PKU1734 Sightseeing trip (CEOI99)

很久没写结题报告了
今天做Sightseeing trip 上来贴个
Ural:1004

Sightseeing trip
Time Limit:1000MS  Memory Limit:65536K
Total Submit:317 Accepted:133 Special Judged

Description
There is a travel agency in Adelton town on Zanzibar island. It has decided to offer its clients, besides many other attractions, sightseeing the town. To earn as much as possible from this attraction, the agency has accepted a shrewd decision: it is necessary to find the shortest route which begins and ends at the same place. Your task is to write a program which finds such a route.

In the town there are N crossing points numbered from 1 to N and M two-way roads numbered from 1 to M. Two crossing points can be connected by multiple roads, but no road connects a crossing point with itself. Each sightseeing route is a sequence of road numbers y_1, ..., y_k, k>2. The road y_i (1<=i<=k-1) connects crossing points x_i and x_{i+1}, the road y_k connects crossing points x_k and x_1. All the numbers x_1,...,x_k should be different.The length of the sightseeing route is the sum of the lengths of all roads on the sightseeing route, i.e. L(y_1)+L(y_2)+...+L(y_k) where L(y_i) is the length of the road y_i (1<=i<=k). Your program has to find such a sightseeing route, the length of which is minimal, or to specify that it is not possible,because there is no sightseeing route in the town.

Input
The first line of input contains two positive integers: the number of crossing points N<=100 and the number of roads M<=10000. Each of the next M lines describes one road. It contains 3 positive integers: the number of its first crossing point, the number of the second one, and the length of the road (a positive integer less than 500).

Output
There is only one line in output. It contains either a string 'No solution.' in case there isn't any sightseeing route, or it contains the numbers of all crossing points on the shortest sightseeing route in the order how to pass them (i.e. the numbers x_1 to x_k from our definition of a sightseeing route), separated by single spaces. If there are multiple sightseeing routes of the minimal length, you can output any one of them.

Sample Input

5 7
1 4 1
1 3 300
3 1 10
1 2 16
2 3 100
2 5 15
5 3 20

 

Sample Output

1 3 5 2

 

点数是100个 题目意思是找一个最小权圈 以任意序输出
从图论的角度上考虑 应该是任意选一条边（枚举） 然后删除边 再以一个点为原点求Dijkstra 找出最小权圈 o(M*N^2)的复杂度 一个更加好的算法是限定枚举的点为圈内序号最大的点 这样就避免了对一个圈的多次枚举(参考程序3)
如果直接搜就是任选一个点开始走回到原点则记录长度 搜的时候必须要先对每个点的边按照边权进行排序 以备后面大量剪枝

程序1

 1 // Solution
 2 // by oyjpArt
 3 // Algorithm:Search
 4 #include  < vector >
 5 #include  < iostream >
 6 #include  < algorithm >
 7 using namespace std;
 8
 9 const   int  N  =   101 ;
10 struct Node { int  x, w; void  set ( int  xx,  int  ww) {x  = xx; w  =  ww; }};
11 vector < Node >  adj[N];
12 int  nv, ne, ans[N], na, S, rec[N];
13 bool chk[N];
14 int  best;
15
16 bool operator < ( const  Node &  a,  const  Node &  b) {
17  return a.w  <  b.w;
18 }
19
20 void search( int  x,  int  sum,  int  depth,  int  father) {
21   int  i;
22   if (x  ==  S  &&  chk[x]) {
23    if (sum  <  best) {
24    best  =  sum;  na  =  depth;
25     for (i  =   0 ; i  <  depth; i ++ ) ans[i]  =  rec[i];
26   }
27   return;
28  }
29  rec[depth]  =  x;
30   for (i  =   0 ; i  <  adj[x].size();  ++ i)  if (adj[x][i].x ! =  father)  if (!chk[adj[x][i].x] || adj[x][i].x  ==  S) {
31   chk[adj[x][i].x]  =   1 ;
32    if (sum  +  adj[x][i].w  <  best) search(adj[x][i].x, sum  +  adj[x][i].w, depth + 1 , x);
33   chk[adj[x][i].x]  =   0 ;
34  }
35 } 
36
37 int  main() {
38  scanf( " %d %d " ,  & nv,  & ne);
39   int  i, u, v, w;
40  Node  now ;
41   for (i  =   0 ; i  <  ne; i ++ ) {
42   scanf( " %d %d %d " ,  & u,  & v,  & w);
43    -- u;  -- v;
44    now .set(v, w);
45   adj[u].push_back( now );
46    now .x  =  u;
47   adj[v].push_back( now );
48  }
49   for (i  =   0 ; i  <  nv;  ++ i) 
50   sort(adj[i].begin(), adj[i].end());
51  
52  best  =   123456789 ;
53   for (i  =   0 ; i  <  nv;  ++ i) {
54   memset(chk,  0 , nv  *  sizeof(bool));
55   S  =  i;
56   search(i,  0 ,  0 ,  - 1 );
57  }
58
59   if (best  ==   123456789 ) { printf( " No solution.\n " ); return  0 ; }
60  printf( " %d " , ans[ 0 ] + 1 );
61   for (i  =   1 ; i  <  na;  ++ i) printf( "  %d " , ans[i] + 1 ); putchar( ' \n');
62
63  return  0 ;
64 }
65
66

程序2

 1 // Solution
 2 // by oyjpArt
 3 // Algorithm : Enumerate  +  Dijkstra
 4 #include  < stdio.h >
 5 #include  < string .h >
 6
 7 const   int  N  =   101 , M  =   20001 , MAXINT  =   2000000000 ;
 8 int  ne, nv;
 9 struct E {
10   int  x, w; E *   next ;
11  void  set ( int  xx,  int  ww, E *  nn) {x  =  xx; w  =  ww;  next   =  nn;}
12 }e[M],  *  head[N];
13 int  best, dist[N], q[N], ans[N], pre[N], na;
14 bool chk[N];
15
16 void Dijk( int  st,  int   end ,  int  ow) {
17  memset(chk,  0 , sizeof(chk));
18  memset(dist,  - 1 , sizeof(dist));
19   int  qe  =   1 , qs  =   0 , i;
20  E  *  p;
21   for (i  =   0 ; i  <  nv;  ++ i)  if (i ! =  st) {
22    for (p  =  head[st]; p ! =   NULL ; p  =  p -> next ) {
23     if (p -> x  ==  i  &&  p -> w  >   0   &&  (dist[i]  ==   - 1  || dist[i]  >  p -> w ) ) 
24     dist[i]  =  p -> w;
25   }
26    if (dist[i]  ==   - 1 ) dist[i]  =  MAXINT;
27  }
28  q[ 0 ]  =  st;
29  dist[st]  =   0 ;
30  chk[st]  =   1 ;
31   for (i  =   0 ; i  <  nv;  ++ i) pre[i]  =  st;
32  pre[st]  =   - 1 ;
33   while (qs  <  qe) {
34    int  cur  =  q[qs ++ ];
35   chk[cur]  =   1 ;
36    if (ow  +  dist[cur]  >=  best) return;
37    if (cur  ==   end ) {
38     if (dist[ end ]  +  ow  <  best) {
39     na  =   0 ;
40      for (i  =  cur; i ! =   - 1 ; i  =  pre[i]) ans[na ++ ]  =  i;
41     best  =  dist[ end ]  +  ow;
42    }
43    return;
44   }
45    int  _min  =  MAXINT, mini  =   - 1 ;
46    for (i  =   0 ; i  <  nv; i ++ )  if (!chk[i]) {
47     if (dist[i]  <  _min) {
48     _min  =  dist[i];
49     mini  =  i;
50    }
51   }
52    if (mini  ==   - 1 ) return;
53   q[qe ++ ]  =  mini;
54    for (i  =   0 ; i  <  nv;  ++ i)  if (!chk[i]) {
55     for (p  =  head[mini]; p ! =   NULL ; p  =  p -> next )  if (p -> x  ==  i)  break;
56     if (p  ==   NULL ) continue;
57     if (p -> w  >   0   &&  p -> w  +  dist[mini]  <  dist[i]) {
58     dist[i]  =  p -> w  +  dist[mini];
59     pre[i]  =  mini;
60    }
61   }
62  }
63 }
64
65 int  main() {
66  scanf( " %d %d " ,  & nv,  & ne);
67  memset(head,  NULL , nv  *  sizeof(E * ));
68   int  i, u, v, w;
69   for (i  =   0 ; i  <  ne;  ++ i) {
70   scanf( " %d %d %d " ,  & u,  & v,  & w);
71    -- u;  -- v;
72   e[ 2 * i].set(u, w, head[v]);
73   head[v]  =   & e[ 2 * i];
74   e[ 2 * i + 1 ].set(v, w, head[u]);
75   head[u]  =   & e[ 2 * i + 1 ];
76  }
77  E  *  p,  *  q;
78  best  =  MAXINT;
79   for (i  =   0 ; i  <  nv;  ++ i) {
80    for (p  =  head[i]; p ! =   NULL ; p  =  p -> next ) {
81     int  w  =  p -> w;
82     int  j  =  p -> x;
83     for (q  =  head[i]; q ! =   NULL ; q  =  q -> next )  if (q -> x  ==  j) q -> w  =   - q -> w;
84     for (q  =  head[j]; q ! =   NULL ; q  =  q -> next )  if (q -> x  ==  i) q -> w  =   - q -> w;
85    Dijk(i, j, w);
86     for (q  =  head[i]; q ! =   NULL ; q  =  q -> next )  if (q -> x  ==  j) q -> w  =   - q -> w;
87     for (q  =  head[j]; q ! =   NULL ; q  =  q -> next )  if (q -> x  ==  i) q -> w  =   - q -> w;
88   }
89  }
90   if (best  ==  MAXINT) printf( " No solution.\n " );
91   else  {
92   printf( " %d " , ans[ 0 ]  +   1 );
93    for (i  =   1 ; i  <  na;  ++ i) printf( "  %d " , ans[i]  +   1 ); putchar( ' \n');
94  }
95  return  0 ;
96 }
97 // 唉 不用vector代码量增大好多。。晕倒
98

程序3：
经wywcgs大牛提醒 改写成了Floyd程序 时间锐减

 1 #include  < stdio.h >
 2 #include  < string .h >
 3
 4 const   int  N  =   101 ;
 5 const   int  MAXINT  =   123456789 ;
 6 int  ne, nv;
 7 int  adj[N][N];
 8 int  pre[N][N];
 9 int  conn[N][N];
10 int  na, ans[N];
11 int  best;
12
13 void floyd() {
14      int  i, j, k, tmp, p;
15      for (k  =   0 ; k  <  nv;  ++ k) {
16          for (i  =   0 ; i  <  k;  ++ i) {
17              for (j  =   0 ; j  <  k;  ++ j)  if (conn[i][k]  &&  conn[k][j]  &&  j ! =  i) {
18                  if ( (tmp  =  adj[i][j]  +  conn[k][i]  +  conn[j][k])  <  best) {
19                     best  =  tmp;
20                     na  =   1 ; ans[ 0 ]  =  k; p  =  i;
21                      while (p ! =   - 1 ) {
22                         ans[na ++ ]  =  p;
23                         p  =  pre[p][j];
24                     }
25                 }
26             } 
27         }
28          for (i  =   0 ; i  <  nv;  ++ i) 
29              for (j  =   0 ; j  <  nv;  ++ j) {
30                  if (adj[i][j]  >  adj[i][k]  +  adj[k][j]) {
31                     adj[i][j]  =  adj[i][k]  +  adj[k][j];
32                     pre[i][j]  =  pre[i][k];
33                 }
34             }
35     }
36 }
37
38 int  main() {
39      int  i, j, u, v, w;
40     memset(pre,  - 1 , sizeof(pre));
41     scanf( " %d %d " ,  & nv,  & ne);
42      for (i  =   0 ; i  <  nv;  ++ i) {
43          for (j  =  i + 1 ; j  <  nv;  ++ j) 
44             adj[i][j]  =  adj[j][i]  =  MAXINT;
45         adj[i][i]  =   0 ;
46     }
47      for (i  =   0 ; i  <  ne;  ++ i) {
48         scanf( " %d %d %d " ,  & u,  & v,  & w);
49          -- u;  -- v;
50          if (w  <  adj[u][v])     
51             conn[u][v]  =  conn[v][u]  =  adj[u][v]  =  adj[v][u]  =  w;
52         pre[u][v]  =  v, pre[v][u]  =  u;
53     }
54     best  =  MAXINT;
55     floyd();
56      if (best  ==  MAXINT) printf( " No solution.\n " );
57      else  {
58          for (i  =   0 ; i  <  na;  ++ i) {
59             printf( " %d " , ans[i]  +   1 );
60              if (i ! =  na - 1 ) putchar( '  ');
61              else  putchar( ' \n');
62         }
63     }
64
65     return  0 ;
66 }
67