给一个有向图，顶点数少于50，有Q条询问（Q<=100000），对于每条询问（x1 y1），输出他们之间的最小密度路径的密度（长度比边数）
两种实践方法，一个是FLOYD，再就是二维SPFA，注意数据有环，SPFA时要控制边数少于n才行，至于为什么，是困扰几代人的烦恼了。
d[i,j,k]表示从i到j走过k条路径的最小路径长度
floyd
View Code
1 program path(input,output);
2 var
3 i,j,k,l,m,n,w,p,x,y:longint;
4 max,tmp:real;
5 f:array[0..51,0..51,0..51] of longint;
6 begin
7 assign(input,'path.in');
8 reset(input);
9 assign(output,'path.out');
10 rewrite(output);
11 readln(n,m);
12 fillchar(f,sizeof(f),255);
13 for i:=1 to m do
14 begin
15 readln(x,y,w);
16 if (f[x,y,1]=-1) or (f[x,y,1]>w) then
17 f[x,y,1]:=w;
18 end;
19 for i:=1 to n do f[i,i,0]:=0;
20 for l:=2 to n do
21 for k:=1 to n do
22 for i:=1 to n do
23 for j:=1 to n do
24 if (f[i,k,l-1]>-1) and (f[k,j,1]>-1) and
25 ((f[i,j,l]=-1) or (f[i,k,l-1]+f[k,j,1]<f[i,j,l])) then
26 f[i,j,l]:=f[i,k,l-1]+f[k,j,1];
27 readln(p);
28 for p:=1 to p do
29 begin
30 readln(x,y);
31 max:=100110001;
32 for i:=0 to n do
33 if (f[x,y,i]<>-1) and (f[x,y,i]<max*i) then max:=f[x,y,i]/i;
34 if max<>100110001 then writeln(max:0:3)
35 else writeln('OMG!');
36 end;
37 close(input);
38 close(output);
39 end.
SPFA
View Code
1 program path(input,output);
2 type
3 node = ^link;
4 link = record
5 goal,w : longint;
6 next : node;
7 end;
8 node2 = record
9 now : integer;
10 step : longint;
11 end;
12 var
13 l : array[0..100] of node;
14 v : array[0..60,0..2010] of boolean;
15 d : array[0..60,0..60,0..2010] of longint;
16 g : array[0..60,0..60] of boolean;
17 q : array[0..100010] of node2;
18 answer : array[0..60,0..60] of double;
19 n,m,qq : longint;
20 procedure add(xx,yy,ww: longint );
21 var
22 tt : node;
23 begin
24 new(tt);
25 tt^.w:=ww;
26 tt^.goal:=yy;
27 tt^.next:=l[xx];
28 l[xx]:=tt;
29 end; { add }
30 procedure init;
31 var
32 i,j,k,xxx,yyy,www : longint;
33 begin
34 readln(n,m);
35 for i:=1 to n do
36 l[i]:=nil;
37 for i:=1 to n do
38 for j:=1 to n do
39 for k:=1 to m do
40 d[i,j,k]:=maxlongint>>2;
41 for i:=1 to n do
42 for j:=1 to n do
43 if i=j then
44 answer[i,j]:=0
45 else
46 answer[i,j]:=-1;
47 fillchar(g,sizeof(g),false);
48 for i:=1 to m do
49 begin
50 readln(xxx,yyy,www);
51 add(xxx,yyy,www);
52 g[xxx,yyy]:=true;
53 end;
54 for k:=1 to n do
55 for i:=1 to n do
56 for j:=1 to n do
57 g[i,j]:=g[i,j] or (g[i,k] and g[k,j]);
58 end; { init }
59 procedure spfa(vo :longint );
60 var
61 head,tail : longint;
62 t : node;
63 begin
64 fillchar(v,sizeof(v),false);
65 head:=0;
66 tail:=1;
67 d[vo,vo,0]:=0;
68 v[vo,0]:=true;
69 q[1].step:=0;
70 q[1].now:=vo;
71 while head<tail do
72 begin
73 inc(head);
74 v[q[head].now,q[head].step]:=false;
75 if q[head].step=n then
76 continue;
77 t:=l[q[head].now];
78 while t<>nil do
79 begin
80 if d[vo,q[head].now,q[head].step]+t^.w<d[vo,t^.goal,q[head].step+1] then
81 begin
82 d[vo,t^.goal,q[head].step+1]:=d[vo,q[head].now,q[head].step]+t^.w;
83 if not v[t^.goal,q[head].step+1] then
84 begin
85 inc(tail);
86 v[t^.goal,q[head].step+1]:=true;
87 q[tail].now:=t^.goal;
88 q[tail].step:=q[head].step+1;
89 end;
90 end;
91 t:=t^.next;
92 end;
93 end;
94 end; { spfa }
95 procedure main;
96 var
97 i,j,x1,y1 : longint;
98 begin
99 for i:=1 to n do
100 spfa(i);
101 readln(qq);
102 for i:=1 to qq do
103 begin
104 readln(x1,y1);
105 if not g[x1,y1] then
106 begin
107 writeln('OMG!');
108 continue;
109 end;
110 if answer[x1,y1]>=0 then
111 begin
112 writeln(answer[x1,y1]:0:3);
113 continue;
114 end;
115 answer[x1,y1]:=999999999;
116 for j:=1 to m do
117 if d[x1,y1,j]<(maxlongint>>2) then
118 if (d[x1,y1,j]/j)<answer[x1,y1] then
119 answer[x1,y1]:=d[x1,y1,j]/j;
120 writeln(answer[x1,y1]:0:3);
121 end;
122 end; { main }
123 begin
124 assign(input,'path.in');reset(input);
125 assign(output,'path.out');rewrite(output);
126 init;
127 main;
128 close(input);
129 close(output);
130 end.