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 */

转载于:https://www.cnblogs.com/cmyg/p/10713800.html

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