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QOJ

IDSubmission IDProblemHackerOwnerResultSubmit timeJudge time
#945#394046#7743. Grand Finale0000pncmarherFailed.2024-10-10 14:46:082024-10-10 14:46:09

Details

Extra Test:

Invalid Input

input:

250
100 100
BWBWWWWBBBWWWWBWBWWWWWBWWBWWBWWBWBWBBWWWWWWWWBBBWWWWBWWWBWBWWBWBBWBWBWWWWWWWBWBWWWWBBBWBWWWBBWWWWBWB
GWWWWWWBWWQBQWBWWWBBWBBWBWQWWBWQWWBBWWWQQQWQBWBWWWQQWQWWBWWBBWBWBBWWWQWWQWWWBWWWQBWWQWWBBWQBBWWQWQQB
100 100
WBWWWWBWBWWBBBBWBWWBWBWBBWWBWBWWWWBWWWWWWWWWBWBBWWWWWWBWWBWWWWBBWBWBBWWWBWBBWW...

output:


result:

FAIL Token parameter [name=sp] equals to "GWWWWWWBWWQBQWBWWWBBWBBWBWQWWB...QWWQWWWBWWWQBWWQWWBBWQBBWWQWQQB", doesn't correspond to pattern "[BQW]{1,2500}" (test case 1, stdin, line 4)

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#394046#7743. Grand FinalemarherAC ✓422ms105584kbC++171.5kb2024-04-19 21:37:002024-04-19 21:37:01

answer

#include<bits/stdc++.h>
using namespace std;
const int N=5e3+50,inf=1e9;

int n,m,f[N][N],g[N][N],T,a[N],b[N],w1[N],w2[N],fir[3][N],t1,t2;
char s[N],t[N];

void clear()
{
    for(int i=0;i<=m;i++)
    for(int j=0;j<=n+m;j++)f[i][j]=g[i][j]=0;
    for(int i=0;i<=m;i++)w1[i]=w2[i]=b[i]=0;
    t1=t2=0;
}

int chk(int k)
{
    for(int t=0;t<=m;t++)
    {
        int c1=t1+fir[1][t]-(t-2*(k-n));
        int c2=t2+fir[2][t]-(k-n);
        if(c1<0||c2<0||!g[t][c2]||f[t][c2]>c1)continue;
        return 1;
    }
    return 0;
}

int F(int i,int j)
{
    if(j>n+m||j<0)return inf;
    return f[i][j];
}

void sol()
{
    cin>>n>>m>>(s+1)>>(t+1);
    for(int i=1;i<=n;i++)a[i]=s[i]=='B'?2:(s[i]=='Q'?1:0),t1+=a[i]==1,t2+=a[i]==2;
    for(int i=1;i<=m;i++)b[i]=t[i]=='B'?2:(t[i]=='Q'?1:0),w1[i]=b[i]==1,w2[i]=b[i]==2,fir[1][i]=fir[1][i-1]+w1[i],fir[2][i]=fir[2][i-1]+w2[i];
    for(int i=m-1;i>=0;i--)
    {
        int f1=w1[i+1],f2=w2[i+1];
        for(int j=0;j<=n+m;j++)f[i][j]=min(max(1,F(i+1,j+f2)+1-f1),j?max(0,F(i+2,j-1+f2)-f1):inf);
    }
    g[0][t2]=1;
    for(int i=0;i<m;i++)for(int j=0;j<=t2+fir[2][i];j++)if(g[i][j])
    {
        int c2=j,c1=t1+fir[1][i]-(i-2*(t2+fir[2][i]-j));
        if(c2>0)g[min(m,i+2)][j-1+w2[i+1]+w2[i+2]]=1;if(c1>0)g[i+1][j+w2[i+1]]=1;
    }
    for(int k=n;k<=n+m;k++)if(chk(k)){cout<<k<<'\n';return;}
    puts("IMPOSSIBLE");
}

main()
{
    // freopen("in.txt","r",stdin);
    cin>>T;
    while(T--)sol(),clear();
}