PAT 1097. Deduplication on a Linked List (25)

地址： http://www.patest.cn/contests/pat-a-practise/1097

没什么好说的， 我逗比就是了

  1 #include<iostream>
  2 #include<iomanip>
  3 #include<cstdio>
  4 #include<cstring>
  5 #include<map>
  6 #include<algorithm>
  7 
  8 using namespace std;
  9 
 10 const int MAX_NODES = 100000 + 10;
 11 struct List
 12 {
 13     int next[MAX_NODES];
 14     int last;
 15     int cur;
 16 
 17     List():last(0),cur(0)
 18     {
 19         memset(next, 0xff, sizeof(next));
 20     }
 21 
 22     void add_node(int position)
 23     {
 24         next[position] = next[last];
 25         next[last] = position;
 26         last = position;
 27     }
 28 }source, splited;
 29 
 30 const int MAX_NUM = 10000 + 10;
 31 const int NUL = -1;
 32 bool num_hash[MAX_NUM];
 33 
 34 struct Node
 35 {
 36     int addr;
 37     int key;
 38     int next_addr;
 39 
 40     inline bool operator < (const Node &other) const
 41     {
 42         return (*this).addr < other.addr;
 43     }
 44 } nodes[MAX_NODES];
 45 int addr2int[1000000 + 10];
 46 int head_addr;
 47 int n;
 48 
 49 void split()
 50 {
 51     memset(num_hash, 0, sizeof(num_hash));
 52     int pre = 0;
 53     for (int i = source.next[0]; i != -1; i = source.next[i])
 54     {
 55         int num = abs(nodes[i-1].key);
 56 
 57         if (!num_hash[num])
 58         {
 59             num_hash[num] = true;
 60             pre = i;
 61         }
 62         else
 63         {
 64             source.next[pre] = source.next[i];
 65             splited.add_node(i);
 66             i = pre;
 67         }
 68     }
 69 }
 70 void print()
 71 {
 72     for (int i = source.next[0]; i != -1; i = source.next[i])
 73     {
 74         cout << setw(5) << setfill('0') << nodes[i-1].addr << " ";
 75         cout << nodes[i-1].key << " ";
 76         if (source.next[i] == -1) cout << -1 << endl;
 77         else cout << setw(5) << setfill('0') <<  nodes[source.next[i] - 1].addr << endl;
 78     }
 79     for (int i = splited.next[0]; i != -1; i = splited.next[i])
 80     {
 81         cout << setw(5) << setfill('0') << nodes[i-1].addr << " ";
 82         cout << nodes[i-1].key << " ";
 83         if (splited.next[i] == -1)  cout << -1 << endl; 
 84         else cout << setw(5) << setfill('0') <<  nodes[splited.next[i] - 1].addr << endl;
 85     }
 86 }
 87 
 88 int main()
 89 {
 90     while(cin >> head_addr >> n)
 91     {
 92         for (int i = 0; i < n; ++i)
 93         {
 94             cin >> nodes[i].addr >> nodes[i].key >> nodes[i].next_addr;
 95         }
 96         sort(nodes, nodes + n);
 97         for (int i = 0; i < n; ++i)
 98         {
 99             addr2int[nodes[i].addr] = i + 1;
100         }
101         int position = addr2int[head_addr];
102         for (int i = 0; i < n; ++i)
103         {
104             source.add_node(position);
105             if (nodes[position - 1].next_addr == -1) break;
106             position = addr2int[nodes[position - 1].next_addr];
107         }
108         split();
109         print();
110     }
111     return 0;
112 }

 

posted @ 2015-04-09 22:08  ACSeed  Views(245)  Comments(0)    收藏  举报