1、斜率

可以證明如果兩點之間還有一點的話那麽原來的兩個點連線一定不會是最大斜率

然後我就寫了個沙茶分治…………

其實根據上面的推論只用枚舉相鄰的兩個點，掃一遍就可以了

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <iostream>
 4 using namespace std;
 5 typedef pair<int,int> P;
 6 int n,a,b;
 7 P p[500001];
 8 double ans;
 9 void solve(int l,int r)
10 {
 
11     if (l>=r) return;
12     int mid=(l+r)/2;
13     for (int i=l;i<=r;i++) if (i!=mid)
14     {
15         ans=max(ans,double(p[mid].second-p[i].second)/(p[mid].first-p[i].first));
16     }
17     solve(l,mid-1);solve(mid+1,r);
18 }
19 int main()
20 {
21     freopen("slope.in","r",stdin);
22     freopen("
 
slope.out","w",stdout); 
23     scanf("%d",&n);
24     for (int i=1;i<=n;i++)
25     {
26         scanf("%d%d",&a,&b);
27         p[i]=P(a,b);
28     }
29     sort(p+1,p+n+1);
30     ans=double(p[2].second-p[1].second)/(p[2].first-p[1].first); 
31     solve(1,n);
32     printf("%.3f",ans);
 
33 } 

2、最優路徑

這不就是原來做的過路費嗎……

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <queue>
 4 #include <cstring>
 5 #define ll long long
 6 using namespace std;
 7 typedef pair<int,int> P;
 8 struct E{
 9     int next,to,w;
10 }e[250001];
11 priority_queue<P,vector<P>,greater<P> > que;
12 int n,m,a,b,c,sz=0,head[501],v[501],reg[501],f[501];
13 ll w[501][501];
14 void insert(int a,int b,int w)
15 {
16     sz++;
17     e[sz].next=head[a];
18     head[a]=sz;
19     e[sz].to=b;
20     e[sz].w=w;
21 }
22 void dijkstra(int s)
23 {
24     memset(reg,0,sizeof(reg));
25     memset(f,0x7f,sizeof(f));
26     f[s]=1;
27     que.push(P(0,s));
28     while(!que.empty())
29     {
30         int x=que.top().second;que.pop();
31         if (reg[x]) continue;
32         reg[x]=1;
33         for (int i=head[x];i;i=e[i].next)
34         {
35             if (v[e[i].to]<=v[s]&&f[e[i].to]>max(f[x],e[i].w))
36             {
37                 f[e[i].to]=max(f[x],e[i].w);
38                 que.push(P(f[e[i].to],e[i].to));
39             }
40         }
41     }
42     for (int i=1;i<=n;i++)
43         for (int j=1;j<=n;j++)
44             if (reg[i]&&reg[j]) w[i][j]=min(w[i][j],(ll)max(f[i],f[j])*v[s]);
45 }
46 int main()
47 {
48     freopen("path.in","r",stdin);
49     freopen("path.out","w",stdout);
50     ios::sync_with_stdio(false); 
51     memset(w,0x7f,sizeof(w));
52     cin>>n>>m;
53     for (int i=1;i<=n;i++) cin>>v[i];
54     for (int i=1;i<=m;i++)
55     {
56         cin>>a>>b>>c; 
57         insert(a,b,c);insert(b,a,c);
58     }
59     for (int i=1;i<=n;i++) dijkstra(i);
60     for (int i=1;i<=n;i++)
61     {
62         for (int j=1;j<=n;j++)
63         {
64             if (i==j){
65                 cout<<0<<" ";
66             }else{
67                 if (w[i][j]==w[0][0])
68                     cout<<-1<<" ";
69                 else
70                     cout<<w[i][j]<<" ";
71             }
72         }
73         cout<<endl;
74     }
75 } 

3、小G的線段樹

【亂搞/分治】【最短路】【線段樹】Day 10.24