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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#864606#4773. Piece it togetherrladbekaAC ✓443ms161264kbC++205.3kb2025-01-20 20:10:252025-01-20 20:10:25

Judging History

你现在查看的是最新测评结果

  • [2025-01-20 20:10:25]
  • 评测
  • 测评结果:AC
  • 用时:443ms
  • 内存:161264kb
  • [2025-01-20 20:10:25]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef array<ll, 2> pii;
typedef array<ll, 3> tii;
typedef vector<pii> vpii;
typedef double lf;
typedef string S;
#define V vector
#define PQ priority_queue
#define fastio; cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(false);
#define vcin; for(int i=0; i<n; i++) cin >> v[i];
#define forf(i, s, e) for(ll i=s; i<e; i++)
#define forb(i, s, e) for(ll i=s-1; i>=e; i--)
#define pb push_back
#define sortv(v) sort(v.begin(), v.end())
#define sortc(v, cmp) sort(v.begin(), v.end(), cmp)
#define all(v) v.begin(), v.end()
const ll mod=1e9+7, MOD=998244353;
const ll dir4[4][2]={{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
const ll dir8[8][2]={{0, 1}, {1, 0}, {-1, 0}, {0, -1}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
const ll inf=2147483647, linf=9223372036854775807;
const double pi=acos(-1), E=2.718281828459;
ll gcd(ll a, ll b){return b?gcd(b, a%b):a;}

ll n, e, N, M, Wc, Bc;
vvi edge, redge;
V<bool> vsd, rvsd;
vi vseq;
vvi scc;

void dfs(ll pnt){
    vsd[pnt]=true;
    for(ll i : edge[pnt]){
        if(!vsd[i]) dfs(i);
    }
    vseq.pb(pnt);
}

void rdfs(ll pnt){
    rvsd[pnt]=true;
    for(ll i : redge[pnt]){
        if(!rvsd[i]) rdfs(i);
    }
    scc[scc.size()-1].pb(pnt);
}

bool _2sat(){
    vi sat(n);

    forf(cnt, 0, scc.size())
        for(ll i : scc[cnt])
            sat[i]=cnt+1;

    forf(i, 0, 2*N*M) if(sat[i]==sat[i+2*N*M]) return false;

    return true;
}

bool solve(){
    cin >> N >> M; n=N*M*4; // 0,1 UD / 2,3 LR
    edge.clear(); redge.clear(); vsd.clear(); rvsd.clear();
    vseq.clear(); scc.clear();
    edge.resize(n); redge.resize(n);
    vsd.resize(n); rvsd.resize(n);
    Bc=0; Wc=0;

    V<string> s(N);
    forf(i, 0, N) cin >> s[i];
    forf(i, 0, N){
        forf(j, 0, M){
            Wc+=(s[i][j]=='W');
            Bc+=(s[i][j]=='B');

            if(s[i][j]=='B'){
                bool up=(i>0 && s[i-1][j]=='W');
                bool down=(i<N-1 && s[i+1][j]=='W');
                ll O=M*i+j, U=M*(i-1)+j, L=M*i+(j-1), lr=N*M, Not=2*N*M;
                if(up && down){
                    edge[U+Not].pb(O);
                    edge[O+Not].pb(U);
                }
                else if(up){
                    edge[U+Not].pb(U);
                }
                else if(down){
                    edge[O+Not].pb(O);
                }
                else{
                    return false;
                }

                bool left=(j>0 && s[i][j-1]=='W');
                bool right=(j<M-1 && s[i][j+1]=='W');


                if(left && right){
                    edge[L+Not+lr].pb(O+lr);
                    edge[O+Not+lr].pb(L+lr);
                }
                else if(left){
                    edge[L+Not+lr].pb(L+lr);
                }
                else if(right){
                    edge[O+Not+lr].pb(O+lr);
                }
                else{
                    return false;
                }

            }
        }
    }

    forf(i, 0, N){
        forf(j, 0, M){
            if(s[i][j]!='W') continue;

            if(i<N-1 && j<M-1 && s[i+1][j]=='B' && s[i][j+1]=='B'){
                ll UD=i*M+j, LR=i*M+j+N*M, Not=2*N*M;
                edge[UD].pb(LR+Not);
                edge[LR].pb(UD+Not);
            }
            if(i<N-1 && j>0 && s[i+1][j]=='B' && s[i][j-1]=='B'){
                ll UD=i*M+j, LR=i*M+(j-1)+N*M, Not=2*N*M;
                edge[UD].pb(LR+Not);
                edge[LR].pb(UD+Not);
            }
            if(i>0 && j<M-1 && s[i-1][j]=='B' && s[i][j+1]=='B'){
                ll UD=(i-1)*M+j, LR=i*M+j+N*M, Not=2*N*M;
                edge[UD].pb(LR+Not);
                edge[LR].pb(UD+Not);
            }
            if(i>0 && j>0 && s[i-1][j]=='B' && s[i][j-1]=='B'){
                ll UD=(i-1)*M+j, LR=i*M+(j-1)+N*M, Not=2*N*M;
                edge[UD].pb(LR+Not);
                edge[LR].pb(UD+Not);
            }
            if(i>0 && i<N-1 && s[i-1][j]=='B' && s[i+1][j]=='B'){
                ll UD=(i-1)*M+j, LR=i*M+j, Not=2*N*M;
                edge[UD].pb(LR+Not);
                edge[LR].pb(UD+Not);
            }
            if(j>0 && j<M-1 && s[i][j-1]=='B' && s[i][j+1]=='B'){
                ll UD=i*M+j-1+N*M, LR=i*M+j+N*M, Not=2*N*M;
                edge[UD].pb(LR+Not);
                edge[LR].pb(UD+Not);
            }
        }
    }

    if(Wc!=Bc*2) return false;

    forf(i, 0, n)
        for(ll j : edge[i])
            redge[j].pb(i);

    /*forf(i, 0, n){
        cout << i << ") ";
        for(ll j : edge[i])
            cout << j << " ";
        cout << "\n";
    }*/

    forf(i, 0, n){
        sortv(edge[i]);
        sortv(redge[i]);
    }

    forf(i, 0, n){
        if(!vsd[i]) dfs(i);
    }

    while(vseq.size()){
        ll vt=vseq[vseq.size()-1];
        if(rvsd[vt]){
            vseq.pop_back();
            continue;
        }

        scc.pb({});
        rdfs(vt);
        sortv(scc[scc.size()-1]);
    }

    /*cout << " \n";
    for(vi v : scc){
        for(ll i : v) cout << i <<" ";
        cout <<" / ";
    }
    cout << "\n";*/

    return _2sat();


}

int main(){
    fastio;

    ll t; cin >> t;
    while(t--){
        cout << (solve()?"YES\n":"NO\n");
    }

    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 3712kb

input:

2
3 4
BWW.
WWBW
..WB
3 3
W..
BW.
WBW

output:

YES
NO

result:

ok 2 token(s): yes count is 1, no count is 1

Test #2:

score: 0
Accepted
time: 443ms
memory: 161264kb

input:

70
3 4
BWW.
WWBW
..WB
3 3
W..
BW.
WBW
1 1
B
3 3
...
.W.
...
2 2
W.
BW
2 3
.W.
WBW
1 3
WBW
2 5
.W.W.
WB.BW
2 2
WW
.B
2 2
WB
..
3 3
WWB
BWW
WWB
3 5
.W.WB
WBW.W
...WB
4 5
..W..
.WBW.
WBWBW
.WBW.
3 9
BWW...W..
WWBW..BW.
..WB..WBW
4 12
BWWBWWBWWBWW
WWBWWBWWBWWB
BWWBWWBWWBWW
WWBWWBWWBWWB
7 7
BWWBBWW
WBWWW...

output:

YES
NO
NO
NO
YES
NO
NO
YES
NO
NO
NO
NO
YES
NO
YES
YES
NO
YES
YES
YES
NO
YES
YES
NO
NO
NO
NO
NO
NO
NO
NO
NO
YES
YES
YES
NO
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
NO
YES
YES
YES
NO
YES
NO
YES
NO
NO
NO
NO
NO
YES
NO
YES
YES
NO
NO
NO
NO
NO

result:

ok 70 token(s): yes count is 34, no count is 36