题解 | #太阳之华#

纯BFS，首先最开始模拟的时候我们很容易就会发现，对于红色，必须存在一个连通块和所有蓝色格子四联通，否则平局。 对于蓝色，则是除非全部都是蓝色格子，不然赢不了 所以直接BFS所有红色的连通块查找是否存在一个连通块和所有蓝色格子四联通就行了

代码如下:

//               BggBB                wZPXsv:.         UBgQGv
//               BgEQQ            ,:sJ.    .,.  ::,    rBORZJ
//              .BgggB   .sJrc7r77:.,          .:;75as :BgOMp
//              :BgERB. 2a:          ,:.             :cEBOgMB:
//       wR:    rBODgBKL:  ,:r:     ,srLL;.        .   ,BRggBJ ;       s
//       gR1    sBgDBQi  :csLr;         ,rJ7.       ,,  BMp5QQJ;w:   :rg,
//   JRJipEs    XBRBO   7s;,               :c;        .,BE5OgB  :7L    L
//    cZQDB:    RBBP  :L;.                   :r         GBBEgQ5 .;:w
//       ..    ;DBZ  r7.         ............ :r         rBBpgBr  i.w.
//            pBBR  r;      ........,.....,....:r .        BBGEOZ  r w,
//           BQBB  :,      ..,:.................i:.,.       gBQB6B1;r 5:
//           5BB       ..:,...;.......,.,........:.     .QL  ZBRMXBB,. X.
//            B.     ... 7r ..:,...,.,....:: ....:,.XQr;BBB   EBGMB.    L
//         ,DB:     ..,. :L   ,r  ,.,.,.. :B: ....: 1BHKBQB;   RBgMES:
//     2   GB:   .  ,,.. rg  . w;.........:X2;.::::.    ,SE,    BBRBQBa
//     ;   77,  . .,,... 7S,...,5...,.,.. ;7 1: ..::...     ,.  .BMEGB:
//        sr;R . ..,.... U:S ...rr.....,. s; .Jr  .,.........,.  2QSgr ..
//       K2 X:   ,,.,.:::2 1; ..,::.......X    ri  ....,,.....,.  BJM  ..
//     ,Q7  Q   .,....c;.L .5  ..,...,.. cr .:ri2a   . ;r..,.,.,  P:P:,
//    ;H.  ra  .,.,.. H7:;  ,c   .,,,,,.:U RBBBBBBB:,. .r ....,,  r:p,;r
//         G: .i.... :s::r;2pBL:; .,,,..K. Bss5L7LEMJvrr;...:..,  ,pZ :X:
//         M  .::... 7:.7QBBBBBKBr.....sr    w:,  ,:6.r;,..::.,.  :5:  :Z
//         Q ...i,.. 1sBO::U;; :rvr:::cr     6 :HS, p  X. :;....  :U L  :
//         O ...,r.  BBP  77..,r;c .LS:      w; r;  H  1 ,;....., rJ;H
//         g ....,i, OS   ,X..7HrL   :        1r..:s. L:,;....,:. K2:.L
// .:i;:, .g  .....:;sES   L7 .,,7             ,. ,  r;,:.,..::. iE;: s.
// ;.  .ss:M; :..,...v:17   ;7,.:.  .              .7;.,.. ;7: .rD,:i; 2
//   ,.   : G:6r. ... ;7g.                       rJi:.  ,:J1:.r2,rr::v ;,
//   7v7: ;.7:7UP2i:,:rr,7          .. .::         rJrrcJsc: L;L: J;::7 r
// i....:visP.27rLJLU5r;css         ,.            ,5s,c;:,r. c..c  7;,v,
// : ,. i, ,RO.:       .E;Hw,                   .DBH7;r7.:i  X5LLU. .. r
// , ...BU.  .w:;r1r    .sr,vc:               .pBREDQBBL.;:  QBBBBBXXP7.
// , , :QBB1 ;:7v:vJ:    Jv.i1EgBgJi,     .,::.gDZKPHGBs,;  .BQMDU5GQBBB7
// , .. LDr 7rLLL,U,r.   cDQBQBRGERQBBQ,:;r7s2PDRZZpEDBL,., ,BRpZZZZOOgB7
// , .:. . DQK.r.7; r.   :gBDZpZPEpZ6gMa6RMBBQgEKZ6DDMBr,..::BEZpEGDZPPG
//   si,: :2QBac;..,r     PQgPPPEZOZDGDRRDDEGpZpOOgORBH,: ..:Qg6ZZp2JsO;
//  Q: :J7 7      .,.     6RQQDD6ZpZpZ6Z6ZpZ6gOgDOGRDR  :.  .MMOpX52UKDr
// BB7 1  .:UMOPXr        gQgEgDDZZpE6E6Z6ZPZGOpDRDUSOEHBBL  7BRP5HXXXDJ.
// GXB1  .: UBMQBBg       JB6pKpPG6O6G6EZEpEPp6MGSwPEDRQgMBP  :RPXUX6PQa;
// gXOR .,,pgGE6ODBB,      QQOZ6RMBBBQMgDZZ6ZGE21HOEZpEpEZgB,  :ZZppBQs,v
// gwXBr,;EBEEpGZGGBQ:     .RMgK7r:::i7s1HPZHDHLEGPpKpK6PpZBs .D;g6GMZ  ;

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define ll long long
#define lowbit(x) x&(-x)

const int MOD=1e9 + 7;
const int N=1e6+10;

int qpow(int num,int k) {
    int res=1;
    while(k) {
        if(k%2) res = (res * num) % MOD;
        num = (num * num) % MOD;
        k/=2 ;
    }
    return res % MOD;
}

int inv(int x) {
    return qpow(x,MOD-2);
}
 
int lcm(int x,int y) {
    return x * y / __gcd(x,y);
}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int Base = uniform_int_distribution<>(8e8,9e8)(rng); 

int a[N];
int b[N];

char c[2020][2020];

int dx[4] = {0,1,0,-1};
int dy[4] = {1,0,-1,0};

int vis[2020][2020];

void solve() {
    int n,m;
    cin>>n>>m;

    int sum = 0;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            cin>>c[i][j];
            vis[i][j] = 0;
            if(c[i][j] == '.') sum ++;
        }
    }

    // for(int i=1;i<=n;i++){
    //     for(int j=1;j<=m;j++) cout<<c[i][j];
    //     cout<<"\n";
    // }

    if(sum == n * m){
        cout<<"Blue\n";
        return ;
    }

    int sign = 1;

    //红方要获胜必须要有一个连通块，所有蓝色都和这个连通块的格子四联通

    int counts = 0;
    queue<pair<int,int>>q;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(c[i][j] == '#'){
                counts ++;
                q.push({i,j});
                vis[i][j] = 1;

                int num = 0;
                while(!q.empty()){
                    auto [x,y] = q.front();
                    q.pop();

                    for(int k=0;k<4;k++){
                        int ddx = x + dx[k];
                        int ddy = y + dy[k];

                        if(ddx >= 1 && ddx <= n && ddy >= 1 && ddy <= m){
                            if(c[ddx][ddy] == '#' && !vis[ddx][ddy]){
                                vis[ddx][ddy] = 1;
                                q.push({ddx,ddy});
                            }
                            else if(c[ddx][ddy] == '.' && vis[ddx][ddy] != counts){
                                num ++;
                                vis[ddx][ddy] = counts;
                            }
                        }
                    }
                }

                if(num == sum){
                    cout<<"Red\n";
                    return ;
                }
            }
        }
    }
    cout<<"Draw\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int t=1;
    cin>>t;

    while(t--) {
        solve();
    }
    return 0;
}