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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
#115225#5810. Doubly-sorted GridZesty_Fox30 ✓9762ms173000kbC++202.0kb2023-06-25 10:05:042023-06-25 10:05:07

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-06-25 10:05:07]
  • 评测
  • 测评结果:30
  • 用时:9762ms
  • 内存:173000kb
  • [2023-06-25 10:05:04]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

using u32=unsigned int;
using i64=long long;
using u64=unsigned long long;
using db=double;
using vi=vector<int>;
using pii=pair<int,int>;

template<typename T>
T read(){
    T x=0,f=0;char ch=getchar();
    while(!isdigit(ch)) f|=(ch=='-'),ch=getchar();
    while(isdigit(ch)) x=x*10+(ch-'0'),ch=getchar();
    return f?-x:x;
}

#define rdi read<int>
#define rdi64 read<i64>
#define fi first
#define se second
#define pb push_back

const int N=10,N2=N*2,C=26,MOD=10007;

int popc[1<<N2],kth[1<<N2][N2];
void init(int mx){
    popc[0]=0;
    for(int i=1;i<(1<<mx);i++){
        popc[i]=popc[i>>1]+(i&1);
        kth[i][0]=__builtin_ctz(i);
        for(int j=1;j<popc[i];j++)
            kth[i][j]=kth[i&(i-1)][j-1];
    }
}

template<typename T>
void inc(T &a,T b){
    if((a+=b)>=MOD) a-=MOD;
}

int n,m;
char a[N][N2];
int f[2][N][1<<N2];
void solve(int tid){
    n=rdi(),m=rdi();
    for(int i=0;i<n;i++)
        scanf("%s",a[i]);

    vi st;
    for(int i=0;i<(1<<(n+m));i++)
        if(popc[i]==n) st.pb(i);
    memset(f,0,sizeof(f));
    f[0][0][((1<<n)-1)<<m]=1;
    for(int i=0,nw=0;i<C;i++,nw^=1){
        for(int j=0;j<n;j++){
            for(auto s:st){
                int v=f[nw][j][s]%MOD;
                if(!v) continue;
                int p=kth[s][j],mi=(j?kth[s][j-1]+1:0);
                for(int k=p,fl=1;k>=mi&&fl;k--){
                    int n_s=s^(1<<p)^(1<<k);
                    if(j!=n-1) f[nw][j+1][n_s]+=v;
                    else f[nw^1][0][n_s]+=v;
                    char ch=a[j][m-(k-j)];
                    fl&=(ch=='.'||ch==i+'a');
                }
                f[nw][j][s]=0;
            }
        }
    }
    int ans=f[C&1][0][(1<<n)-1]%MOD;
    printf("Case #%d: %d\n",tid,ans);
}

int main(){
#ifdef LOCAL
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif
    init(N2);
    int T=rdi();
    for(int i=1;i<=T;i++) solve(i);
    return 0;
}

详细

Subtask #1:

score: 10
Accepted

Test #1:

score: 10
Accepted
time: 241ms
memory: 171512kb

input:

40
4 4
....
....
....
....
4 4
abcd
bcde
cdef
defg
4 4
bbcc
ccde
cdef
deef
4 4
....
....
.r..
..e.
3 4
....
....
...g
1 3
f..
2 2
.b
bb
2 3
...
..f
4 4
....
....
..t.
..d.
3 3
.gn
...
...
4 4
...a
...b
...c
...d
4 4
...b
..bb
.bbb
bbbb
4 4
a...
.b..
..c.
...d
3 3
c..
glw
..w
1 4
....
1 3
d.v
4 2
..
...

output:

Case #1: 7285
Case #2: 1
Case #3: 1
Case #4: 0
Case #5: 718
Case #6: 231
Case #7: 2
Case #8: 686
Case #9: 0
Case #10: 3500
Case #11: 330
Case #12: 14
Case #13: 4096
Case #14: 2756
Case #15: 3737
Case #16: 19
Case #17: 8175
Case #18: 3737
Case #19: 7279
Case #20: 256
Case #21: 23
Case #22: 7569
Case ...

result:

ok 40 lines

Subtask #2:

score: 20
Accepted

Test #2:

score: 20
Accepted
time: 9762ms
memory: 173000kb

input:

40
5 10
..........
..........
..........
..........
..........
10 10
....a.....
....c.....
....g.....
....j.....
....l.....
....m.....
....n.....
....q.....
....t.....
....v.....
10 10
..........
..........
..........
..........
..........
..........
..........
..........
...ok.....
..........
9 10
...

output:

Case #1: 8791
Case #2: 7220
Case #3: 0
Case #4: 2510
Case #5: 4299
Case #6: 1663
Case #7: 981
Case #8: 1430
Case #9: 8814
Case #10: 2877
Case #11: 2124
Case #12: 0
Case #13: 4311
Case #14: 4458
Case #15: 8699
Case #16: 3727
Case #17: 14
Case #18: 5478
Case #19: 7922
Case #20: 6789
Case #21: 1252
Cas...

result:

ok 40 lines

Extra Test:

score: 0
Extra Test Passed