joisino's travel-白红宇

joisino's travel

阅读量：6273 次

发布时间：2019-06-22

本文共 5625 字，大约阅读时间需要 18 分钟。

D - joisino's travel

Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

There are N towns in the State of Atcoder, connected by M bidirectional roads.

The i-th road connects Town Ai and Bi and has a length of Ci.

Joisino is visiting R towns in the state, r1,r2,..,rR (not necessarily in this order).

She will fly to the first town she visits, and fly back from the last town she visits, but for the rest of the trip she will have to travel by road.

If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be?

Constraints

2≤N≤200

1≤M≤N×(N−1)⁄2

2≤R≤min(8,N) (min(8,N) is the smaller of 8 and N.)

ri≠rj(i≠j)

1≤Ai,Bi≤N,Ai≠Bi

(Ai,Bi)≠(Aj,Bj),(Ai,Bi)≠(Bj,Aj)(i≠j)

1≤Ci≤100000

Every town can be reached from every town by road.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M Rr1 … rRA1 B1 C1:AM BM CM

Output

Print the distance traveled by road if Joisino visits the towns in the order that minimizes it.

Sample Input 1

Copy

3 3 31 2 31 2 12 3 13 1 4

Sample Output 1

Copy

For example, if she visits the towns in the order of 1, 2, 3, the distance traveled will be 2, which is the minimum possible.

Sample Input 2

Copy

3 3 21 32 3 21 3 61 2 2

Sample Output 2

Copy

The shortest distance between Towns 1 and 3 is 4. Thus, whether she visits Town 1 or 3 first, the distance traveled will be 4.

Sample Input 3

Copy

4 6 32 3 41 2 42 3 34 3 11 4 14 2 23 1 6

Sample Output 3

Copy

//题意，n 个点， m 条边，，R 个需要去的地方，可以从任意一点出发，终于任意一点，但必须走完 R 个点，问最小路径为多少？

//首先，用Floyd求出最短路，然后暴力枚举要走的点的顺序即可 8！也就1千万

1 # include 
        
           2 # include 
         
            3 # include 
          
             4 # include 
           
              5 # include 
            
               6 # include 
             
               7 # include 
              
                8 # include 
                9 # include 
                
                  10 # include 
                 
                   11 # include 
                  
                    12 # include 
                   
                     13 # include 
                    
                      14 # pragma comment(linker,"/STACK:102400000,102400000") 15 using namespace std; 16 # define LL long long 17 # define pr pair 18 # define mkp make_pair 19 # define lowbit(x) ((x)&(-x)) 20 # define PI acos(-1.0) 21 # define INF 0x3f3f3f3f3f3f3f3f 22 # define eps 1e-8 23 # define MOD 1000000007 24 25 inline int scan() { 26 int x=0,f=1; char ch=getchar(); 27 while(ch<'0'||ch>'9'){ if(ch=='-') f=-1; ch=getchar();} 28 while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} 29 return x*f; 30 } 31 inline void Out(int a) { 32 if(a<0) {putchar('-'); a=-a;} 33 if(a>=10) Out(a/10); 34 putchar(a%10+'0'); 35 } 36 # define MX 205 37 /**************************/ 38 39 int n,m,R; 40 int ans; 41 int goal[MX]; 42 int mp[MX][MX]; 43 int G[10][10]; 44 45 void floyd() 46 { 47 for (int k=1;k<=n;k++) 48 for (int i=1;i<=n;i++) 49 for (int j=1;j<=n;j++) 50 if (mp[i][j]>mp[i][k]+mp[k][j]) 51 mp[i][j] = mp[i][k]+mp[k][j]; 52 } 53 54 int prim() 55 { 56 int dis[10]; 57 int vis[10]; 58 memset(dis,0x3f,sizeof(dis)); 59 memset(vis,0,sizeof(vis)); 60 dis[1] = 0; 61 62 int all = 0; 63 for (int i=1;i<=R;i++) 64 { 65 int dex, mmm = INF; 66 for (int j=1;j<=R;j++) 67 { 68 if (!vis[j]&&dis[j]
                     
                      G[dex][j]) 79 dis[j]=G[dex][j]; 80 } 81 } 82 return all; 83 } 84 85 int vis[10]; 86 void dfs(int s,int pos,int far) 87 { 88 if (s>R) 89 { 90 ans = min(ans,far); 91 return ; 92 } 93 for (int i=1;i<=R;i++) 94 { 95 if (vis[i]) continue; 96 vis[i] = 1; 97 dfs(s+1,i,far+G[pos][i]); 98 vis[i]=0; 99 }100 }101 102 int main()103 {104 scanf("%d%d%d",&n,&m,&R);105 106 for (int i=1;i<=R;i++)107 goal[i] = scan();108 109 memset(mp,0x3f,sizeof(mp));110 for (int i=1;i<=m;i++)111 {112 int a,b,c;113 scanf("%d%d%d",&a,&b,&c);114 mp[a][b] = mp[b][a] = c;115 }116 117 floyd();118 for (int i=1;i<=R;i++)119 for (int j=1;j<=R;j++)120 G[i][j] = mp[ goal[i] ][ goal[j] ];121 122 ans = INF;123 for (int i=1;i<=R;i++)124 {125 vis[i]=1;126 dfs(2,i,0);127 vis[i]=0;128 }129 printf("%d\n",ans);130 }

View Code

//如果R再大点(16左右)，还可以状态压缩一下 dp[i][j] 表在 i 点，状态为 j 时的最小路径

1 # include 
        
          2 # include 
         
           3 # include 
          
            4 # include 
           
             5 # include 
            
              6 # include 
             
               7 # include 
              
                8 # include 
                9 # include 
                
                 10 # include 
                 
                  11 # include 
                  
                   12 # include 
                   
                    13 # include 
                    
                     14 using namespace std;15 # define LL long long16 # define pr pair17 # define mkp make_pair18 # define lowbit(x) ((x)&(-x))19 # define PI acos(-1.0)20 # define INF 0x3f3f3f3f21 # define eps 1e-822 # define MOD 100000000723 24 inline int scan() {25 int x=0,f=1; char ch=getchar();26 while(ch<'0'||ch>'9'){ if(ch=='-') f=-1; ch=getchar();}27 while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}28 return x*f;29 }30 inline void Out(int a) {31 if(a<0) {putchar('-'); a=-a;}32 if(a>=10) Out(a/10);33 putchar(a%10+'0');34 }35 # define MX 20536 /**************************/37 38 int n,m,R;39 int ans;40 int goal[MX];41 int mp[MX][MX];42 int G[25][25];43 int dp[16][(1<<16)];44 45 void floyd()46 {47 for (int k=1;k<=n;k++)48 for (int i=1;i<=n;i++)49 for (int j=1;j<=n;j++)50 if (mp[i][j]>mp[i][k]+mp[k][j])51 mp[i][j] = mp[i][k]+mp[k][j];52 }53 54 int main()55 {56 scanf("%d%d%d",&n,&m,&R);57 for (int i=1;i<=R;i++)58 goal[i] = scan();59 memset(mp,0x3f,sizeof(mp));60 for (int i=1;i<=m;i++)61 {62 int a,b,c;63 scanf("%d%d%d",&a,&b,&c);64 mp[a][b] = mp[b][a] = c;65 }66 67 floyd();68 for (int i=1;i<=R;i++)69 for (int j=1;j<=R;j++)70 G[i][j] = mp[ goal[i] ][ goal[j] ];71 72 memset(dp,0x3f,sizeof(dp));73 for (int i=1;i<=R;i++)74 dp[i-1][(1<<(i-1))] = 0;75 76 for (int i=0;i<(1<