Codeforces Round #283 (Div. 2)CDE题解

Posted 2022-10-14 ccsu-kid

C. Removing Columns

https://codeforc.es/contest/496/problem/C

从前往后标记那些前面已经确定字典序合法的行，对于字典序不合法又没被标记的直接删除该列。

 1 #define bug(x) cout<<#x<<" is "<<x<<endl
 2 #define IO std::ios::sync_with_stdio(0)
 3 #include <bits/stdc++.h>
 4 #define iter ::iterator
 5 #define pa pair<int,int>
 6 using namespace  std;
 7 #define ll long long
 8 #define mk make_pair
 9 #define pb push_back
10 #define se second
11 #define fi first
12 #define ls o<<1
13 #define rs o<<1|1
14 ll mod=998244353;
15 const int N=1e2+5;
16 int n,m;
17 char s[N][N];
18 int vis[N];
19 int main()
20     IO;
21     cin>>n>>m;
22     for(int i=1;i<=n;i++)
23         cin>>s[i]+1;
24     
25     int ans=0;
26     for(int i=1;i<=m;i++)
27         int f=0;
28         for(int j=2;j<=n;j++)
29             if(s[j][i]<s[j-1][i]&&!vis[j])f=1;
30         
31         if(f)ans++;
32         else
33             for(int j=2;j<=n;j++)
34                 if(s[j][i]>s[j-1][i])vis[j]=1;
35             
36         
37  
38     
39     cout<<ans<<endl;
40 

D. Tennis Game

https://codeforc.es/contest/496/problem/D

预处理各自胜场数对应的编号，枚举t，从前往后判断每轮是由谁赢下，

结合最后一个场是谁赢的谁就要赢判断是否合法就做完了。

 1 #define bug(x) cout<<#x<<" is "<<x<<endl
 2 #define IO std::ios::sync_with_stdio(0)
 3 #include <bits/stdc++.h>
 4 #define iter ::iterator
 5 #define pa pair<int,int>
 6 using namespace  std;
 7 #define ll long long
 8 #define mk make_pair
 9 #define pb push_back
10 #define se second
11 #define fi first
12 #define ls o<<1
13 #define rs o<<1|1
14 ll mod=998244353;
15 const int N=1e5+5;
16 int n;
17 int aid[N],bid[N];
18 int as[N],bs[N];
19 vector<pa>ans;
20 int main()
21     IO;
22     cin>>n;
23     int h1=0,h2=0;
24     for(int i=1;i<=n;i++)
25         int x;
26         cin>>x;
27         if(x==1)
28             h1++;
29             aid[h1]=i;
30         
31         else
32             h2++;
33             bid[h2]=i;
34         
35         as[i]=h1;
36         bs[i]=h2;
37     
38     for(int t=1;t<=n;t++)
39         int p=0,w1=0,w2=0,w3=0;
40         while(p<n)
41             int c1=as[p],c2=bs[p];
42             if(c1+t>h1&&c2+t>h2)break;
43             if(c1+t<=h1&&c2+t<=h2)
44                 if(aid[c1+t]<bid[c2+t])
45                     w1++;
46                     w3=1;
47                     p=aid[c1+t];
48                 
49                 else
50                     w2++;
51                     w3=2;
52                     p=bid[c2+t];
53                 
54             
55             else if(c1+t<=h1)
56                 w1++;
57                 w3=1;
58                 p=aid[c1+t];
59             
60             else
61                 w2++;
62                 w3=2;
63                 p=bid[c2+t];
64             
65         
66         if(p==n)
67             if(w3==1&&w1>w2)ans.pb(mk(w1,t));
68             if(w3==2&&w2>w1)ans.pb(mk(w2,t));
69         
70     
71     sort(ans.begin(),ans.end());
72     cout<<ans.size()<<endl;;
73     for(auto tmp :ans)
74         cout<<tmp.fi<<" "<<tmp.se<<endl;
75     
76 

E. Distributing Parts

https://codeforc.es/contest/496/problem/E

贪心。

把每个演员分配给音乐，在满足演员的r2>=音乐的r1的情况下，取l1尽量小的音乐。

用set和struct搞一搞很方便，注意struct里的x和id都必须安排顺序，否则insert时会失败。

 1 #define bug(x) cout<<#x<<" is "<<x<<endl
 2 #define IO std::ios::sync_with_stdio(0)
 3 #include <bits/stdc++.h>
 4 #define iter ::iterator
 5 #define pa pair<int,int>
 6 using namespace  std;
 7 #define ll long long
 8 #define mk make_pair
 9 #define pb push_back
10 #define se second
11 #define fi first
12 #define ls o<<1
13 #define rs o<<1|1
14 ll mod=998244353;
15 const int N=1e5+5;
16 int n,m;
17 struct node
18     int l,r,k,id;
19     bool operator <(const node &t)const
20         return r<t.r;
21     
22 a[N],b[N];
23 bool cmp(node n1,node n2)
24     return n1.r<n2.r;
25 
26 struct point
27     int x,id;
28     point(int x,int id):x(x),id(id)
29     bool operator <(const point &t)const
30         if(x==t.x)return id<t.id;
31         return x<t.x;
32     
33 ;
34 set<point>s;
35 int d[N];
36 int main()
37     IO;
38     cin>>n;
39     for(int i=1;i<=n;i++)
40         cin>>a[i].l>>a[i].r;
41         a[i].id=i;
42     
43     cin>>m;
44     for(int i=1;i<=m;i++)
45         cin>>b[i].l>>b[i].r>>b[i].k;
46         b[i].id=i;
47     
48     sort(a+1,a+1+n);
49     sort(b+1,b+1+m);
50     int ans=0;
51     int j=1;
52     for(int i=1;i<=m;i++)
53         while(j<=n&&a[j].r<=b[i].r)
54             s.insert(point(a[j].l,a[j].id));
55             j++;
56         
57         set<point>iter it;
58         for(int k=1;k<=b[i].k;k++)
59             it=s.lower_bound(point(b[i].l,-1));
60             if(it==s.end())break;
61             point tmp=*it;
62             d[tmp.id]=b[i].id;
63             s.erase(tmp);
64             ans++;
65         
66     
67     if(ans<n)cout<<"NO"<<endl;
68     else
69         cout<<"YES"<<endl;
70         for(int i=1;i<=n;i++)
71             cout<<d[i]<<" ";
72         
73     
74 
75 /*
76 3
77 1 2
78 1 2
79 1 2
80 2
81 1 2 1
82 1 2 2
83 */