请利用Kernighan-Lin算法求解MaxCut问题,并编写C++语言代码
时间: 2024-03-23 11:39:12 浏览: 16
Kernighan-Lin算法是一种贪心算法,用于解决MaxCut问题。它的基本思想是不断地将节点划分为两个子集,使得子集之间的边权和最大化。
以下是Kernighan-Lin算法的C++代码实现:
```c++
#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
using namespace std;
// 计算节点的度数
vector<int> calc_degrees(const vector<vector<int>>& graph) {
vector<int> degrees(graph.size(), 0);
for (int i = 0; i < graph.size(); ++i) {
degrees[i] = accumulate(graph[i].begin(), graph[i].end(), 0);
}
return degrees;
}
// 计算节点的初始划分
vector<int> calc_initial_partition(const vector<int>& degrees) {
vector<int> partition(degrees.size(), 0);
int sum_degrees = accumulate(degrees.begin(), degrees.end(), 0);
int half_sum_degrees = sum_degrees / 2;
int curr_sum_degrees = 0;
for (int i = 0; i < degrees.size(); ++i) {
curr_sum_degrees += degrees[i];
if (curr_sum_degrees > half_sum_degrees) {
partition[i] = 1;
}
}
return partition;
}
// 计算节点之间的边权和
int calc_cut_size(const vector<vector<int>>& graph, const vector<int>& partition) {
int cut_size = 0;
for (int i = 0; i < graph.size(); ++i) {
for (int j = 0; j < graph[i].size(); ++j) {
if (partition[i] != partition[graph[i][j]]) {
cut_size += 1;
}
}
}
return cut_size;
}
// 计算节点的移动增益
vector<int> calc_gain(const vector<vector<int>>& graph, const vector<int>& partition, const vector<int>& degrees) {
vector<int> gain(graph.size(), 0);
for (int i = 0; i < graph.size(); ++i) {
int cut_size_diff = 0;
for (int j = 0; j < graph[i].size(); ++j) {
if (partition[i] != partition[graph[i][j]]) {
cut_size_diff -= 1;
} else {
cut_size_diff += 1;
}
}
gain[i] = cut_size_diff - 2 * degrees[i] * (partition[i] - 0.5);
}
return gain;
}
// 交换节点的划分
void swap_partition(vector<int>& partition, int i, int j) {
int tmp = partition[i];
partition[i] = partition[j];
partition[j] = tmp;
}
// Kernighan-Lin算法
void kernighan_lin(const vector<vector<int>>& graph, vector<int>& partition) {
vector<int> degrees = calc_degrees(graph);
int cut_size = calc_cut_size(graph, partition);
bool improved = true;
while (improved) {
improved = false;
vector<int> gain = calc_gain(graph, partition, degrees);
vector<int> moved(graph.size(), 0);
for (int i = 0; i < graph.size(); ++i) {
if (moved[i]) {
continue;
}
int max_gain = 0;
int max_gain_idx = -1;
for (int j = 0; j < graph[i].size(); ++j) {
int k = graph[i][j];
if (!moved[k] && gain[k] > max_gain) {
max_gain = gain[k];
max_gain_idx = k;
}
}
if (max_gain_idx != -1) {
swap_partition(partition, i, max_gain_idx);
cut_size += max_gain;
moved[i] = true;
moved[max_gain_idx] = true;
improved = true;
}
}
}
}
// 测试代码
int main() {
vector<vector<int>> graph = {{1, 2}, {0, 2, 3}, {0, 1, 3}, {1, 2}};
vector<int> partition = calc_initial_partition(calc_degrees(graph));
kernighan_lin(graph, partition);
cout << calc_cut_size(graph, partition) << endl;
return 0;
}
```
在上面的代码中,我们首先计算了每个节点的度数,然后根据节点的度数计算了初始的划分。接着,我们不断交换节点的划分,直到无法继续改进为止。在交换节点的划分时,我们选择移动增益最大的节点。最后,我们计算了最终的划分对应的边权和。
使用Kernighan-Lin算法可以在多项式时间内解决MaxCut问题。