Bzoj3036: Home of the green frog & wikioi2488: Home of the green frog

3036: Time Limit: 2 sec memory limit: 128 MB
Submit: 108 solved: 73
[Submit] [Status] Description

With the release of the new version of Baidu space, the blog pet green frog has completed its mission to find its new home.

A directed acyclic graph is provided. The starting point is 1 and the ending point is N, and each edge has a length. The green frog starts from the starting point and ends.
When each vertex is reached, if K roads leave the vertex, the green frog can choose any road to leave the vertex, and the probability of going to each path is 1/K.
What is the total length of the path from the start point to the end point?

Input

The first line: two integers n m, representing n vertices and m edges in the graph.
Row 2 to row 1st + M: each row has three integers, a B c, representing a directed edge with a length of C from A to B.

Output

The expected value of the total length from the start point to the end point path is rounded to two decimal places.

 

Sample input4 4
1 2 1
1 3 2
2 3 3
3 4 4

Sample output7.00hint

 

For 100% of data, n <= 100000, m <= 2 * n

 

Source

Poetize3

Question: A simple expectation for DP is to push the code back from the end point:

 1 {$M 1000000000,0,maxlongint} 2 const maxn=100000+1000; 3 type node=record 4       next,go,w:longint; 5       end; 6 var e:array[0..2*maxn] of node; 7     outp,head:array[0..maxn] of longint; 8     f:array[0..maxn] of double; 9     i,n,m,x,y,z,j,tot:longint;10 procedure insert(x,y,z:longint);11  begin12    inc(tot);13    e[tot].go:=y;e[tot].w:=z;e[tot].next:=head[x];head[x]:=tot;14  end;15 procedure init;16  begin17    readln(n,m);18    for i:=1 to m do19     begin20       readln(x,y,z);insert(x,y,z);inc(outp[x]);21     end;22  end;23 procedure dfs(x:longint);24  var i,y:longint;25  begin26    if f[x]<>-1.0 then exit;27    f[x]:=0.0;28    i:=head[x];29    while i<>0 do30     begin31       y:=e[i].go;32       dfs(y);33       f[x]:=f[x]+f[y]+e[i].w;34       i:=e[i].next;35     end;36    if outp[x]<>0 then f[x]:=f[x]/outp[x];37  end;38 39 procedure main;40  begin41   for i:=1 to n do f[i]:=-1.0;42   dfs(1);43   writeln(f[1]:0:2);44  end;45 begin46   assign(input,‘input.txt‘);assign(output,‘output.txt‘);47   reset(input);rewrite(output);48   init;49   main;50   close(input);close(output);51 end.                            

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