NOIP 2012 提高组第二试模拟赛 Solution

Posted 2020-10-07

第一题

题意

 

数据范围

 

Solution

三分求下凹函数最值

 

 1 #include <cstdio>
 2 #include <queue>
 3 #include <iostream>
 4 using namespace std;
 5 inline void read(int &k)
 6 {
 7     k=0;int f=1;char c=getchar();
 8     while (c<‘0‘||c>‘9‘)c==‘-‘&&(f=-1),c=getchar();
 9     while (c>=‘0‘&&c<=‘9‘)k=k*10+c-‘0‘,c=getchar();
10     k*=f;
11 }
12 const int maxn=1e5+100;
13 int n;
14 double a[maxn],b[maxn],c[maxn];
15 double l,r,lmid,rmid,tmpl,tmpr,tmplm,tmprm; 
16 inline double max(double a,double b)
17 {
18     if (a<b)return b;
19     return a;
20 }
21 inline double f(double x)
22 {
23     double ans=a[1]*x*x+b[1]*x+c[1];
24     for (int i=2;i<=n;i++)
25     ans=max(a[i]*x*x+b[i]*x+c[i],ans);
26     return ans;
27 }
28 inline double abs(double a)
29 {
30     return a>0?a:-a;
31 }
32 int main()
33 {
34     freopen("curves.in","r",stdin);
35     freopen("curves.out","w",stdout);
36     read(n);
37     for (int i=1;i<=n;i++)
38     {
39         scanf("%lf%lf%lf",&a[i],&b[i],&c[i]);
40     }
41     l=0.0;r=1000.0;
42     do
43     {
44         lmid=l+(r-l)/3.0;rmid=l+(r-l)/3.0*2.0;
45         tmplm=f(lmid);tmprm=f(rmid);
46         if (f(lmid)>f(rmid))l=lmid;else r=rmid;
47     }while (abs(l-r)>1e-10);
48     printf("%.3lf\\n",f(lmid));
49 }

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第二题

数据范围

 

 

Solution

优先队列/堆模拟

 1 #include <cstdio>
 2 #include <queue>
 3 #include <iostream>
 4 using namespace std;
 5 priority_queue<int> qmax;
 6 priority_queue<int,vector<int>,greater<int> > qmin;
 7 inline void read(int &k)
 8 {
 9     k=0;int f=1;char c=getchar();
10     while (c<‘0‘||c>‘9‘)c==‘-‘&&(f=-1),c=getchar();
11     while (c>=‘0‘&&c<=‘9‘)k=k*10+c-‘0‘,c=getchar();
12     k*=f;
13 }
14 int l,n,tmp,cur=0,head=1;long long ans=0;
15 int main()
16 {
17     freopen("cate.in","r",stdin);
18     freopen("cate.out","w",stdout);
19     read(l);read(n);
20     for (int i=1;i<=n;i++)
21     {
22         read(tmp);
23         if (!tmp)
24         {
25             read(tmp);
26             if (tmp>=cur)qmin.push(tmp);else qmax.push(tmp);
27         }
28         else 
29         {
30             if (head)
31             {
32                 if (qmax.empty()&&qmin.empty())continue;
33                 if (qmax.empty())
34                 {
35                     ans+=qmin.top()-cur;
36                     cur=qmin.top();
37                     qmin.pop();
38                     continue;
39                 }
40                 if (qmin.empty())
41                 {
42                     ans+=cur-qmax.top();
43                     cur=qmax.top();
44                     qmax.pop();
45                     head=0;
46                     continue;
47                 }
48                 if (cur-qmax.top()==qmin.top()-cur)
49                 {
50                     ans+=qmin.top()-cur;
51                     cur=qmin.top();
52                     qmin.pop();
53                 }
54                 else
55                 {
56                     if (qmin.top()-cur<cur-qmax.top())ans+=qmin.top()-cur,cur=qmin.top(),qmin.pop();
57                     else head=0,ans+=cur-qmax.top(),cur=qmax.top(),qmax.pop();
58                 }
59             }
60             else
61             {
62                 if (qmax.empty()&&qmin.empty())continue;    
63                 if (qmax.empty())
64                 {
65                     ans+=qmin.top()-cur;
66                     cur=qmin.top();
67                     qmin.pop();
68                     head=1;
69                     continue;
70                 }
71                 if (qmin.empty())
72                 {
73                     ans+=cur-qmax.top();
74                     cur=qmax.top();
75                     qmax.pop();
76                     continue;
77                 }
78                 if (cur-qmax.top()==qmin.top()-cur)
79                 {
80                     ans+=cur-qmax.top();
81                     cur=qmax.top();
82                     qmax.pop();
83                 }
84                 else
85                 {
86                     if (qmin.top()-cur>cur-qmax.top())ans+=cur-qmax.top(),cur=qmax.top(),qmax.pop();
87                     else head=1,ans+=qmin.top()-cur,cur=qmin.top(),qmin.pop();
88                 }
89             }
90         }
91     }
92     printf("%lld\\n",ans);
93 }

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第三题

题意

 

数据范围

Solution

O(n^2)/60%：dfs（遍历整颗树）

 

 1 #include <cstdio>
 2 #include <queue>
 3 #include <iostream>
 4 #include <cstring>
 5 #define ll long long 
 6 using namespace std;
 7 inline void read(ll &k)
 8 {
 9     k=0;ll f=1;char c=getchar();
10     while (c<‘0‘||c>‘9‘)c==‘-‘&&(f=-1),c=getchar();
11     while (c>=‘0‘&&c<=‘9‘)k=k*10+c-‘0‘,c=getchar();
12     k*=f;
13 }
14 const int maxn=300100*2;
15 bool v[maxn];
16 ll n,x,y,z,ans,la,tot,last[maxn],next[maxn],value[maxn],to[maxn],col[maxn];
17 void dfs(int now,ll sum,int dep,int la)
18 {
19     v[now]=1;
20     for (int cur=last[now];cur;cur=next[cur])
21     if ((col[cur]!=la)&&(!v[to[cur]]))
22         dfs(to[cur],sum+value[to[cur]],dep+1,col[cur]);
23     if (dep)
24     ans+=sum;
25 }
26 int main()
27 {
28     freopen("gorgeous.in","r",stdin);
29     freopen("gorgeous.out","w",stdout);
30     read(n);
31     for (int i=1;i<=n;i++)
32     read(value[i]);
33     for (int i=1;i<n;i++)
34     {
35         read(x);read(y);read(z);
36         next[++tot]=last[x];
37         to[tot]=y;
38         col[tot]=z;
39         last[x]=tot;
40         next[++tot]=last[y];
41         to[tot]=x;
42         col[tot]=z;
43         last[y]=tot;
44     }
45     for (int i=1;i<=n;i++)
46     {
47         memset(v,0,sizeof(v));
48         dfs(i,value[i],0,0);
49     }
50     printf("%lld\\n",ans/2);
51 }

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