QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#483367#9114. Black or White 2ucup-team1525#WA 460ms24672kbC++205.8kb2024-07-18 16:20:142024-07-18 16:20:14

Judging History

你现在查看的是最新测评结果

  • [2024-07-18 16:20:14]
  • 评测
  • 测评结果:WA
  • 用时:460ms
  • 内存:24672kb
  • [2024-07-18 16:20:14]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
int t,n,m,k;
const int N=1505;
int a[N][N];
// int f[12][100][1<<8],fr[10][100][1<<8];
// int n,m,k;
// int calc(int x,int y)
// {
//     int sum=0;
//     for (int i=0;i<n-1;++i)
//     {
//         int o=(x>>i)&1,p=(x>>(i+1))&1,q=(y>>i)&1,r=(y>>(i+1))&1;
//         if (o+p+q+r==2)
//             ++sum;
//     }
//     return sum;
// }
const int inf=1e9+7;
int f[65][70][1<<9],fr[65][70][1<<9];
int prea[9][9][70][9][9];
void prework(int n,int m,int k,int a[9][9])
{
    for (int i=1;i<=n*m;++i)
        for (int j=0;j<=k;++j)
            for (int l=0;l<(1<<(n+1));++l)
                f[i][j][l]=inf;
    for (int i=0;i<(1<<n);++i)
        if (__builtin_popcount(i)<=k)
            f[n][k-__builtin_popcount(i)][i]=0;
    int mask=(1<<(n+1))-1;
    for (int i=2;i<=m;++i)
    {
        for (int j=1;j<=n;++j)
        {
            for (int l=0;l<=k;++l)
                for (int o=0;o<(1<<(n+1));++o)
                {
                    int id=(i-1)*n+j;
                    if (f[id-1][l][o]<inf)
                    {
                        int tmp=o&1;
                        if ((o>>n)&1) ++tmp;
                        if ((o>>(n-1)&1)) ++tmp;
                        if (j==1)
                            tmp=-1;
                        if (l!=0)
                        {
                            if (f[id][l-1][((o<<1)&mask)|1]>f[id-1][l][o]+(tmp==1?1:0))
                            {
                                f[id][l-1][((o<<1)&mask)|1]=f[id-1][l][o]+(tmp==1?1:0);
                                fr[id][l-1][((o<<1)&mask)|1]=o;
                            }
                        }
                        if (f[id][l][((o<<1)&mask)]>f[id-1][l][o]+(tmp==2?1:0))
                        {
                            f[id][l][((o<<1)&mask)]=f[id-1][l][o]+(tmp==2?1:0);
                            fr[id][l][((o<<1)&mask)]=o;
                        }
                    }
                }
            // for (int l=0;l<=k;++l)
            //     for (int o=0;o<(1<<n);++o)
        }
    }
    int mnp=0;
    for (int i=1;i<(1<<(n+1));++i)
        if (f[n*m][0][i]<f[n*m][0][mnp])
            mnp=i;
    int s=0;
    for (int i=m;i>=2;--i)
        for (int j=n;j>=1;--j)
        {
            a[i][j]=mnp&1;
            mnp=fr[(i-1)*n+j][s][mnp];
            s+=a[i][j];
        }
    for (int i=n;i>=1;--i)
    {
        a[1][i]=mnp&1;
        mnp>>=1;
    }
}
void solve()
{
    scanf("%d%d%d",&n,&m,&k);
    int inv=0,trp=0;
    if (k>n*m/2)
    {
        k=n*m-k;
        inv=1;
    }
    if (n>m)
    {
        swap(n,m);
        trp=1;
    }
    for (int i=1;i<=n;++i)
        for (int j=1;j<=m;++j)
            a[i][j]=0;
    if (m<=8)
    {
        for (int i=1;i<=n;++i)  
            for (int j=1;j<=m;++j)
                a[i][j]=prea[n][m][k][i][j];
    }
    else
    {
        if (k>=2*n)
        {
            int p=(k-(n+1)/2)/n+1;
            int sum=p*n+(n+1)/2;
            if (p==3&&k<3*n)
            {
                k-=(n+1)/2;
                for (int j=1;j<=n;j+=2)
                    a[j][m]=1;
                p=(k-(n+1)/2)/n+1;
                sum=p*n+(n+1)/2;
            }
            for (int i=1;i<=n;++i)
                for (int j=1;j<=p;++j)
                {
                    a[i][j]=1;
                    // if (sum>=k&&i%2==1&&j%2==1)
                    // {
                    //     --sum;
                    //     a[i][j]=0;
                    // }
                }   
            for (int j=1;j<=p;++j)
                for (int i=1;i<=n;++i)
                    if (sum>k&&i%2==1&&j%2==1)
                    {
                        --sum;
                        a[i][j]=0;
                    }
            for (int i=1;i<=n;++i)
                if (i%2==1)
                    a[i][p+1]=1;
        }
        else
        {
            int sum=0;
            for (int i=1;i<=n;++i)
                for (int j=1;j<=m;++j)
                    if (i%2==1&&j%2==1&&k!=sum)
                    {
                        ++sum;
                        a[i][j]=1;
                    }
            // if (sum!=k)
            // {u
            //     int inf=1e9+7;u
                // for (int i=1;i<=um;++i)
                //     for (int j=0;j<=k;++j)
                //         for (int l=0;l<(1<<n);++l)
                //             f[i][j][l]=inf;
                // for (int i=0;i<(1<<n);++i)
                //     if (__builtin_popcount(i)<=k)
                //         f[1][k-__builtin_popcount(i)][i]=0;
            //     for (int i=1;i<m;++i)
            //         for (int j=0;j<=k;++j)
            //             for (int l=0;l<(1<<n);++l)
            //                 if (f[i][j][l]<inf)
            //                     for 
            // }
            // for (int i=1;i<=n;++i)
            //     for (int j=1;j<=m;++j)
            //         if (a[i][j]==0&&k!=0)
            //         {
            //             --k;
            //             a[i][j]=1;
            //         }
        }
    }
    if (inv)
    {
        for (int i=1;i<=n;++i)
            for (int j=1;j<=m;++j)
                a[i][j]=!a[i][j];
    }
    if (trp)
    {
        for (int i=1;i<=m;++i)
        {
            for (int j=1;j<=n;++j)
                printf("%d",a[j][i]);
            puts("");
        }
    }
    else
    {
        for (int i=1;i<=n;++i)
        {
            for (int j=1;j<=m;++j)
                printf("%d",a[i][j]);
            puts("");
        }
    }
}
int main()
{
    for (int i=2;i<=8;++i)
        for (int j=i;j<=8;++j)
            for (int k=0;k<=i*j;++k)
                prework(i,j,k,prea[i][j][k]);
    scanf("%d",&t);
    while (t--)
        solve();
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 311ms
memory: 22464kb

input:

2
2 2 2
2 3 0

output:

10
01
000
000

result:

ok Output is valid. OK.

Test #2:

score: -100
Wrong Answer
time: 460ms
memory: 24672kb

input:

27520
2 2 0
2 2 1
2 2 2
2 2 3
2 2 4
2 3 0
2 3 1
2 3 2
2 3 3
2 3 4
2 3 5
2 3 6
3 2 0
3 2 1
3 2 2
3 2 3
3 2 4
3 2 5
3 2 6
3 3 0
3 3 1
3 3 2
3 3 3
3 3 4
3 3 5
3 3 6
3 3 7
3 3 8
3 3 9
2 4 0
2 4 1
2 4 2
2 4 3
2 4 4
2 4 5
2 4 6
2 4 7
2 4 8
3 4 0
3 4 1
3 4 2
3 4 3
3 4 4
3 4 5
3 4 6
3 4 7
3 4 8
3 4 9
3 4 10...

output:

00
00
10
00
10
01
01
11
11
11
000
000
100
000
100
000
110
100
011
111
011
111
111
111
00
00
00
10
00
00
10
00
00
11
10
00
01
11
11
01
11
11
11
11
11
000
000
000
100
000
000
101
000
000
110
100
000
111
010
000
000
101
111
001
011
111
010
111
111
011
111
111
111
111
111
0000
0000
1000
0000
1000
0000
1...

result:

wrong answer The number of black cell is not K