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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#419062#6534. Peg SolitaireXiaoretaW#AC ✓1ms3732kbC++202.0kb2024-05-23 17:25:462024-05-23 17:25:47

Judging History

你现在查看的是最新测评结果

  • [2024-05-23 17:25:47]
  • 评测
  • 测评结果:AC
  • 用时:1ms
  • 内存:3732kb
  • [2024-05-23 17:25:46]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define sz(a) ((int)a.size())
#define all(a) a.begin(), a.end()
#define rep(i, l, r) for (int i = l; i < r; ++i)
#define per(i, r, l) for (int i = r-1; i >= l; --i)
typedef long long ll;
typedef pair<int, int> PI;
template<typename T> bool setmax(T &a, T b) { return (a < b ? a = b, 1 : 0); }
template<typename T> bool setmin(T &a, T b) { return (a > b ? a = b, 1 : 0); }

int dir[4][2]={
    {1,0},{-1,0},{0,1},{0,-1}
};


void dfs(vector<vector<bool>>& broad,int& ans){
    bool f=1;
    int n=broad.size();
    int m=broad[0].size();
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            if(broad[i][j]==0)continue;
            for(int d=0;d<4;d++){
                int x1=i+dir[d][0];
                int x2=i+2*dir[d][0];
                int y1=j+dir[d][1];
                int y2=j+2*dir[d][1];
                if(x1>=0&&x1<n&&x2>=0&&x2<n&&y1>=0&y1<m&&y2>=0&&y2<m&&broad[x1][y1]==1&&broad[x2][y2]==0){
                    f=0;
                    broad[i][j]=0;
                    broad[x1][y1]=0;
                    broad[x2][y2]=1;
                    dfs(broad,ans);
                    broad[i][j]=1;
                    broad[x1][y1]=1;
                    broad[x2][y2]=0;
                }
            }
        }
    }
    if(f==1){
        int cnt=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(broad[i][j])cnt++;
            }
        }
        ans=min(ans,cnt);
    }
}

void solve() {
    int n,m,k;
    cin>>n>>m>>k;
    vector<vector<bool>> broad;
    broad.resize(n,vector<bool>(m,0));
    for(int i=0;i<k;i++){
        int x,y;
        cin>>x>>y;
        broad[x-1][y-1]=1;
    }
    int ans=36;
    dfs(broad,ans);
    cout<<ans<<'\n';
}
int main() {
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    int tt; cin >> tt;
    while (tt--) solve();

    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3488kb

input:

3
3 4 5
2 2
1 2
1 4
3 4
1 1
1 3 3
1 1
1 2
1 3
2 1 1
2 1

output:

2
3
1

result:

ok 3 number(s): "2 3 1"

Test #2:

score: 0
Accepted
time: 0ms
memory: 3484kb

input:

20
2 1 2
1 1
2 1
5 1 3
3 1
2 1
4 1
3 3 6
1 2
2 2
1 1
2 3
3 1
3 2
4 4 4
2 3
3 1
3 2
1 2
1 1 1
1 1
5 2 6
3 2
4 1
2 1
5 2
2 2
5 1
1 3 1
1 2
1 5 1
1 5
4 6 5
4 6
4 4
2 3
4 3
1 6
6 6 3
2 4
1 3
2 1
2 2 2
2 1
1 1
5 3 4
2 2
5 1
4 3
3 2
6 5 6
5 5
6 5
2 4
2 1
3 4
1 4
2 6 5
1 6
2 1
1 4
2 3
1 3
3 5 6
2 1
3 3
1 5...

output:

2
2
3
1
1
2
1
1
3
3
2
1
3
3
2
1
3
1
2
2

result:

ok 20 numbers

Test #3:

score: 0
Accepted
time: 0ms
memory: 3492kb

input:

20
2 1 1
2 1
4 3 2
4 3
1 1
6 4 6
4 3
5 4
4 2
3 2
3 4
2 3
3 6 4
3 1
2 6
2 4
3 5
6 6 6
3 3
3 6
4 2
3 2
4 1
1 4
3 3 1
1 1
2 3 2
2 2
2 1
3 2 2
1 2
2 2
3 4 5
2 1
3 4
3 1
2 2
3 2
2 5 5
1 5
2 3
2 5
1 4
2 2
5 6 6
5 1
2 2
1 4
5 3
4 4
1 1
2 3 4
1 1
1 2
2 2
2 1
2 3 4
1 1
2 1
1 2
2 2
6 3 6
4 2
4 1
1 3
6 2
3 3
2...

output:

1
2
2
4
4
1
1
1
2
2
6
2
2
3
4
1
1
1
3
2

result:

ok 20 numbers

Test #4:

score: 0
Accepted
time: 0ms
memory: 3732kb

input:

20
1 5 3
1 2
1 3
1 1
5 6 6
5 4
2 4
4 4
4 6
2 5
3 3
5 2 4
4 1
3 1
4 2
5 2
6 3 6
3 1
5 2
3 2
4 1
4 3
3 3
5 3 5
5 2
2 2
3 1
4 3
3 2
6 1 4
3 1
5 1
1 1
4 1
3 5 6
1 1
2 1
3 4
1 5
2 2
3 3
2 5 2
2 5
2 1
5 4 6
1 3
4 3
2 2
1 2
3 3
2 3
2 3 3
2 3
1 1
1 3
1 1 1
1 1
1 1 1
1 1
1 1 1
1 1
6 4 5
2 1
6 2
3 2
1 2
4 2
5...

output:

2
1
2
2
1
2
3
2
2
3
1
1
1
3
2
2
2
3
2
1

result:

ok 20 numbers

Test #5:

score: 0
Accepted
time: 1ms
memory: 3540kb

input:

20
6 6 6
2 4
3 4
3 3
4 3
4 2
5 2
6 6 6
2 4
3 4
3 3
4 3
4 2
5 2
6 6 6
2 4
3 4
3 3
4 3
4 2
5 2
6 6 6
2 4
3 4
3 3
4 3
4 2
5 2
6 6 6
2 4
3 4
3 3
4 3
4 2
5 2
6 6 6
2 4
3 4
3 3
4 3
4 2
5 2
6 6 6
2 4
3 4
3 3
4 3
4 2
5 2
6 6 6
2 4
3 4
3 3
4 3
4 2
5 2
6 6 6
2 4
3 4
3 3
4 3
4 2
5 2
6 6 6
2 4
3 4
3 3
4 3
4 2
5...

output:

2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2

result:

ok 20 numbers