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 }
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 }
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 }
时间： 2024-10-03 22:51:33