贪心总结 ###K

总结acwing上的贪心问题

记下结论即可

左端点升序应该对应右端点降序左端点降序对应右端点升序

所以只需要猜左端点/ 右端点升序即可而不用猜四种

1.区间选点问题，选最少的点使得每个区间至少包含一个点

思路：右端点升序排序，没有被覆盖的右端点全部选上

题目链接：https://www.acwing.com/problem/content/907/

题目链接：https://www.acwing.com/problem/content/114/ 可以将问题转换成区间选点问题，只需要考虑每个岛能被覆盖的线段是那一段即可

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e5+10;
 4 const int mod=1e9+7;
 5 #define ll long long
 6 #define pi pair<ll,ll>
 7 #define fi first
 8 #define sc second
 9 #define pb push_back
10 pi a[maxn];
11 
12 bool cmp(pi a,pi b)
13 {
14     return a.sc<b.sc;
15 }
16 
17 
18 int main()
19 {
20     ios::sync_with_stdio(0);
21     cin.tie(0);
22     int n;
23     cin>>n;
24     for(int i=1;i<=n;i++) cin>>a[i].fi>>a[i].sc;
25     sort(a+1,a+1+n,cmp);
26     int now=a[1].sc;
27     int ans=1;
28     for(int i=2;i<=n;i++)
29     {
30         if(a[i].fi<=now&&now<=a[i].sc) continue;
31         ans++;
32         now=a[i].sc;
33     }
34     cout<<ans<<'\n';
35 
36 
37 
38 
39 }

View Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e3+10;
 4 const int mod=1e9+7;
 5 const double eps=1e-9;
 6 #define ll long long
 7 #define pi pair<double,double>
 8 #define fi first
 9 #define sc second
10 #define pb push_back
11 
12 struct ac
13 {
14     double l,r;
15     bool operator<(ac a)
16     {
17         return r<a.r;
18     }
19 };
20 ac a[maxn];
21 
22 
23 
24 int main()
25 {
26     ios::sync_with_stdio(0);
27     cin.tie(0);
28     int n,d;
29     cin>>n>>d;
30     int f=0;
31     int tot=0;
32     for(int i=1;i<=n;i++)
33     {
34         int x,y;
35         cin>>x>>y;
36         if(y>d) f=1;
37         double t=sqrt((double)d*d-y*y);
38         double l=-x-t,r=-x+t;
39         a[++tot]={l,r};
40     }
41     if(f)
42     {
43         cout<<-1<<'\n';
44         return 0;
45     }
46     sort(a+1,a+1+n);
47     int ans=1;
48     double now=a[1].r;
49     for(int i=2;i<=n;i++)
50     {
51         if(a[i].l<now+eps) continue;
52         ans++;
53         now=a[i].r;
54     }
55     cout<<ans<<'\n';
56 
57 
58 
59 
60 
61 }

View Code

或者左端点降序排序一直选左端点即可

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define pi pair<int,int>
 5 #define pb push_back
 6 #define fi first
 7 #define sc second
 8 #define ull unsigned long long
 9 const int maxn=1e5+10;
10 const int mod=1e9+7;
11 
12 pi a[maxn];
13 
14 
15 int main()
16 {
17     ios::sync_with_stdio(0);
18     cin.tie(0);
19     int n;
20     cin>>n;
21     for(int i=1;i<=n;i++)
22     {
23         cin>>a[i].fi>>a[i].sc;
24     }
25     sort(a+1,a+1+n);
26     reverse(a+1,a+1+n);
27     int ans=0;
28     int now=2e9;
29     for(int i=1;i<=n;i++)
30     {
31         if(now>=a[i].fi&&now<=a[i].sc) continue;
32         ans++;
33         now=a[i].fi;
34     }
35     cout<<ans<<'\n';
36 
37 
38 
39 
40 
41 }

View Code

2.最大不相交区间数量，问如何选择尽量多的区间，使得区间没有交集

思路：右端点升序排序，当前的区间如果左端点大于右端点那么就选上，更新now （右端点）

题目链接：https://www.acwing.com/problem/content/910/

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e5+10;
 4 const int mod=1e9+7;
 5 #define ll long long
 6 #define pi pair<int,int>
 7 #define fi first
 8 #define sc second
 9 #define pb push_back
10 pi a[maxn];
11 
12 bool cmp(pi a,pi b)
13 {
14     return a.sc<b.sc;
15 }
16 
17 
18 int main()
19 {
20     ios::sync_with_stdio(0);
21     cin.tie(0);
22     int n;
23     cin>>n;
24     for(int i=1;i<=n;i++) cin>>a[i].fi>>a[i].sc;
25     sort(a+1,a+1+n,cmp);
26     int now=a[1].sc;
27     int ans=1;
28     for(int i=2;i<=n;i++)
29     {
30         if(a[i].fi<=now) continue;
31         now=a[i].sc;
32         ans++;
33     }
34     cout<<ans<<'\n';
35 
36 
37 
38 
39 }

View Code

或者左端点降序排序

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define pi pair<int,int>
 5 #define pb push_back
 6 #define fi first
 7 #define sc second
 8 #define ull unsigned long long
 9 const int maxn=1e5+10;
10 const int mod=1e9+7;
11 
12 pi a[maxn];
13 
14 
15 
16 int main()
17 {
18     ios::sync_with_stdio(0);
19     cin.tie(0);
20     int n;
21     cin>>n;
22     for(int i=1;i<=n;i++) cin>>a[i].fi>>a[i].sc;
23     sort(a+1,a+1+n);
24     reverse(a+1,a+1+n);
25     int now=2e9;
26     int ans=0;
27     for(int i=1;i<=n;i++)
28     {
29         int l=a[i].fi,r=a[i].sc;
30         if(r<now) now=l,ans++;
31     }
32     cout<<ans<<'\n';
33 
34 
35 
36 
37 
38 
39 }

View Code

3.区间分组问题，问如何将每个区间分组，使得每组区间内部两两没有交集

思路:以左端点升序排序，然后用小根堆维护每次区间的右端点的值，当前区间的左端点大于某一组中的最小值即可以划分进这一组

用小根堆维护最小值，每次可以放进新的组就删除再插入，否则只有插入同理右端点降序排序也可以

题目链接：https://www.acwing.com/problem/content/description/908/

题目链接：https://www.acwing.com/problem/content/113/ 一样区间分组不过多一个给每个区间编号

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e5+10;
 4 const int mod=1e9+7;
 5 #define ll long long
 6 #define pi pair<int,int>
 7 #define fi first
 8 #define sc second
 9 #define pb push_back
10 struct ac
11 {
12     int l,r;
13     bool operator<(ac a)
14     {
15         return l<a.l;
16     }
17 };
18 ac a[maxn];
19 
20 
21 
22 int main()
23 {
24     ios::sync_with_stdio(0);
25     cin.tie(0);
26     int n;
27     cin>>n;
28     for(int i=1;i<=n;i++) cin>>a[i].l>>a[i].r;
29     sort(a+1,a+1+n);
30     int ans=0;
31     priority_queue<pi>q;
32     for(int i=1;i<=n;i++)
33     {
34         int u=a[i].l;
35         if(!q.empty()&&q.top().sc<u)
36         {
37             q.pop();
38             q.push({-a[i].r,a[i].r});
39             continue;
40         }
41         q.push({-a[i].r,a[i].r});
42         ans++;
43     }
44     cout<<ans<<'\n';
45 
46 
47 
48 
49 
50 }

View Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=5e4+10;
 4 const int mod=1e9+7;
 5 #define ll long long
 6 #define pi pair<int,int>
 7 #define fi first
 8 #define sc second
 9 #define pb push_back
10 struct ac
11 {
12     int v,p;
13     bool operator<(ac a)const
14     {
15         return v>a.v;
16     }
17 };
18 int ans[maxn];
19 
20 struct bc
21 {
22     int l,r,p;
23     bool operator<(bc a)
24     {
25         return l<a.l;
26     }
27 };
28 bc a[maxn];
29 
30 
31 
32 int main()
33 {
34     ios::sync_with_stdio(0);
35     cin.tie(0);
36     int n;
37     cin>>n;
38     priority_queue<ac>q;
39     for(int i=1;i<=n;i++) cin>>a[i].l>>a[i].r,a[i].p=i;
40     sort(a+1,a+1+n);
41     int tot=0;
42     for(int i=1;i<=n;i++)
43     {
44         int id=a[i].p;
45         if(!q.empty()&&q.top().v<a[i].l)
46         {
47             ac u=q.top();
48             q.pop();
49             ans[id]=u.p;
50             q.push({a[i].r,u.p});
51         }
52         else
53         {
54             q.push({a[i].r,++tot});
55             ans[id]=tot;
56         }
57     }
58     cout<<tot<<'\n';
59     for(int i=1;i<=n;i++)
60     {
61         cout<<ans[i]<<'\n';
62     }
63 
64 
65 
66 
67 }

View Code

4.区间覆盖问题给定一段目标区间，给定n个区间，从n个区间中选最少的区间覆盖掉目标区间

思路：左端点升序排序每次找包含当前st 的区间的右端点最大的，然后更新st为这个最大值继续找

题目链接：https://www.acwing.com/problem/content/909/

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define pi pair<int,int>
 5 #define pb push_back
 6 #define fi first
 7 #define sc second
 8 #define ull unsigned long long
 9 const int maxn=1e5+10;
10 const int mod=5000011;
11 
12 
13 pi a[maxn];
14 int n;
15 int s,t;
16 
17 
18 
19 int main()
20 {
21     ios::sync_with_stdio(0);
22     cin.tie(0);
23     cin>>s>>t;
24     cin>>n;
25     for(int i=1;i<=n;i++) cin>>a[i].fi>>a[i].sc;
26     sort(a+1,a+1+n);
27     int now=s;
28     int ans=0;
29     for(int i=1;i<=n;i++)
30     {
31         int j=i;
32         int tmp=-2e9;
33         ans++;
34         while(j<=n&&a[j].fi<=now)
35         {
36             tmp=max(tmp,a[j].sc);
37             j++;
38         }
39         if(tmp<=now)
40         {
41             ans=-1;
42             break;
43         }
44         now=tmp;
45         if(now>=t) break;
46         i=j-1;
47     }
48     if(now<t) ans=-1;
49     cout<<ans<<'\n';
50 
51 
52 
53 }

View Code

5. 所有数到中位数的距离之和最小

题目链接：https://www.acwing.com/problem/content/106/

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e5+10;
 4 const int mod=1e9+7;
 5 const double eps=1e-9;
 6 #define ll long long
 7 #define pi pair<int,int>
 8 #define fi first
 9 #define sc second
10 #define pb push_back
11 int a[maxn];
12 
13 
14 
15 
16 int main()
17 {
18     ios::sync_with_stdio(0);
19     cin.tie(0);
20     int n;
21     cin>>n;
22     for(int i=1;i<=n;i++) cin>>a[i];
23     sort(a+1,a+1+n);
24     ll ans=0;
25     int x=a[(n+1)/2];
26     for(int i=1;i<=n;i++) ans+=abs(x-a[i]);
27     cout<<ans<<'\n';
28 
29 
30 
31 }

View Code

6 耍杂技的牛特别一点的贪心，根据两个值的排序来贪心类似题是国王游戏

题目链接：https://www.acwing.com/problem/content/127/

考虑 1 2 放和2 1 这样放我们想让 1 2 更小 w2-s1<w1-s2 整理得 s1+w2>s2+w2 于是猜测按照这样排序即可

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e5+10;
 4 const int mod=1e9+7;
 5 const double eps=1e-9;
 6 #define ll long long
 7 #define pi pair<int,int>
 8 #define fi first
 9 #define sc second
10 #define pb push_back
11 
12 struct ac
13 {
14     int w,s;
15     bool operator<(ac a)
16     {
17         return w+s>a.w+a.s;
18     }
19 };
20 ac a[maxn];
21 
22 
23 int main()
24 {
25     ios::sync_with_stdio(0);
26     cin.tie(0);
27     int n;
28     cin>>n;
29     for(int i=1;i<=n;i++) cin>>a[i].w>>a[i].s;
30     sort(a+1,a+1+n);
31     ll ans=-1e18;
32     ll sum=0;
33     for(int i=n;i>=1;i--)
34     {
35         ans=max(ans,sum-a[i].s);
36         sum+=a[i].w;
37     }
38     cout<<ans<<'\n';
39 
40 
41 
42 }

View Code

7.题目链接：https://www.acwing.com/problem/content/description/112/

思路：左端点降序排序每次选最大的能满足的或者右端点升序排序每次选能满足的最小的防晒霜

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define pi pair<int,int>
 5 #define pb push_back
 6 #define fi first
 7 #define sc second
 8 #define ull unsigned long long
 9 const int maxn=3e3+10;
10 const int mod=1e9+7;
11 
12 
13 struct ac
14 {
15     int l,r;
16     bool operator<(ac a)
17     {
18         return r<a.r;
19     }
20 };
21 ac a[maxn];
22 pi b[maxn];
23 
24 
25 int main()
26 {
27     ios::sync_with_stdio(0);
28     cin.tie(0);
29     int n,m;
30     cin>>n>>m;
31     for(int i=1;i<=n;i++)
32     {
33         int l,r;
34         cin>>l>>r;
35         a[i]={l,r};
36     }
37     for(int i=1;i<=m;i++)
38     {
39         cin>>b[i].fi>>b[i].sc;
40     }
41     sort(a+1,a+1+n);
42     sort(b+1,b+1+m);
43     int ans=0;
44     for(int i=1;i<=n;i++)
45     {
46         for(int j=1;j<=m;j++)
47         {
48             int x=b[j].fi;
49             if(b[j].sc&&x<=a[i].r&&x>=a[i].l)
50             {
51                 ans++;
52                 b[j].sc--;
53                 break;
54             }
55         }
56     }
57     cout<<ans<<'\n';
58 
59 
60 
61 
62 
63 }

View Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define pi pair<int,int>
 5 #define pb push_back
 6 #define fi first
 7 #define sc second
 8 #define ull unsigned long long
 9 const int maxn=3e3+10;
10 const int mod=5000011;
11 
12 
13 struct ac
14 {
15     int l,r;
16     bool operator<(ac a)
17     {
18         return l>a.l;
19     }
20 };
21 ac a[maxn];
22 pi b[maxn];
23 
24 
25 int main()
26 {
27     ios::sync_with_stdio(0);
28     cin.tie(0);
29     int n,m;
30     cin>>n>>m;
31     for(int i=1;i<=n;i++)
32     {
33         int l,r;
34         cin>>l>>r;
35         a[i]={l,r};
36     }
37     sort(a+1,a+1+n);
38     for(int i=1;i<=m;i++)
39     {
40         cin>>b[i].fi>>b[i].sc;
41     }
42     sort(b+1,b+1+m);
43     reverse(b+1,b+1+m);
44     int ans=0;
45     for(int i=1;i<=n;i++)
46     {
47         for(int j=1;j<=m;j++)
48         {
49             int x=b[j].fi;
50             if(x>=a[i].l&&x<=a[i].r&&b[j].sc)
51             {
52                 b[j].sc--;
53                 ans++;
54                 break;
55             }
56         }
57     }
58     cout<<ans<<'\n';
59 
60 
61 
62 
63 }

View Code

posted @ 2021-02-15 22:02 canwinfor 阅读(9) 评论(0) 收藏举报

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贪心总结 ###K

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贪心 总结 ###K

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贪心总结 ###K