2050 Programming Competition (CCPC)
Pro&Sol
链接: https://pan.baidu.com/s/17Tt3EPKEQivP2-3OHkYD2A 提取码: wbnu 复制这段内容后打开百度网盘手机App，操作更方便哦
6491 时间间隔
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <iostream> 11 using namespace std; 12 13 #define ll long long 14 15 const int maxn=1e5+10; 16 const int inf=1e9; 17 const double eps=1e-8; 18 19 char s[maxn],str[5]="2050"; 20 21 int main() 22 { 23 int t,len,i; 24 scanf("%d",&t); 25 while (t--) 26 { 27 scanf("%s",s); 28 len=strlen(s); 29 for (i=0;i<len;i++) 30 if (s[i]!=str[i%4]) 31 break; 32 if (i==len && len%4==0) 33 printf("Yes\n"); 34 else 35 printf("No\n"); 36 } 37 return 0; 38 } 39 /* 40 41 */
6491 时间间隔
S距离2050年1月1日0点0时0分多少秒
2050_value-this_value 而不是差值绝对值模100 (不过感觉有点牵强)
负数的结果进行相应处理
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <iostream> 11 using namespace std; 12 13 #define ll long long 14 15 const int maxn=1e4+10; 16 const int inf=1e9; 17 const double eps=1e-8; 18 19 20 21 int main() 22 { 23 int t,a,b,c,d,e,f; 24 scanf("%d",&t); 25 while (t--) 26 { 27 scanf("%d-%d-%d %d:%d:%d",&a,&b,&c,&d,&e,&f); 28 printf("%d\n",((-e*60-f)%100+100)%100); 29 } 30 return 0; 31 } 32 /* 33 7 34 2023-12-01 03:12:12 35 68 36 2024-05-01 03:12:12 37 68 38 3000-03-01 23:00:59 39 41 40 6000-03-01 00:00:00 41 0 42 6661-01-12 13:34:45 43 15 44 6661-01-12 13:34:46 45 14 46 2049-12-31 23:59:59 47 1 48 */
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <iostream> 11 using namespace std; 12 13 #define ll long long 14 15 const int maxn=1e4+10; 16 const int inf=1e9; 17 const double eps=1e-8; 18 19 int mon[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 20 21 ll cal(int a,int b,int c,int d,int e,int f) 22 { 23 ll sum=0; 24 int i; 25 for (i=1;i<a;i++) 26 if (i%4==0 && (i%400==0 || i%100!=0)) 27 sum+=1ll*366*24*60*60; 28 else 29 sum+=1ll*365*24*60*60; 30 if (i%4==0 && (i%400==0 || i%100!=0)) 31 mon[2]=29; 32 for (i=1;i<b;i++) 33 sum+=1ll*24*60*60*mon[i]; 34 sum+=1ll*24*60*60*(c-1)+d*60*60+e*60+f; 35 return sum; 36 } 37 38 int main() 39 { 40 int t,a,b,c,d,e,f; 41 ll tot=0; 42 tot=cal(2050,1,1,0,0,0); 43 scanf("%d",&t); 44 while (t--) 45 { 46 scanf("%d-%d-%d %d:%d:%d",&a,&b,&c,&d,&e,&f); 47 printf("%lld\n",(tot-cal(a,b,c,d,e,f)%100+100)%100); 48 } 49 return 0; 50 } 51 /* 52 53 */
1 import java.text.SimpleDateFormat; 2 import java.util.*; 3 4 public class Main { 5 public static void main(String[] args) throws Exception { 6 //upper lower 7 SimpleDateFormat tf=new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); 8 Date t1=tf.parse("2050-01-01 00:00:00"); 9 Scanner in=new Scanner(System.in); 10 String s; 11 int t; 12 t=in.nextInt(); 13 s=in.nextLine(); 14 while (t-->0) { 15 s=in.nextLine(); 16 Date t2=tf.parse(s); 17 //System.out.println(t2); //test 18 //System.out.println(t2.getTime()/1000); 19 //ms 20 System.out.println(( Math.abs(t1.getTime()-t2.getTime()) )/1000%100); 21 // System.out.println(dif); 22 } 23 in.close(); 24 } 25 } 26 /* 27 7 28 2023-12-01 03:12:12 29 2024-05-01 03:12:12 30 2000-03-01 23:00:59 31 2049-03-01 00:00:00 32 2048-01-12 13:34:45 33 2023-01-12 13:34:46 34 2036-12-31 23:59:59 35 36 68 37 68 38 41 39 0 40 15 41 14 42 1 43 */
6492 分宿舍
贪心，参见下方注释
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <iostream> 11 using namespace std; 12 13 #define ll long long 14 15 const int maxn=1e5+10; 16 const int inf=1e9; 17 const double eps=1e-8; 18 19 ll n,m,k,a,b,c; 20 21 ll cal(ll g) 22 { 23 /* 24 替代 2*3=3*2 25 两人间 <=2 26 三人间 <=1 27 */ 28 ll tot=1e18; 29 int i; 30 for (i=0;i<=2;i++) 31 tot=min(tot,i*a+(g-i*2+2)/3*b); 32 for (i=0;i<=1;i++) 33 tot=min(tot,(g-i*3+1)/2*a+i*b); 34 return tot; 35 } 36 37 int main() 38 { 39 ll tot; 40 int t,i; 41 scanf("%d",&t); 42 while (t--) 43 { 44 scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&k,&a,&b,&c); 45 tot=1e18; 46 for (i=k;i>=0;i--) 47 tot=min(tot,c*i+cal(n+k-i)+cal(m+k-i)); 48 49 printf("%lld\n",tot); 50 } 51 return 0; 52 } 53 /* 54 1 55 1000 1000 1000 1000000000 1000000000 1000000000 56 57 1 58 1 1 0 3 2 1 59 */
预处理f[x]，x个人，需要花费的最小代码。对于每个k对应的n+k和m+k，O(1)得到答案。
\( f[i]=min(f[i-2]+a,f[i-3]+b) \\ f[j]=0 j<=0 \)
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <iostream> 11 using namespace std; 12 13 #define ll long long 14 15 const int maxn=2e3+10; 16 const int inf=1e9; 17 const double eps=1e-8; 18 19 ll n,m,k,a,b,c,f[maxn]; 20 21 int main() 22 { 23 ll tot; 24 int t,i,lim; 25 scanf("%d",&t); 26 while (t--) 27 { 28 scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&k,&a,&b,&c); 29 memset(f,sizeof(0x3f),sizeof(f)); 30 lim=max(n+k,m+k); 31 f[0]=0; 32 for (i=0;i<=lim;i++) 33 if (i<2) 34 f[i]=min(f[i],a); 35 else 36 f[i]=min(f[i],f[i-2]+a); 37 for (i=0;i<=lim;i++) 38 if (i<3) 39 f[i]=min(f[i],b); 40 else 41 f[i]=min(f[i],f[i-3]+b); 42 43 tot=1e18; 44 for (i=k;i>=0;i--) 45 tot=min(tot,c*i+f[n+k-i]+f[m+k-i]); 46 printf("%lld\n",tot); 47 } 48 return 0; 49 } 50 /* 51 1 52 1000 1000 1000 1000000000 1000000000 1000000000 53 54 1 55 1 1 0 3 2 1 56 */
暴力，不会超时
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std;
#define ll long long
const int maxn=1e5+10;
const int inf=1e9;
const double eps=1e-8;
int main()
{
int t;
ll n,m,k,a,b,c,i,j,tot,sum,x,y;
scanf("%d",&t);
while (t--)
{
scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&k,&a,&b,&c);
tot=1e18;
for (i=k;i>=0;i--)
{
sum=c*i;
x=n+k-i;
y=1e18;
for (j=0;j<=(x+1)/2;j++)
y=min(y,j*a+(x-j*2+2)/3*b);
sum+=y;
x=m+k-i;
y=1e18;
for (j=0;j<=(x+1)/2;j++)
y=min(y,j*a+(x-j*2+2)/3*b);
sum+=y;
tot=min(tot,sum);
}
printf("%lld\n",tot);
}
return 0;
}
/*
1
1000 1000 1000 1000000000 1000000000 1000000000
1
1 1 0 3 2 1
*/
以下方法是错误的
6493 PASS
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cmath>
4 #include <cstring>
5 #include <string>
6 #include <algorithm>
7 #include <set>
8 #include <map>
9 #include <queue>
10 #include <iostream>
11 using namespace std;
12
13 #define ll long long
14
15 const int maxn=1e4+10;
16 const int inf=1e9;
17 const double eps=1e-8;
18
19 int g[maxn],a[maxn];
20
21 int main()
22 {
23 int t,n,m,k,i,tot;
24 scanf("%d",&t);
25 while (t--)
26 {
27 scanf("%d%d%d",&n,&m,&k);
28 memset(g,0,sizeof(g));
29 for (i=1;i<=n;i++)
30 {
31 scanf("%d",&a[i]);
32 g[a[i]]++;
33 }
34 for (i=1;i<=m;i++)
35 g[i]/=k;
36 tot=0;
37 for (i=1;i<=n/2;i++)
38 {
39 g[a[i]]--;
40 if (g[a[i]]>=0)
41 tot++;
42 }
43 printf("%d\n",tot);
44 }
45 return 0;
46 }
47 /*
48
49 */
1005球赛
一开始猜测贪心有可能是错的，所以选择使用dp
\( \begin{equation*}\begin{split}&condition \quad (a,b) \quad -> \quad condition (x,y) \quad + \quad c(+1/+0)\\&(10,k)/(k,10) \quad -> \quad (0,0) \quad [(11,k)/(k,11)] \quad + \quad 1 \quad k<=9\\&(10,11)/(11,10) \quad -> \quad (0,0) \quad [(10,12)/(12,10)] \quad +1\\&(10,11)/(11,10) \quad -> \quad (10,10) \quad [(11,11)]\\&f[x][y]=max(f[a][b]+c)\end{split}\end{equation*} \)
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std;
#define ll long long
const int maxn=1e4+10;
const int inf=1e9;
const double eps=1e-8;
char str[maxn];
int f[maxn][12][12];
int main()
{
int T,g,len,i,j,k,s,t;
scanf("%d",&T);
while (T--)
{
scanf("%s",str+1);
len=strlen(str+1);
for (i=0;i<=len;i++)
for (j=0;j<12;j++)
for (k=0;k<12;k++)
f[i][j][k]=-1;
f[0][0][0]=0;
for (i=1;i<=len;i++)
{
for (j=0;j<12;j++)
for (k=0;k<12;k++)
if (f[i-1][j][k]!=-1)
{
if (str[i]=='A' || str[i]=='?')
{
s=j+1;
t=k;
if (s==11 && t==11)
s=10,t=10;
if ((s>=11 && s-t>=2) || (t>=11 && t-s>=2))
f[i][0][0]=max(f[i][0][0],f[i-1][j][k]+1);
else
f[i][s][t]=max(f[i][s][t],f[i-1][j][k]);
}
if (str[i]=='B' || str[i]=='?')
{
s=j;
t=k+1;
if (s==11 && t==11)
s=10,t=10;
if ((s>=11 && s-t>=2) || (t>=11 && t-s>=2))
f[i][0][0]=max(f[i][0][0],f[i-1][j][k]+1);
else
f[i][s][t]=max(f[i][s][t],f[i-1][j][k]);
}
}
}
g=0;
for (i=0;i<12;i++)
for (j=0;j<12;j++)
g=max(g,f[len][i][j]);
printf("%d\n",g);
}
return 0;
}
/*
1
AAAAAAAAAAA
??????????????????????
?????
*/
6495 冰水挑战
无贪心方法等。应该往dp想。c不应该作为一维。应该很自然就能想到。。。
\( 前i个挑战中，选择j个挑战，剩余的最大体力\\ \begin{equation*}\begin{split}&f[i][j]=f[i-1][j]+c_{i}\\&f[i][j]=max(f[i][j],min(f[i-1][j-1],b_{i})-a_{i}+c_{i})) \quad if \, min(f[i-1][j-1],b_{i})-a_{i}>0 \, , \, i>0 \\\end{split}\end{equation*} \)
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <iostream> 11 using namespace std; 12 13 #define ll long long 14 15 const int maxn=1e3+10; 16 const ll inf=1e18; 17 const double eps=1e-8; 18 19 ll f[maxn][maxn]; 20 21 int main() 22 { 23 int t,n,i,j; 24 ll m,a,b,c; 25 scanf("%d",&t); 26 while (t--) 27 { 28 scanf("%d%lld",&n,&m); 29 for (i=0;i<=n;i++) 30 for (j=1;j<=n;j++) 31 f[i][j]=-inf; 32 f[0][0]=m; 33 for (i=1;i<=n;i++) 34 { 35 scanf("%lld%lld%lld",&a,&b,&c); 36 f[i][0]=f[i-1][0]+c; 37 for (j=1;j<=n;j++) 38 { 39 f[i][j]=f[i-1][j]+c; 40 if (min(f[i-1][j-1],b)-a>0) 41 f[i][j]=max(f[i][j],min(f[i-1][j-1],b)-a+c); 42 } 43 } 44 for (j=n;j>=1;j--) 45 if (f[n][j]>0) 46 break; 47 printf("%d\n",j); 48 } 49 return 0; 50 } 51 /* 52 53 */
6496 大厦
1.新方法
用下面的方法求矩形
对于每个横坐标小块i，求纵坐标上下限[ d[i] , u[i] ]。
若对于某个纵坐标，横坐标小块为[x1,x2]，则往上，[x3,x4]，有x1<=x3 , x4<=x2。往下，同理。(里大，外小)
对于横坐标小块[x1,x2]，求[x1,x2]的纵坐标上下限([ max(d[x1],d[x1+1],...,d[x2]) , min(u[x1],u[x1+1],...,u[x2]) ])
st/线段树处理。
O(T*(n*m+n*n/2))
横坐标小块12；13；14；..；23；24；...；n-1 n
1*n 个数n+(n-1)+(n-2)+...+1 = n*(n+1)/2
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cmath>
4 #include <cstring>
5 #include <string>
6 #include <algorithm>
7 #include <set>
8 #include <map>
9 #include <queue>
10 #include <iostream>
11 using namespace std;
12
13 #define ll long long
14
15 const int maxn=1e3+10;
16 const int inf=1e9;
17 const double eps=1e-8;
18 const ll mod=1e9+7;
19
20 int c1[maxn],c2[maxn],d[maxn],u[maxn];
21
22 int main()
23 {
24 ll sum;
25 int t,w,h,n,m,i,j,down,up;
26 scanf("%d",&t);
27 while (t--)
28 {
29 scanf("%d%d%d%d",&w,&h,&n,&m);
30 for (i=1;i<=n;i++)
31 scanf("%d",&c1[i]);
32 for (i=1;i<=m;i++)
33 scanf("%d",&c2[i]);
34 memset(d,0x3f,sizeof(d));
35 memset(u,0,sizeof(u));
36
37 sort(c1+1,c1+n+1);
38 sort(c2+1,c2+m+1);
39 for (i=1;i<=n;i++)
40 ///其实用二分更快
41 for (j=1;j<=m;j++)
42 if (0<=c1[i]+c2[j] && c1[i]+c2[j]<=w*2
43 && 0<=c1[i]-c2[j] && c1[i]-c2[j]<=h*2) ///w*2 h*2 int range
44 d[i]=min(d[i],j),u[i]=max(u[i],j);
45
46 sum=0;
47 for (i=1;i<n;i++)
48 {
49 down=d[i],up=u[i];
50 for (j=i+1;j<=n;j++)
51 {
52 down=max(down,d[j]),up=min(up,u[j]);
53 if (up>=down)
54 sum=(sum+1ll*(up-down)*(up-down+1)/2)%mod;
55 }
56 }
57 printf("%lld\n",sum); ///ll
58 }
59 return 0;
60 }
61 /*
62
63 */
2.bitset(题解)
time(/32)
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <iostream> 11 #include <bitset> 12 using namespace std; 13 14 #define ll long long 15 16 const int maxn=1e3+10; 17 const int maxm=1e3+10; 18 const int inf=1e9; 19 const double eps=1e-8; 20 const ll mod=1e9+7; 21 22 int c1[maxn],c2[maxn]; 23 bitset<maxm> f[maxn]; 24 25 int main() 26 { 27 ll sum; 28 int t,w,h,n,m,i,j,g; 29 scanf("%d",&t); 30 while (t--) 31 { 32 scanf("%d%d%d%d",&w,&h,&n,&m); 33 for (i=1;i<=n;i++) 34 scanf("%d",&c1[i]); 35 for (i=1;i<=m;i++) 36 scanf("%d",&c2[i]); 37 38 for (i=1;i<=n;i++) 39 f[i].reset(); 40 41 sort(c1+1,c1+n+1); 42 sort(c2+1,c2+m+1); 43 for (i=1;i<=n;i++) 44 for (j=1;j<=m;j++) 45 if (0<=c1[i]+c2[j] && c1[i]+c2[j]<=w*2 46 && 0<=c1[i]-c2[j] && c1[i]-c2[j]<=h*2) ///w*2 h*2 int range 47 f[i].set(j,1); 48 49 sum=0; 50 for (i=1;i<n;i++) 51 for (j=i+1;j<=n;j++) 52 { 53 g=(f[i]&f[j]).count(); 54 sum=(sum+1ll*g*(g-1)/2)%mod; 55 } 56 printf("%lld\n",sum); ///ll 57 } 58 return 0; 59 } 60 /* 61 62 */
6497 骑行
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <iostream> 11 using namespace std; 12 13 #define ll long long 14 15 const int maxn=1e3+10; 16 const int inf=1e9; 17 const double eps=1e-8; 18 19 double v,t; 20 double w[maxn],s[maxn],a[maxn],lim[maxn]; 21 22 void work(double v1,double v2,double v0,double len,double a) 23 { 24 double dist=(v0*v0-v1*v1)/2/a + (v0*v0-v2*v2)/2/a; 25 double dist1=(v2*v2-v1*v1)/2/a; 26 if (dist<=len) 27 { 28 t+=(v0-v1)/a + (v0-v2)/a + (len-dist)/v0; 29 v=v2; 30 } 31 else if (v2>v1 && dist1>len) 32 { 33 v=sqrt(2*a*len + v1*v1); 34 t+=(v-v1)/a; 35 } 36 else 37 { 38 double v3=sqrt((2*a*len + v1*v1 + v2*v2)/2); 39 t+=(v3-v1)/a + (v3-v2)/a; 40 v=v2; 41 } 42 } 43 44 int main() 45 { 46 double temp; 47 int T,n,i; 48 scanf("%d",&T); 49 while (T--) 50 { 51 scanf("%d",&n); 52 for (i=1;i<=n;i++) 53 scanf("%lf%lf%lf",&w[i],&s[i],&a[i]); 54 lim[n]=s[n]; 55 for (i=n-1;i>=1;i--) 56 { 57 temp=sqrt(2*a[i+1]*w[i+1]+lim[i+1]*lim[i+1]); 58 lim[i]=min(min(s[i],s[i+1]),temp); 59 } 60 61 v=0,t=0; 62 for (i=1;i<=n;i++) 63 work(v,lim[i],s[i],w[i],a[i]); 64 printf("%.10f\n",t); 65 } 66 return 0; 67 } 68 /* 69 */
6498 跨洋飞行
一个点x到另外一个点y，到达点y时，要预留min(y to others)的油。参见题解。
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cmath>
4 #include <cstring>
5 #include <string>
6 #include <algorithm>
7 #include <set>
8 #include <map>
9 #include <queue>
10 #include <iostream>
11 using namespace std;
12
13 #define ll long long
14
15 const int maxn=1e3+10;
16 const int inf=1e9;
17 const double eps=1e-10;
18
19 double x[maxn],y[maxn],dist[maxn],road[maxn][maxn],mr[maxn];
20 bool vis[maxn];
21
22 int main()
23 {
24 double a,b,c,d,u;
25 int t,n,i,j,p;
26 scanf("%d",&t);
27 while (t--)
28 {
29 scanf("%d%lf%lf%lf%lf%lf",&n,&a,&b,&c,&d,&u);
30 a+=b;
31 for (i=1;i<=n;i++)
32 scanf("%lf%lf",&x[i],&y[i]);
33 for (i=1;i<=n;i++)
34 mr[i]=1.0e18;
35 for (i=1;i<n;i++)
36 for (j=i+1;j<=n;j++)
37 {
38 road[i][j]=road[j][i]=sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]) );
39 mr[i]=min(mr[i],road[i][j]);
40 mr[j]=min(mr[j],road[i][j]);
41 }
42 mr[1]=0; ///
43 for (i=0;i<=n;i++)
44 dist[i]=1.0e18;
45 dist[1]=0;
46 memset(vis,0,sizeof(vis));
47 for (i=1;i<n;i++)
48 {
49 p=0;
50 for (j=1;j<=n;j++)
51 if (!vis[j] && dist[p]>dist[j])
52 p=j;
53 if (p==n)
54 break;
55 vis[p]=1;
56 for (j=1;j<=n;j++)
57 if (!vis[j] && mr[j]+road[p][j]<u+eps && dist[j]>dist[p] +a + road[p][j]*d + (mr[j]+road[p][j]-mr[p])*c)
58 dist[j]=dist[p] +a + road[p][j]*d + (mr[j]+road[p][j]-mr[p])*c;
59 }
60 printf("%.10f\n",dist[n]);
61 }
62 return 0;
63 }
64 /*
65
66 */
错误想法：
下界提高，x到y，至少dist（x，y）+min y to others（z）。而miny to others是有用的，x到y，剩余z，而y到其它点，至少要使用z，不浪费。除了最后一个点，浪费min n to others。但倒数第二个点到点n，而倒数第二个点到其它点的最短距离有可能不是该点到点n的距离。
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cmath>
4 #include <cstring>
5 #include <string>
6 #include <algorithm>
7 #include <set>
8 #include <map>
9 #include <queue>
10 #include <iostream>
11 using namespace std;
12
13 #define ll long long
14
15 const int maxn=1e3+10;
16 const int inf=1e9;
17 const double eps=1e-10;
18
19 double x[maxn],y[maxn],dist[maxn],road[maxn][maxn],mr[maxn];
20 bool vis[maxn];
21
22 int main()
23 {
24 double a,b,c,d,u;
25 int t,n,i,j,p;
26 scanf("%d",&t);
27 while (t--)
28 {
29 scanf("%d%lf%lf%lf%lf%lf",&n,&a,&b,&c,&d,&u);
30 a+=b;
31 for (i=1;i<=n;i++)
32 scanf("%lf%lf",&x[i],&y[i]);
33 for (i=1;i<=n;i++)
34 mr[i]=1.0e18;
35 for (i=1;i<n;i++)
36 for (j=i+1;j<=n;j++)
37 {
38 road[i][j]=road[j][i]=sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]) );
39 mr[i]=min(mr[i],road[i][j]);
40 mr[j]=min(mr[j],road[i][j]);
41 }
42 for (i=1;i<=n;i++)
43 for (j=1;j<=n;j++)
44 if (i!=j && road[i][j]+mr[i]>u+eps)
45 road[i][j]=-1;
46 else
47 road[i][j]=road[i][j]*(c+d)+a;
48 for (i=0;i<=n;i++)
49 dist[i]=1.0e18;
50 dist[1]=0;
51 memset(vis,0,sizeof(vis));
52 for (i=1;i<n;i++)
53 {
54 p=0;
55 for (j=1;j<=n;j++)
56 if (!vis[j] && dist[p]>dist[j])
57 p=j;
58 if (p==n)
59 break;
60 vis[p]=1;
61 for (j=1;j<=n;j++)
62 if (!vis[j] && road[p][j]!=-1 && dist[j]>dist[p]+road[p][j])
63 dist[j]=dist[p]+road[p][j];
64 }
65 printf("%.10f\n",dist[n]+mr[n]*c);
66 }
67 return 0;
68 }
69 /*
70
71 */
