Poj 2531 network saboteur solution report (randomization algorithm)

Source: poj 2531 network saboteur

Http://acm.pku.edu.cn/JudgeOnline/problem? Id = 2531

 

Solution Type: randomizationAlgorithm

 

Author: Liu yaning

 

Question:

Divide a complete graph into two parts to maximize the Edge Weight of the two parts.

 

Solution:

Randomly change the position of a point, calculate the weight, repeat 200000 times, and take the maximum value output.

 

Submission:

1. Wrong answer multiple times: the number of random times is insufficient.
2. Time limit exceeded multiple times. Because the positions of all vertices are randomly allocated each time, the time-out will occur when the number of random times is too large.

 

Note:

This method is usually used. It can be used on poj and zoj ......

This question can also be done using dynamic planning or clever application pruning, but I have not found a solution ......

 

SourceProgram:

# Include <iostream>

 

Using namespace STD;

 

Int map [30] [30], part [30]; // The part array is used to separate points.

 

Int main ()

{

Long N, I, j,;

While (CIN> N)

{

For (I = 1; I <= N; I ++)

For (j = 1; j <= N; j ++)

Cin> map [I] [J];

Memset (part, 0, sizeof (part ));

Long max =-1, t, sum = 0;

T = 200000; // random number of times

While (t --)

{

A = rand () % N + 1; // randomly generate a number and obtain the location of the corresponding vertex.

If (part [a]) part [a] = 0; // change the position of this Vertex

Else part [a] = 1;

For (I = 1; I <= N; I ++) // The right to update the changed Graph

{

If (part [I] & part [a] &! = I) sum-= map [a] [I];

If (! Part [I] &! Part [a] &! = I) sum-= map [a] [I];

If (part [I] &! Part [a]) sum + = map [a] [I];

If (! Part [I] & part [a]) sum + = map [a] [I];

}

If (sum> MAX) max = sum; // update the maximum value

}

Cout <max <Endl;

}

Return 0;

}