[POJ2253 Frogger]

[题目来源]：POJ2253

[关键字]：最短路径

[题目大意]：给出一个图，求从一号点到第二号点所走过的路径中最长的一条最短可以是多少。

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[分析]：其实就是最短路径的变形。只需把松弛条件改为d[j] > max(d[p],map[p,j])p为新加入的点，j为可更新的点，map[p,j]为p到j的路径长度，d[p]为到p所走过的最大路径。

[代码]：

View Code

 1 {
 2 PROB:POJ2253
 3 DATE:2011\10\15
 4 }
 5 program project1;
 6 var
 7   n, tc: longint;
 8   x, y: array[0..210] of longint;
 9   map: array[0..210,0..210] of real;
10   d: array[0..210] of real;
11   b: array[0..210] of boolean;
12 
13 procedure init;
14 var
15   i, j: longint;
16 begin
17   readln(n);
18   if n = 0 then halt;
19   for i := 1 to n do
20     readln(x[i],y[i]);
21   readln;
22   fillchar(map,sizeof(map),0);
23   for i := 1 to n do
24     for j := 1 to n do
25       map[i,j] := sqrt(sqr(x[i]-x[j])+sqr(y[i]-y[j]));
26 end;
27 
28 function max(x, y: real):real;
29 begin
30   if x > y then exit(x) else exit(y);
31 end;
32 
33 procedure dij;
34 var
35   i, j, p: longint;
36   min: real;
37 begin
38   fillchar(d,sizeof(d),100);
39   fillchar(b,sizeof(b),false);
40   d[1] := 0;
41   for i := 1 to n do
42     begin
43       min := maxlongint;
44       for j := 1 to n do
45         if (not b[j]) and (min > d[j]) then
46           begin
47             min := d[j];
48             p := j;
49           end;
50       b[p] := true;
51       for j := 1 to n do
52         if (not b[j]) and (map[p,j] <> 0) then
53           if d[j] > max(map[p,j],d[p]) then d[j] := max(map[p,j],d[p]);
54     end;
55 end;
56 
57 begin
58 //  assign(input,'d:\1.in');reset(input);
59 //  assign(output,'d:\1.out');rewrite(output);
60   tc := 0;
61   while 1 = 1 do
62     begin
63       inc(tc);
64       init;
65       dij;
66       writeln('Scenario #',tc);
67       writeln('Frog Distance = ',d[2]:0:3);
68       writeln;
69     end;
70 //  close(input);
71 //  close(output);
72 end.