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  • 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
");
34         else
35             printf("No
");
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
",((-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
",(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
",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
",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
",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
",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

    ( egin{equation*}egin{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
",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\ egin{equation*}egin{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
",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
",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
",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
",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
",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
",dist[n]+mr[n]*c);
66     }
67     return 0;
68 }
69 /*
70 
71 */
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  • 原文地址：https://www.cnblogs.com/cmyg/p/10713800.html
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