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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
#605659#7733. Cool, It’s Yesterday Four Times MoreGodwangRE 9ms6156kbC++237.5kb2024-10-02 18:21:442024-10-02 18:21:49

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你现在查看的是最新测评结果

  • [2024-10-02 18:21:49]
  • 评测
  • 测评结果：RE
  • 用时：9ms
  • 内存：6156kb
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  • [2024-10-02 18:21:44]
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answer

#include <iostream>
using namespace std;
#include <set>
#include <algorithm>
#include <cmath>
#include <map>
#include <cstdio>
#include <string>
#include <cstring>
#include <string.h>
#include <stdlib.h>
#include <iomanip>
#include <fstream>
#include <stdio.h>
#include <stack>
#include <queue>
#include <ctype.h>
#include <vector>
#include <random>
#include<list> 
#define ll long long
#define ull unsigned long long
#define pb push_back
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define pii pair<int, int>
#define pli pair<ll, int>
#define pil pair<int, ll>
#define pll pair<ll, ll>
#define lowbit(x) ((x)&(-x))
ll extend_gcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll d = extend_gcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}
ll fastpow(ll a, ll n, ll mod)
{
    ll ans = 1;
    a %= mod;
    while (n)
    {
        if (n & 1)
            ans = (ans * a) % mod; //% mod
        a = (a * a) % mod;         //% mod
        n >>= 1;
    }
    return ans;
}

inline void write(__int128 x)
{
    if (x > 9)
    {
        write(x / 10);
    }
    putchar(x % 10 + '0');
}
__int128 sqrt(__int128 m)
{
    __int128 leftt = 0, rightt = ((__int128)1) << 51, ret = -1, mid;
    while (leftt < rightt)
    {
        mid = (leftt + rightt) / 2;
        if (mid * mid > m)
        {
            rightt = mid;
        }    
        else
        {
            leftt = mid + 1;
            ret = mid;
        }
    }
    return ret;
}

const double eps = 1e-6;
int sgn(double x)
{
    if(fabs(x)<eps)
    {
        return 0;
    }
    else return x<0?-1:1;
}

struct Point
{
    double x,y;
    Point()
    {

    }
    Point(double x,double y):x(x),y(y)
    {

    }
    Point operator + (Point B)
    {
        return Point(x+B.x,y+B.y);
    }
    Point operator - (Point B)
    {
        return Point(x-B.x,y-B.y);
    }
    bool operator == (Point B)
    {
        return sgn(x-B.x)==0&&sgn(y-B.y)==0;
    }
    bool operator < (Point B)
    {
        return sgn(x-B.x)<0||(sgn(x-B.x)==0&&sgn(y-B.y)<0);
    }
};
typedef Point Vector;
double Cross(Vector A,Vector B)//叉积
{
    return A.x*B.y-A.y*B.x;
}
double Distance(Point A,Point B)
{
    return hypot(A.x-B.x,A.y-B.y);
}
int Convex_hull(Point *p,int n,Point *ch)
{
    n=unique(p,p+n)-p;
    sort(p,p+n);
    int v=0;

    for(int i=0;i<n;i++)
    {
        while (v>1&&sgn(Cross(ch[v-1]-ch[v-2],p[i]-ch[v-1]))<=0)
        {
            v--;
        }
        ch[v++]=p[i];
    }

    int j=v;

    for(int i=n-2;i>=0;i--)
    {
        while (v>j&&sgn(Cross(ch[v-1]-ch[v-2],p[i]-ch[v-1]))<=0)
        {
            v--;
        }
        ch[v++]=p[i];
    }
    if(n>1)
    {
        v--;
    }
    return v;
}

int kmp(string s, string p)
{
    int ans = 0, lastt = -1;
    int lenp = p.size();
    vector<int > Next(lenp+3,0);
    rep(i, 1, lenp - 1)
    {
        int j = Next[i];
        while (j && p[j] != p[i])
        {
            j = Next[j];
        }
        if (p[j] == p[i])
        {
            Next[i + 1] = j + 1;
        }
        else
        {
            Next[i + 1] = 0;
        }
    }
    int lens = s.size();
    int j = 0;
    rep(i, 0, lens - 1)
    {
        while (j && s[i] != p[j])
        {
            j = Next[j];
        }
        if (s[i] == p[j])
        {
            j++;
        }
        if (j == lenp)
        {
            ans++;
        }
    }
    return ans;
}

int dir[4][2] =
    {
        {-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 左右上下
// int dir[8][2]={
//         {-1, 0}, {0, 1}, {1, 0}, {0, -1},{-1,-1},{-1,1},{1,-1},{1,1}
// };
        
#define endl '\n'//交互题请删除本行
const ll inf = 1000000000000000000ll;
const ll mod1 = 998244353ll, P1 = 131, mod2 = 1e9 + 7ll, P2 = 13331;
ll inverse(ll x)
{
    return fastpow(x,mod1-2,mod1);
}

const int N = 1e4 + 10, M = 2e6 + 10;

///////////////////////////////////

int tt;

int n,m;

string s[N];

int win[M];

bool vis[M];



//map<node,bool > ma;

int ans;

int son;

queue<vector<int > > q;

bool flag;

///////////////////////////////////

ll get(int i,int j,int ii,int jj)
{
    return i*(m+1)*(n+1)*(m+1)+j*(n+1)*(m+1)+ii*(m+1)+jj;
}

void dfs(int i,int j,int ii,int jj)
{
    //
   // cout<<i<<" "<<j<<" "<<ii<<" "<<jj<<endl;
    if(i<1||i>n||j<1||j>m||s[i][j]=='O')
    {
        return;
    }
    if(s[i][j]=='.'&&(  (ii<1||ii>n||jj<1||jj>m)      )|| s[ii][jj]=='O' )
    {
        flag=1;
        //
        if(flag)
        {
          //  
        }
        return;
    }

    if(vis[get(i,j,ii,jj)])
    {
        return;
    }
    //
    
    vis[get(i,j,ii,jj)]=1;
    vector<int > v;
    v.pb(i);
    v.pb(j);
    v.pb(ii);
    v.pb(jj);
    
    q.push(v);
    

    rep(k,0,3)
    {
        int x1=i+dir[k][0],y1=j+dir[k][1],x2=ii+dir[k][0],y2=jj+dir[k][1];
        //
        //cout<<dir[i][0]<<" "<<dir[i][1]<<endl;
        
        dfs(x1,y1,x2,y2);
    }
    
}

///////////////////////////////////

void init()
{
    ans=0;
    son=0;
    rep(i,1,(n+3)*(m+1)*(n+1)*(m+1))
    {
        win[i]=0;
        vis[i]=0;
    }
}

///////////////////////////////////

int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//交互题请删除本行
   // freopen("ain.txt", "r", stdin); freopen("aout.txt", "w", stdout);
    
    cin>>tt;
    while (tt--)
    {
        cin>>n>>m;
        init();
        rep(i,1,n)
        {
            cin>>s[i];
            s[i]=' '+s[i];
            rep(j,1,m)
            {
                if(s[i][j]=='.')
                {
                    son++;
                }
            }
        }
        rep(i,1,n)
        {
            rep(j,1,m)
            {
                if(s[i][j]=='O')
                {
                    continue;
                }
                int ok=0;
                rep(ii,1,n)
                {
                    rep(jj,1,m)
                    {
                        if(s[ii][jj]=='O'||ii==i&&jj==j)
                        {
                            continue;
                        }

                        flag=0;
                        if(vis[get(i,j,ii,jj)])
                        {
                            if(win[get(i,j,ii,jj)])
                            {
                                ok++;
                            }
                            continue;
                        }
                        
                        dfs(i,j,ii,jj);

                        while (q.size())
                        {
                            vector<int >  arr=q.front();
                            q.pop();
                            win[get(arr[0],arr[1],arr[2],arr[3])]=flag;
                        }
                        
                        //
                        
                        if(win[get(i,j,ii,jj)])
                        {
                          //  cout<<i<<" "<<j<<" "<<ii<<" "<<jj<<" OK"<<endl;
                            ok++;
                        }
                    }
                }

                if(ok+1==son)
                {
                    ans++;
                }

            }
        }
        cout<<ans<<endl;


    }
    


    return 0;
}
源文件, 语言: C++23

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 5996kb

input:

4
2 5
.OO..
O..O.
1 3
O.O
1 3
.O.
2 3
OOO
OOO

output:

3
1
0
0

result:

ok 4 lines

Test #2:

score: 0
Accepted
time: 2ms
memory: 5988kb

input:

200
2 4
OOO.
OO..
2 3
OOO
.O.
3 3
O.O
OOO
OO.
4 1
.
.
O
O
1 2
.O
1 1
.
2 5
.OO..
.O.O.
2 1
O
O
1 1
O
1 3
.OO
5 1
O
O
.
O
.
5 2
O.
..
O.
.O
..
5 3
...
...
.OO
..O
OOO
3 5
..O.O
.O.O.
.OO.O
5 2
.O
OO
O.
O.
..
2 1
O
O
3 5
.O.OO
O...O
..OO.
1 5
.....
5 1
O
.
O
.
.
5 3
OOO
OO.
.OO
OO.
O.O
2 1
O
.
5 2
O.
...

output:

3
0
0
2
1
1
3
0
0
1
0
7
9
4
4
0
6
5
2
0
1
6
4
5
2
0
0
5
3
3
1
4
1
0
7
5
2
3
7
3
0
6
2
2
2
0
4
6
6
3
3
2
3
5
2
1
0
3
3
4
4
2
2
0
7
6
4
8
5
3
2
5
2
1
2
1
4
0
0
2
5
1
4
6
6
1
6
2
2
3
4
5
2
1
0
1
9
3
4
11
0
3
2
1
0
0
4
3
1
4
3
8
3
0
3
6
2
5
1
3
3
4
0
2
11
2
2
4
0
4
4
6
2
1
2
3
0
5
0
16
4
3
2
6
0
8
3
3
1...

result:

ok 200 lines

Test #3:

score: 0
Accepted
time: 9ms
memory: 6156kb

input:

50
10 9
OOOO.O...
O...O.OOO
.....O...
..OO..O.O
...O..O.O
..OOO..O.
..OOO...O
.OO.O..OO
.O.O.OO..
.O..O.O.O
10 10
.O.OOO.OO.
...OOOOOOO
...O.O..O.
.O.O..O...
.O.O.OO..O
..OO.O..OO
O....O..OO
OO...OOOOO
OO.OO.O..O
.O.O.OOOOO
10 8
O..OOO.O
O.OOOO..
O..OO.OO
OO..OO..
.OOO..O.
.OO.OO.O
OOO..OO.
..O..OO....

output:

31
40
13
25
40
37
27
29
20
26
25
29
21
29
21
31
32
31
33
34
25
31
18
25
41
28
20
45
20
29
18
21
27
28
35
13
20
17
32
29
28
23
23
23
24
18
28
17
35
24

result:

ok 50 lines

Test #4:

score: -100
Runtime Error

input:

5
1 1000
....O..OO..O..O..OO...OOO.OOOO.O...O....OOOOO.....OO..OOOO.O..O....OOOO..OO..OOOO......O.O.O..O..OO.OO.OO.O....O.O.O.O.O.OO..O.O.OO..O..OO..O.OOO...O...O.O.O..O....O.OO...O..O...O.OOO..OO.O..O.OO.O.O..OOOO..O.OO.O.O....O.OO.......OOOO....O.O.O.O..OOO.O.OO.OOO..O...O.O.O.O.OO.O.OOOO...O.OO.....

output:

result:
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