2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest Union of Doubly Linked Lists
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题目
题意: 已知每个节点i的前驱和后继(0表示没有), 合并成双向链表。
思路: 如果是双向链表必然有一个节点没有前驱,任意选一个没有前驱的点, 每次找到其路径上的尾部, 将下一个位置的头加入尾部。
#include <stdio.h>#include <string.h>#include <iostream>#include <vector>using namespace std;const int MAXN = 110;int l[MAXN], r[MAXN];vector <int> a;vector <int> e;int main(){ int n; scanf("%d", &n); a.clear(); e.clear(); for(int i=1;i<=n;++i) { scanf("%d %d", &l[i], &r[i]); if(l[i]==0) a.push_back(i); } for(int i=0;i<(int)a.size()-1;i++) { int k=a[i]; while(r[k]!=0) k=r[k]; r[k]=a[i+1]; l[a[i+1]]=k; } for(int i=1;i<=n;++i) printf("%d %d\n", l[i], r[i]); return 0;}
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