i, j))
                    res|=1<<IDX(4-i, j);
        val=res;
    }
    inline void rotate()
    {
        fdiag();
        fhori();
    }
    inline void fix()
    {
        while (1)
        {
            bool ok=1;
            for (ll i=0; i<5; i++)
                if (BIT(val, 0, i))
                    {ok=0; break;}
            if (!ok) break;
            val>>=5;
        }
        while (1)
        {
            bool ok=1;
            for (ll i=0; i<5; i++)
                if (BIT(val, i, 0))
                    {ok=0; break;}
            if (!ok) break;
            val>>=1;
        }
    }

    void print()
    {
        cout << h << " " << w << "\n";
        for (ll i=0; i<5; i++)
            for (ll j=0; j<5; j++)
                cout << BIT(val, i, j) << " \n"[j==4];
    }
};

const ll all=(1<<25)-1, inf=1e9, cap=5000;
vector <pair<pll, pll>> best[1005];
vector <piece> I, L, O, T, Z;
ll hori[5], veri[5];
vector <ll> t[26];

struct state
{
    ll val=0;
    state(){}
    state(ll val): val(val){}
    ll evaluate(){return __builtin_popcount(val);}

    bool Move(ll move)
    {
        if (val&move) return 0;
        ll ne=val|move, re=0;
        for (ll k=0; k<5; k++)
        {
            if ((ne|hori[k])==ne) 
                re|=hori[k];
            if ((ne|veri[k])==ne) 
                re|=veri[k];
        }
        val=ne^re;
        return 1;
    }

    ll findbest(ll c)
    {
        pll best={inf, 0};
        for (ll move:t[c]) 
        {
            state ne=*this;
            if (!ne.Move(move)) continue;
            best=min(best, mp(ne.evaluate(), move));
        }
        return best.se;
    }

    void act(ll move)
    {
        for (ll i=0; i<5; i++) for (ll j=0; j<5; j++)
        {
            if (BIT(val, i, j)) cout << '#';
            else if (BIT(move, i, j)) cout << '*';
            else cout << '.'; if (j==4) cout << "\n";
        }
        cout << endl, Move(move);
    }
};

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    for (ll i=0; i<5; i++)
        for (ll j=0; j<5; j++)
        {
            hori[i]|=1<<IDX(i, j);
            veri[j]|=1<<IDX(i, j);
        }
    auto init=[&](vector <piece> &a, char s)
    {
        a.resize(8), a[0]=piece(s);
        for (ll i=1; i<8; i++)
        {
            if (i<4) a[i]=a[i-1], a[i].rotate();
            else a[i]=a[i-4], a[i].fhori();
            a[i].fix();
        }
        map <ll, bool> Map;
        for (ll i=0; i<8; i++) for (ll j=0; j<=5-a[i].h; j++) for (ll k=0; k<=5-a[i].w; k++)
        {
            ll val=a[i].val<<IDX(j, k);
            if (!Map[val]) Map[val]=1, t[s-'A'].pb(val);
        }
    }; init(I, 'I'), init(L, 'L'), init(O, 'O'), init(T, 'T'), init(Z, 'Z');
    string type; cin >> type;
    if (type=="prepare")
    {
        ll n; cin >> n;
        best[0].pb({{0, 0}, {-1, -1}});
        for (ll i=1; i<=n; i++)
        {
            string c; cin >> c;
            for (ll id=0; id<best[i-1].size(); id++)
            {
                pair<pll, pll> s=best[i-1][id];
                for (ll j=0; j<2; j++)
                {
                    state cr(s.fi.se);
                    cr.Move(cr.findbest(c[j]-'A'));
                    best[i].pb({{cr.evaluate(), cr.val}, {id, c[j]-'A'}});
                }
            }
            sort(best[i].begin(), best[i].end());
            if (best[i].size()>cap) best[i].resize(cap);
        }
        string ans="";
        for (ll i=n, j=0; best[i][j].se.fi!=-1; j=best[i][j].se.fi, i--)
            ans=char(best[i][j].se.se+'A')+ans;
        cout << ans;
    }
    else
    {
        state cr; ll n; cin >> n;
        for (ll i=0; i<n; i++)
        {
            char c; cin >> c; c-='A';
            pll best={inf, 0};
            for (ll move:t[c]) 
            {
                state ne=cr;
                if (!ne.Move(move)) continue;
                best=min(best, mp(ne.evaluate(), move));
            }
            if (best.fi==inf) assert(0);
            cr.act(best.se);
        }
    }
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 3588kb

First Run Input

prepare
6
TO LO ZI LI LT LT

First Run Output

OLILLL

Second Run Input

play
6
O
L
I
L
L
L

Second Run Output

**...
**...
.....
.....
.....

##***
##*..
.....
.....
.....

****.
###..
.....
.....
.....

####*
###.*
...**
.....
.....

.***.
###*#
...##
.....
.....

.###.
*....
***##
.....
.....


result:

ok good job!

Test #2:

score: 100
Accepted
time: 0ms
memory: 3636kb

First Run Input

prepare
1
ZI

First Run Output

I

Second Run Input

play
1
I

Second Run Output

****.
.....
.....
.....
.....


result:

ok good job!

Test #3:

score: 100
Accepted
time: 1ms
memory: 3912kb

First Run Input

prepare
10
LO TZ ZO OI OL ZL ZO TL OZ LO

First Run Output

OZZIOLOLOO

Second Run Input

play
10
O
Z
Z
I
O
L
O
L
O
O

Second Run Output

**...
**...
.....
.....
.....

##.**
##**.
.....
.....
.....

##.##
####*
...**
...*.
.....

##*##
..*..
..*##
..*#.
.....

.....
**#..
**###
..##.
.....

.....
###**
....*
..##*
.....

.....
.....
**..#
**###
.....

..*..
..*..
##**#
.....
.....

**#..
**#..
.....
.....
.....

###**
###**
.....
......

result:

ok good job!

Test #4:

score: 0
Stage 2: Program answer Runtime Error

First Run Input

prepare
29
ZT OT ZL LT ZI LO IZ LT OI ZT ZT OL ZI LT LZ TZ IL ZI OT LZ LT LZ OT OI IO OZ IZ ZL ZL

First Run Output

ZOLTZOZLOZTOZTZTIIOLLLTIIZILL

Second Run Input

play
29
Z
O
L
T
Z
O
Z

Second Run Output

.**..
**...
.....
.....
.....

.##**
##.**
.....
.....
.....

.####
##*##
***..
.....
.....

.####
...*.
###**
...*.
.....

*####
**.#.
.*...
...#.
.....

.....
##.#.
.#...
**.#.
**...


result: