ZOJ 3332Strange Country II(图论)

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http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3332

给出一个有向图，要求输出一条路可以一次经过每个点，每个点只能经过一次。乍一看以为是欧拉回路，仔细想想用dfs的递归层数来判断这条路上最长的线路，用ans保存路径

#include<iostream>#include<cstdio>#include<set>#include<string>#include<string.h>#include<cstring>#include<vector>#include<map>#include<queue>#include<stack>#include<cctype>#include<algorithm>#include<sstream>#include<utility>#include<cmath>#define mt(a) memset(a,0,sizeof (a))#define fl(a,b,c) fill(a,b,c)#define SWAP(a,b,t) (t=a,a=b,b=t)#define inf 1000000000+7using namespace std;typedef long long ll;vector<int>G[120];int vis[120];int n;int dis[120];int cot = 0;void  dfs(int node, int k){cot = max(cot, k);dis[k] = node;if (cot == n)return;vector<int>::iterator po;for (po = G[node].begin(); po != G[node].end(); po++){if (!vis[*po]){vis[*po] = 1;dfs(*po, k + 1);if (cot == n)return;vis[*po] = 0;}}}int main(){int T;cin >> T;while (T--){scanf("%d", &n);for (int i = 0; i < 120; i++)G[i].clear();mt(dis);int m = (n - 1)*n / 2;for (int i = 0; i < m; i++){int x, y;scanf("%d %d", &x, &y);G[x].push_back(y);}int flag = 0;for (int i = 1; i <= n; i++){cot = 1;memset(vis, 0, sizeof vis);vis[i] = 1;dfs(i, 1);if (cot == n){flag = 1;break;}}if (!flag)printf("Impossible\n");else{for (int i = 1; i <= n; i++){if (i != 1)printf(" ");printf("%d", dis[i]);}printf("\n");}}return 0;}

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