[POJ1364 King]
[题目来源]：POJ1364
[关键字]：差分约束系统
[题目大意]：给出一些不等式，判断能否同时成立。
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[分析]：依据已给出的不等式构建差分约束系统，然后Bellman—Folyd判断是否有环。
[代码]：
View Code
1 program Project1;
2 type
3 rec = record
4 x, y, d: longint;
5 end;
6 var
7 n, m, st: longint;
8 d: array[0..300] of longint;
9 e: array[0..300] of rec;
10
11 procedure make(i, x, y, d: longint);
12 begin
13 e[i].x := x;
14 e[i].y := y;
15 e[i].d := d;
16 end;
17
18 procedure init;
19 var
20 i, x, y, d: longint;
21 ch, ch2: char;
22 begin
23 st := 9999999;
24 for i := 1 to m do
25 begin
26 read(x,y);
27 read(ch,ch2,ch,ch);
28 readln(d);
29 //writeln(x,' ',ch2,' ',y,' ',d);
30 case ch2 of
31 'g':make(i,x-1,x+y,d+1);
32 'l':make(i,x+y,x-1,-d+1);
33 end;
34 if st > x-1 then st := x-1;
35 end;
36 end;
37
38 function bellman:boolean;
39 var
40 i, j: longint;
41 f: boolean;
42 begin
43 fillchar(d,sizeof(d),100);
44 d[st] := 0;
45 for i := 1 to n do
46 begin
47 f := true;
48 for j := 1 to m do
49 if d[e[j].y] < d[e[j].x]+e[j].d then
50 begin
51 d[e[j].y] := d[e[j].x]+e[j].d;
52 f := false;
53 end;
54 if f then break;
55 end;
56 for i := 1 to m do
57 if d[e[i].y] < d[e[i].x]+e[i].d then exit(false);
58 exit(true);
59 end;
60
61 begin
62 while 1 = 1 do
63 begin
64 read(n);
65 if n = 0 then break;
66 readln(m);
67 init;
68 if bellman then writeln('lamentable kingdom') else writeln('successful conspiracy');
69 end;
70 end.