King Arthur's Knights —— 结题报告！
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http://acm.hdu.edu.cn/showproblem.php?pid=4337
King Arthur's Knights
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Special Judge
Problem Description
I am the bone of my sword. Steel is my body, and the fire is my blood.
- from Fate / Stay Night
You must have known the legend of King Arthur and his knights of the round table. The round table has no head, implying that everyone has equal status. Some knights are close friends with each other, so they prefer to sit next to each other.
Given the relationship of these knights, the King Arthur request you to find an arrangement such that, for every knight, his two adjacent knights are both his close friends. And you should note that because the knights are very united, everyone has at least half of the group as his close friends. More specifically speaking, if there are N knights in total, every knight has at least (N + 1) / 2 other knights as his close friends.
- from Fate / Stay Night
You must have known the legend of King Arthur and his knights of the round table. The round table has no head, implying that everyone has equal status. Some knights are close friends with each other, so they prefer to sit next to each other.
Given the relationship of these knights, the King Arthur request you to find an arrangement such that, for every knight, his two adjacent knights are both his close friends. And you should note that because the knights are very united, everyone has at least half of the group as his close friends. More specifically speaking, if there are N knights in total, every knight has at least (N + 1) / 2 other knights as his close friends.
Input
The first line of each test case contains two integers N (3 <= N <= 150) and M, indicating that there are N knights and M relationships in total. Then M lines followed, each of which contains two integers ai and bi (1 <= ai, bi <= n, ai != bi), indicating that knight ai and knight bi are close friends.
Output
For each test case, output one line containing N integers X1, X2, ..., XN separated by spaces, which indicating an round table arrangement. Please note that XN and X1 are also considered adjacent. The answer may be not unique, and any correct answer will be OK. If there is no solution exists, just output "no solution".
Sample Input
3 31 22 31 34 41 42 42 31 3
Sample Output
1 2 31 4 2 3
Source
2012 Multi-University Training Contest 4
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zhoujiaqi2010
其他大牛的结题报告，我到略看了下。
http://www.cnblogs.com/jiai/archive/2012/08/03/2621040.html
http://blog.sina.com.cn/s/blog_6ffc3bde01012lyj.html
看了大牛的结题报告，说是深搜，再略看了下，就自己敲了下。敲的途中，没看正确 的 结题报告，一路敲下来。样例过了，提交了几次，也wa 了几次，原因也找到了。
题目会做，就觉得不过如此，一个深搜，根本是是H图。
happy中！
加油↖(^ω^)↗！
//King Arthur's Knights#include <cstdio>#include <cstring>#include <vector>#include <iostream>#include <cstdio>using namespace std;vector <vector <int> >g;int a[200];int pos;int n, m;bool use[200];void init(){ g.clear(); g.resize(n + 1); pos = 0; memset(use, false, sizeof (use));}bool f(int x){ if (pos == n) { for (int j = g[1].size() - 1; j >= 0; j--) if (g[1][j] == x) return true; return false; } else { for (int j = g[x].size() - 1; j >= 0; j--) { if ( !use[ g[x][j] ]) { use[ g[x][j]] = true; a[pos++] = g[x][j]; if (f( g[x][j] )) return true; else { use[ g[x][j]] = false; pos--; } } } return false; }}int main(){ while (cin >> n >> m) { init(); { int u, v; for (int j = 0; j < m; j++) { scanf("%d%d", &u, &v); g[u].push_back(v); g[v].push_back(u); } pos = 1; use[1] = true; if (f(1)) { printf("1"); for(int j = 1; j < n; j++) printf(" %d", a[j]); printf("\n"); } else printf("no solution\n"); } } return 0;}
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