Codeforces Round #279 (Div. 2) B. Queue（链表）

仔细观察可以发现，用链表的方法我们可以很容易的将偶数位和奇数位的答案补充完整。   只需要记录一个数的前驱和后驱就行了。

细节参见代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<numeric>
#include<functional>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<cassert>
#include<complex>
#include<iomanip>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int maxn = 200000+5;
int T,n,a[maxn],b[maxn],ans[maxn],vis[1000000+5],m,pre[1000000+5],last[1000000+5];
int main() {
    while(~scanf("%d",&n)) {
        memset(last,-1,sizeof(last));
        memset(pre,-1,sizeof(pre));
        for(int i=0;i<n;i++) {
            scanf("%d%d",&a[i],&b[i]);
            last[a[i]] = b[i];
            pre[b[i]] = a[i];
        }
        if(n % 2 == 0) {
            int v = last[0], res = 2;
            while(v > 0) {
                ans[res] = v;
                v = last[v];
                res+=2;
            }
            v = pre[0]; res = n-1;
            while(v > 0) {
                ans[res] = v;
                v = pre[v];
                res -= 2;
            }
        }
        else {
            memset(vis,0,sizeof(vis));
            int v = last[0], res = 2;
            vis[0] = 1;
            while(v > 0) {
                vis[v] = 1;
                ans[res] = v;
                v = last[v];
                res+=2;
            }
            for(int i=0;i<n;i++) {
                if(pre[a[i]] == -1) {
                    v = a[i]; res = 1; break;
                }
            }
            while(v > 0) {
                ans[res] = v;
                v = last[v];
                res += 2;
            }
        }
        printf("%d",ans[1]);
        for(int i=2;i<=n;i++) {
            printf(" %d",ans[i]);
        }
        printf("\n");
    }
    return 0;
}