[POJ2060 Taxi Cab Scheme]

[题目来源]：POJ2060

[关键字]：二分图最小路径覆盖

[题目大意]：给出一系列出租车客人的时间表，问最少需多少人才能满足每一位顾客。

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[分析]：如果以每个顾客时间看成一个点，一个出租车在它从出发地出发到到达下一个顾客开始第所用时间满足约束条件，则就在这两个任务间连一条边。此时需要求的就是要覆盖所有的点所需的最小的路径数=最大独立集=所有点-最大匹配。

[代码]：

View Code

  1 type
  2   rec = record
  3     sx, sy, ex, ey: longint;
  4   end;
  5 var
  6   tc, n: longint;
  7   map: array[0..1010,0..1010] of boolean;
  8   b: array[0..1000] of boolean;
  9   t: array[0..1000] of string[30];
 10   link: array[0..1000] of longint;
 11   a: array[0..2020] of rec;
 12 
 13 function cha(s1, s2: string):longint;
 14 var
 15   h1, h2, m1, m2, sum: longint;
 16 begin
 17   h1 := (ord(s1[1])-48)*10+ord(s1[2])-48;
 18   h2 := (ord(s2[1])-48)*10+ord(s2[2])-48;
 19   m1 := (ord(s1[4])-48)*10+ord(s1[5])-48;
 20   m2 := (ord(s2[4])-48)*10+ord(s2[5])-48;
 21   //writeln(s1,' ',s2,' ',h1,' ',h2,' ',m1,' ',m2);
 22   sum := (h2-h1)*60+m2-m1;
 23   exit(sum);
 24 end;
 25 
 26 procedure init;
 27 var
 28   i, j, t1, t2: longint;
 29   s: string;
 30 begin
 31   readln(n);
 32   for i := 1 to n do
 33     begin
 34       readln(s);
 35       t[i] := copy(s,1,pos(' ',s)-1);
 36       delete(s,1,pos(' ',s));
 37       val(copy(s,1,pos(' ',s)-1),a[i].sx);
 38       delete(s,1,pos(' ',s));
 39       val(copy(s,1,pos(' ',s)-1),a[i].sy);
 40       delete(s,1,pos(' ',s));
 41       val(copy(s,1,pos(' ',s)-1),a[i].ex);
 42       delete(s,1,pos(' ',s));
 43       val(s,a[i].ey);
 44       //writeln(t[i],' ',a[i].sx,' ',a[i].sy,' ',a[i].ex,' ',a[i].ey);
 45     end;
 46   fillchar(map,sizeof(map),false);
 47   for i := 1 to n do
 48     for j := i+1 to n do
 49       begin
 50         t1 := cha(t[i],t[j]);
 51         t2 := abs(a[i].sx-a[i].ex)+abs(a[i].sy-a[i].ey);
 52         t2 := t2+abs(a[i].ex-a[j].sx)+abs(a[i].ey-a[j].sy);
 53         //writeln(i,' ',j,' ',t1,' ',t2);
 54         if t2 <= t1-1 then
 55           begin
 56             map[i,j] := true;
 57            // map[j,i] := true;
 58           end;
 59       end;
 60   {for i := 1 to n do
 61    for j := 1 to n do
 62      writeln(i,' ',j,' ',map[i,j]);}
 63 end;
 64 
 65 function dfs(k: longint):boolean;
 66 var
 67   i: longint;
 68 begin
 69   for i := 1 to n do
 70     if (map[k,i]) and (not b[i]) then
 71       begin
 72         b[i] := true;
 73         if (link[i] = 0) or (dfs(link[i])) then
 74           begin
 75             link[i] := k;
 76             exit(true);
 77           end;
 78       end;
 79   exit(false);
 80 end;
 81 
 82 procedure work;
 83 var
 84   i, max: longint;
 85 begin
 86   max := 0;
 87   fillchar(link,sizeof(link),0);
 88   for i := 1 to n do
 89     begin
 90       fillchar(b,sizeof(b),false);
 91       if dfs(i) then inc(max);
 92     end;
 93   //writeln(n,' ',max);
 94   writeln(n-max);
 95 end;
 96 
 97 begin
 98   readln(tc);
 99   while tc > 0 do
100     begin
101       init;
102       work;
103       dec(tc);
104     end;
105 end.