CF补题记录

Posted on Edited on

此文章用于记录codeforces正赛/VP时未AC的题目

1
2
3
#unfixed
https://codeforces.com/contest/2032/problem/D
https://codeforces.com/contest/2036/problem/B

STL_using

Sort

div3B.Startup

被sort坑惨了，距离AC只差一个等号

[STL]为什么sort的自定义cmp函数中必须使用严格弱序(strict weak order)

赛时代码和这份代码只差一个 =

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
#include<cstdio>
#include<algorithm>
// #define int long long
#pragma GCC optimize(2)

using namespace std;

inline int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0' && ch<='9')
        x=x*10+ch-'0',ch=getchar();
    return x*f;
}

const int N = 2e5+10;
int n , k;
struct B
{
   int bin , c;
}a[N];

bool cmp( B x , B y ){
    return x.c > y.c;
}

signed main()
{
    int t = read();
    int b , bb;
    while( t -- ){
        n = read() , k = read();
        int cnt = 0;
        int maxn = 0;
        for(int i = 1;i <= k;i ++){
            b = read() , bb = read();
            if( a[b].bin == 0 ) cnt ++;
            a[b].bin ++;
            a[b].c += bb;
            maxn = max( maxn , b );
        }
        sort( a + 1 , a + maxn + 1 , cmp );
        int ans = 0 , minn = min( cnt , n );
        for(int i = 1;i <= minn;i ++){
            ans += a[i].c;
        }
        printf("%d\n",ans);
        for(int i = 1;i <= cnt;i ++){
            a[i].bin = a[i].c = 0;
        }
    }
    return 0;
}

Set

div2D. QED's Favorite Permutation

使用一个差异数组 来记录，点 是否需要移动

如果出现 s[x] == 'L' && s[x+1] == 'R' && diff[x]，则这个节点会导致输出 NO

使用set来维护所有不能从 点 移动到点 的点，如果为空则输出 YES

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
#include<iostream>
#include<vector>
#include<set>
#define int long long

using namespace std;

signed main(){
    int t;cin >> t;while( t -- ){
        int n , q;cin >> n >> q;
        vector<int> p(n);
        for(int i = 0;i < n;i ++){cin >> p[i];p[i] --;}
        vector<int> diff(n);//用于记录差异数组，不等于0则表示需要移动
        for(int i = 0;i < n;i ++){
            diff[min(i,p[i])] ++;
            diff[max(i,p[i])] --;
        }
        for(int i = 1;i < n;i ++) diff[i] += diff[i-1];
        string s;cin >> s;
        set<int> bad;//用于记录坏节点，即因为这个点则无法YES的
        for(int i = 0;i < n;i ++)
            if( s[i] == 'L' && s[i+1] == 'R' && diff[i] )
                bad.insert(i);
        while( q -- ){
            int x;cin >> x;x --;//因为从0开始，所以-1
            if( s[x] == 'L' ) s[x] = 'R';//按照题目要求，翻转
            else s[x] = 'L';
            if( s[x-1] == 'L' && s[x] == 'R' && diff[x-1] )
                bad.insert( x - 1 );
            else bad.erase( x - 1 );
            if( s[x] == 'L' && s[x+1] == 'R' && diff[ x ] )
                bad.insert( x );
            else bad.erase( x );
            if( bad.size() )
                cout << "NO" << endl;
            else
                cout << "YES" << endl;
        }
    }
    return 0;
}

MISC

Misunderstand

div2B. Medians

误认为是数组必须等长划分导致错误

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
#include<iostream>
#define int long long
using namespace std;

int n , k;

signed main()
{
    ios::sync_with_stdio( false );
    int t;cin >> t;
    while( t -- ){
        cin >> n >> k;
        if( n == 1 ){
            cout << "1\n1\n";
            continue;
        }
        if( k == 1 || k == n ){
            cout << "-1\n";
            continue;
        }
        if( k % 2 ){
            cout << 3 << '\n';
            cout << "1 " << k-1 << " " << k+2 << '\n';
        }
        else{
            cout << 3 << '\n';
            cout << "1 " << k << " " << k+1 << '\n';
        }
    }
    return 0;
}