[POJ2513 Colored Sticks]

[题目来源]：POJ2513

[关键字]：字典树 并查集 欧拉路

[题目大意]：给定许多根木棒，两边分别涂有不同颜色，问能否将他们连成一条直线。规定只能将相同颜色的两端相连。

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[分析]：把木棒两端看成节点，将木棒看成边便构成了一个无向图，题目要求便转换成了问这个图里是否有一条欧拉路。可以先用字典树求出各个颜色出现的次数，即节点的度。然后判断欧拉路成立的前提条件：奇数度的节点只能有0或2个。再用并查集判断是否是一个连通图，即所有点在并查集的祖先唯一。根据这些即可判断欧拉路是否存在。

[代码]：

View Code

  1 program Project1;
  2 type
  3   rec = record
  4     t: boolean;
  5     num: longint;
  6     next: array['a'..'z'] of longint;
  7   end;
  8   arr = array[0..250000] of string[10];
  9   arrr = array[0..500000] of longint;
 10 var
 11   numx, numy, f, r: arrr;
 12   x, y: arr;
 13   tree: array[0..1000000] of rec;
 14   tn, tot, n: longint;
 15 
 16 procedure print(x: longint);
 17 begin
 18   case x of
 19     1: writeln('Possible');
 20     -1: writeln('Impossible');
 21   end;
 22   halt;
 23 end;
 24 
 25 function get(k: longint):longint;
 26 begin
 27   if f[k] = k then exit(k);
 28   f[k] := get(f[k]);
 29   get := f[k];
 30 end;
 31 
 32 procedure init;
 33 var
 34   s: string;
 35   i: longint;
 36 begin
 37   n := 0;
 38   while not seekeof do
 39     begin
 40       inc(n);
 41       readln(s);
 42       x[n] := copy(s,1,pos(' ',s)-1);
 43       delete(s,1,pos(' ',s));
 44       y[n] := s;
 45     end;
 46   if n = 0 then print(1);
 47   //for i := 1 to n do writeln(x[i],' ',y[i]);
 48 end;
 49 
 50 procedure work;
 51 var
 52   i, j, now, temp, xx, yy, root: longint;
 53 begin
 54   fillchar(numx,sizeof(numx),0);
 55   fillchar(numy,sizeof(numy),0);
 56   for i := 1 to n do
 57     begin
 58       now := 1;
 59       for j := 1 to length(x[i]) do
 60         if tree[now].next[x[i][j]] <> 0 then now := tree[now].next[x[i][j]]
 61           else
 62             begin
 63               inc(tn);
 64               tree[now].next[x[i][j]] := tn;
 65               now := tn;
 66             end;
 67       if tree[now].t then
 68         begin
 69           numx[i] := tree[now].num;
 70           inc(r[numx[i]]);
 71         end
 72       else
 73         begin
 74           tree[now].t := true;
 75           inc(tot);
 76           numx[i] := tot;
 77           tree[now].num := tot;
 78           r[tot] := 1;
 79         end;
 80     end;
 81   for i := 1 to n do
 82     begin
 83       now := 1;
 84       for j := 1 to length(y[i]) do
 85         if tree[now].next[y[i][j]] <> 0 then now := tree[now].next[y[i][j]]
 86           else
 87             begin
 88               inc(tn);
 89               tree[now].next[y[i][j]] := tn;
 90               now := tn;
 91             end;
 92       if tree[now].t then
 93         begin
 94           numy[i] := tree[now].num;
 95           inc(r[numy[i]]);
 96         end
 97       else
 98         begin
 99           tree[now].t := true;
100           inc(tot);
101           numy[i] := tot;
102           tree[now].num := tot;
103           r[tot] := 1;
104         end;
105     end;
106   {writeln(tot);
107   for i := 1 to n do writeln(numx[i],' ',numy[i]);
108   for i := 1 to tot do writeln(r[i]);}
109 //=========================================
110   temp := 0;
111   for i := 1 to tot do
112     if odd(r[i]) then inc(temp);
113   if not ((temp = 0) or (temp = 2)) then print(-1);
114   for i := 1 to tot do f[i] := i;
115   for i := 1 to n do
116     begin
117       xx := get(numx[i]);
118       yy := get(numy[i]);
119       if xx <> yy then f[xx] := yy;
120     end;
121 //=========================================
122   root := get(1);
123   for i := 1 to tot do
124     if get(i) <> root then print(-1);
125   print(1);
126 end;
127 
128 begin
129   //assign(input,'d:\1.in');reset(input);
130   //assign(output,'d:\1.out');rewrite(output);
131   init;
132   work;
133   //close(input);
134   //close(output);
135 end.