[POJ3274 Gold Balanced Lineup]

[题目来源]：POJ3274

[关键字]：hash

[题目大意]：用一个十进制整数的二进制代表每个奶牛的特征（右往左数第i为为1是有0没有），给出一个奶牛序列找到一个最长的连续满足：此序列中所有奶牛的各个特征和相等。

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[分析]：以样例为例：

111

110

111

010

001

100

010

累加后

111

221

332

342

343

443

453

到此还比较好想接着，都减去最右边的数

000

110<=i

110

120

010

110<=j

120

到此可以看出规律，现求出以上序列，只要查找两个序列相的且使j-i最大，就是答案——hash查找BKDRHash。

[代码]：

View Code

  1 program Project1;
  2 type
  3   arr = array[0..30] of longint;
  4   rec = record
  5     dat: arr;
  6     num: longint;
  7   end;
  8 var
  9   n, m, ans: longint;
 10   a, c: array[0..100010] of arr;
 11   num: array[0..15988] of longint;
 12   h: array[0..15988,0..10] of rec;
 13 
 14 function hash(a: arr):longint;
 15 var
 16   seek, i: longint;
 17 begin
 18   seek := 131;
 19   hash := 0;
 20   for i := 1 to m do
 21     hash := (hash*seek+a[i]) and $FFFFFFFF;
 22   hash := hash and $7FFFFFFF;
 23   hash := hash mod 15988;
 24   if hash < 0 then hash := hash+15988;
 25 end;
 26 
 27 function change_two(x: longint):arr;
 28 var
 29   temp, i: longint;
 30   b: arr;
 31 begin
 32   temp := 0;
 33   while x <> 0 do
 34     begin
 35       inc(temp);
 36       b[temp] := x mod 2;
 37       x := x div 2;
 38     end;
 39   for i := temp+1 to m do b[i] := 0;
 40   for i := 1 to m div 2 do
 41     begin
 42       temp := b[i];
 43       b[i] := b[m-i+1];
 44       b[m-i+1] := temp;
 45     end;
 46   exit(b);
 47 end;
 48 
 49 function cleck(a, b: arr):boolean;
 50 var
 51   i: longint;
 52 begin
 53   for i := 1 to m do
 54     if a[i] <> b[i] then exit(false);
 55   exit(true);
 56 end;
 57 
 58 procedure init;
 59 var
 60   i, j, k, p, r: longint;
 61 begin
 62   readln(n,m);
 63   ans := 0;
 64   for i := 1 to m do
 65     c[0][i] := 0;
 66   k := hash(c[0]);
 67   p := 0;
 68   for j := 1 to num[k] do
 69     if cleck(c[0],h[k,j].dat) then
 70       if 0-h[k,j].num > p then p := 0-h[k,j].num;
 71   if (p <> 0) and (p > ans) then ans := p;
 72   if p = 0 then
 73     begin
 74       inc(num[k]);
 75       h[k,num[k]].dat := c[0];
 76       h[k,num[k]].num :=0;
 77     end;
 78   for i := 1 to n do
 79     begin
 80       readln(r);
 81       a[i] := change_two(r);
 82       {for j := 1 to m do
 83         write(a[i][j]);
 84       writeln;}
 85       for j := 1 to m do
 86         begin
 87           c[i][j] := c[i-1][j]+a[i][j]-a[i][m];
 88           //write(c[i][j]);
 89         end;
 90       k := hash(c[i]);
 91       p := 0;
 92       for j := 1 to num[k] do
 93         if cleck(c[i],h[k,j].dat) then
 94           if i-h[k,j].num > p then p := i-h[k,j].num;
 95       if (p <> 0) and (p > ans) then ans := p;
 96       if p = 0 then
 97         begin
 98           inc(num[k]);
 99           h[k,num[k]].dat := c[i];
100           h[k,num[k]].num := i;
101         end;
102     end;
103   writeln(ans);
104   //readln;
105   //readln;
106 end;
107 
108 begin
109   init;
110 end.