Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=2181
The question of Hamilton's detour around the world
Time limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 2362 Accepted Submission (s): 1490
Problem description A regular solid 12-face body, its 20 vertices mark the world-famous 20 cities, you start from a city after each city just once back to the departure city.
Line I of the first 20 lines of input has 3 numbers representing 3 cities adjacent to the first city. After line 20th, there are 1 numbers m,m<=20,m>=1.m=0 exit.
Output outputs from the first M cities go through every city 1 times and back to m all routes, if there are multiple routes, output in dictionary order, 1 lines per line. The first output per line is the number of routes. Then one: After listing the cities that pass. See sample output
Sample INPUT2 5 201 3 122 4 103 5 81 4 65 7 196 8 174 7 98 10 163 9 1110 12 152 11 1312 14 2013 15 1811 14 169 15 177 16 1 814 17 196 18 201 13 1950
Sample Output1:5 1 2 3 4 8 7 17 18 14 15 16 9 10 11 12 13 20 19 6 52:5 1 2 3 4 8 9 10 11 12 13 20 19 18 14 15 16 17 7 6 53:5 1 2 3 10 9 16 17 18 14 15 11 12 13 20 19 6 7 8 4 54:5 1 2 3 10 11 12 13 20 19 6 7 17 18 14 15 16 9 8 4 55:5 1 2 12 11 10 3 4 8 9 16 15 14 13 20 19 18 17 7 6 56:5 1 2 12 11 15 14 13 20 19 18 17 16 9 10 3 4 8 7 6 57:5 1 2 12 11 15 16 9 10 3 4 8 7 17 18 14 13 20 19 6 58:5 1 2 12 11 15 16 17 18 14 13 20 19 6 7 8 9 10 3 4 59:5 1 2 12 13 20 19 6 7 8 9 16 17 18 14 15 11 10 3 4 510:5 1 2 12 13 20 19 18 14 15 11 10 3 4 8 9 16 17 7 6 511:5 1 20 13 12 2 3 4 8 7 17 16 9 10 11 15 14 18 19 6 512:5 1 20 13 12 2 3 10 11 15 14 18 19 6 7 17 16 9 8 4 513:5 1 20 13 14 15 11 12 2 3 10 9 16 17 18 19 6 7 8 4 5 14:5 1 20 13 14 15 16 9 10 11 12 2 3 4 8 7 17 18 19 6 515:5 1 20 13 14 15 16 17 18 19 6 7 8 9 10 11 12 2 3 4 516:5 1 20 13 14 18 19 6 7 17 16 15 11 12 2 3 10 9 8 4 517:5 1 20 19 6 7 8 9 10 11 15 16 17 18 14 13 12 2 3 4 518:5 1 20 19 6 7 17 18 14 13 12 2 3 10 11 15 16 9 8 4 519:5 1 20 19 18 14 13 12 2 3 4 8 9 10 11 15 16 17 7 6 520:5 1 20 19 18 17 16 9 10 11 15 14 13 12 2 3 4 8 7 6 521:5 4 3 2 1 20 13 12 11 10 9 8 7 17 16 15 14 18 19 6 522:5 4 3 2 1 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 523:5 4 3 2 12 11 10 9 8 7 6 19 18 17 16 15 14 13 20 1 524:5 4 3 2 12 13 14 18 17 16 15 11 10 9 8 7 6 19 20 1 525:5 4 3 10 9 8 7 6 19 20 13 14 18 17 16 15 11 12 2 1 526:5 4 3 10 9 8 7 17 16 15 11 12 2 1 20 13 14 18 19 6 527:5 4 3 10 11 12 2 1 20 13 14 15 16 9 8 7 17 18 19 6 528:5 4 3 10 11 15 14 13 12 2 1 20 19 18 17 16 9 8 7 6 529:5 4 3 10 11 15 14 18 17 16 9 8 7 6 19 20 13 12 2 1 530:5 4 3 10 11 15 16 9 8 7 17 18 14 13 12 2 1 20 19 6 531:5 4 8 7 6 19 18 17 16 9 10 3 2 12 11 15 14 1 3 20 1 532:5 4 8 7 6 19 20 13 12 11 15 14 18 17 16 9 10 3 2 1 533:5 4 8 7 17 16 9 10 3 2 1 20 13 12 11 15 14 18 19 6 534 : 5 4 8 7 17 18 14 13 12 11 15 16 9 10 3 2 1 20 19 6 535:5 4 8 9 10 3 2 1 20 19 18 14 13 12 11 15 16 17 7 6 536:5 4 8 9 10 3 2 12 11 15 16 17 76 19 18 14 13 20 1 537:5 4 8 9 16 15 11 10 3 2 12 13 14 18 17 7 6 19 20 1 538:5 4 8 9 16 15 14 13 12 11 10 3 2 1 20 19 1 8 17 7 6 539:5 4 8 9 16 15 14 18 17 7 6 19 20 13 12 11 10 3 2 1 540:5 4 8 9 16 17 7 6 19 18 14 15 11 10 3 2 12 13 20 1 5 41:5 6 7 8 4 3 2 12 13 14 15 11 10 9 16 17 18 19 20 1 542:5 6 7 8 4 3 10 9 16 17 18 19 20 13 14 15 11 12 2 1 543:5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 544:5 6 7 8 9 16 17 18 19 20 1 2 12 13 14 15 11 10 3 4 545:5 6 7 17 16 9 8 4 3 10 11 15 14 18 19 20 13 12 2 1 546:5 6 7 17 16 15 11 10 9 8 4 3 2 12 13 14 18 19 20 1 547:5 6 7 17 16 15 11 12 13 14 18 19 20 1 2 3 10 9 8 4 548:5 6 7 17 16 15 14 18 19 20 13 12 11 10 9 8 4 3 2 1 549:5 6 7 17 18 19 20 1 2 3 10 11 12 13 14 15 16 9 8 4 550:5 6 7 17 18 19 20 13 14 15 16 9 8 4 3 10 11 12 2 1 551:5 6 19 18 14 13 20 1 2 12 11 15 16 17 7 8 9 10 3 4 552:5 6 19 18 14 15 11 10 9 16 17 7 8 4 3 2 12 13 20 1 553:5 6 19 18 14 15 11 12 13 20 1 2 3 10 9 16 17 7 8 4 554: 5 6 19 18 14 15 16 17 78 9 10 11 12 13 20 1 2 3 4 555:5 6 19 18 17 7 8 4 3 2 12 11 10 9 16 15 14 13 20 1 556:5 6 19 18 17 7 8 9 16 15 14 13 20 1 2 12 11 10 3 4 557:5 6 19 20 1 2 3 10 9 16 15 11 12 13 14 18 17 7 8 4 558:5 6 19 20 1 2 12 13 14 18 17 7 8 9 16 15 11 10 3 4 559:5 6 19 20 13 12 11 10 9 16 15 14 18 17 7 8 4 3 2 1 560:5 6 19 20 13 14 18 17 7 8 4 3 10 9 16 15 11 12 2 1 5 : DFS record path to use test instructions skillfully, this problem gives a fixed number of points, then the statistics of the time direct for (int i = 0; i <; i++) can record 20 points of the father, if you do not know how the path on the last point, Then, by other features to judge the end of the regression from the parent node and then the dictionary order, the matrix storage edge is the best way to automatically use the dictionary order
1#include <cstdio>2#include <cstring>3#include <string>4#include <iostream>5#include <algorithm>6 using namespacestd;7 #defineN 258 intMp[n][n];9 BOOLVis[n];Ten intFa[n]; One intAns[n]; A voidDfsintTmintSintSumint&CNT) - { - if(sum== - ){ the if(!mp[s][tm])return; -printf"%d:", CNT); - intf =s; - //int c = 1; + for(inti =1; i< -; i++){ -Ans[i] =Fa[f]; +f =Fa[f]; A } at for(inti = +; i >0; i--){ -printf"%d", Ans[i]); - } -printf"%d%d\n", S,TM); -CNT = cnt+1; - return; in } - for(inti =1; I <= -; i++){ to if(!vis[i]&&Mp[s][i]) { + //puts ("haha"); -Vis[i] =1; theFa[i] =s; *DFS (tm,i,sum+1, CNT); $Vis[i] =0;Panax Notoginseng } - } the return; + } A intMain () the { +Memset (MP,0,sizeof(MP)); - for(inti =1; I <= -; i++) $ { $ intx, y, Z; -scanf"%d%d%d",&x,&y,&z); -MP[I][X] = Mp[x][i] =1; theMp[i][y] = Mp[y][i] =1; -MP[I][Z] = Mp[z][i] =1;Wuyi } the ints; - while(~SCANF ("%d",&s)) Wu { - if(s==0)return 0; Aboutmemset (Vis,0,sizeof(Vis)); $memset (FA,0,sizeof(FA)); -memset (ans,0,sizeof(ans)); - intCNT =1; -Vis[s] =1; AFa[s] =s; +DFS (S,s,1, CNT); the } - return 0; $}
Hamiltonian bypass World problem (dfs+ record path)
