给出christofides算法的c++源码
时间: 2023-12-15 21:07:13 浏览: 25
以下是一个简单的C++实现,仅供参考:
```c++
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 1005; // 最大点数
struct Point {
double x, y;
Point() {}
Point(double _x, double _y) : x(_x), y(_y) {}
};
double dist(Point a, Point b) {
return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2));
}
int n; // 点的数量
double g[MAXN][MAXN]; // 图的邻接矩阵
double ans; // 最小生成树的长度
vector<Point> points; // 点的坐标
// 生成图的邻接矩阵
void build_graph() {
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
g[i][j] = g[j][i] = dist(points[i], points[j]);
}
}
}
// prim算法求最小生成树
// 返回生成树的边权和
double prim() {
vector<double> lowcost(n, g[0][1]);
vector<int> closest(n, 0);
vector<bool> used(n, false);
used[0] = true;
double sum = 0.0;
for (int i = 1; i < n; i++) {
int v = -1;
for (int j = 0; j < n; j++) {
if (!used[j] && (v == -1 || lowcost[j] < lowcost[v])) {
v = j;
}
}
sum += lowcost[v];
used[v] = true;
for (int j = 0; j < n; j++) {
if (!used[j] && g[v][j] < lowcost[j]) {
lowcost[j] = g[v][j];
closest[j] = v;
}
}
}
return sum;
}
// 求欧拉回路
vector<int> euler_circuit() {
vector<int> circuit;
int start = 0;
circuit.push_back(start);
while (true) {
bool found = false;
for (int i = 0; i < n; i++) {
if (g[circuit.back()][i] > 0) {
circuit.push_back(i);
g[circuit[circuit.size() - 2]][i] = g[i][circuit[circuit.size() - 2]] = 0;
found = true;
break;
}
}
if (!found) {
for (int i = circuit.size() - 2; i >= 0; i--) {
for (int j = 0; j < n; j++) {
if (g[circuit[i]][j] > 0) {
circuit.insert(circuit.begin() + i + 1, j);
g[circuit[i]][j] = g[j][circuit[i]] = 0;
found = true;
break;
}
}
if (found) break;
}
}
if (!found) break;
}
return circuit;
}
// christofides算法
double christofides() {
build_graph();
ans = prim();
vector<int> odd;
for (int i = 0; i < n; i++) {
int deg = 0;
for (int j = 0; j < n; j++) {
if (g[i][j] > 0) deg++;
}
if (deg % 2) odd.push_back(i);
}
sort(odd.begin(), odd.end(), [&](int a, int b) {
return points[a].x < points[b].x || (points[a].x == points[b].x && points[a].y < points[b].y);
});
for (int i = 0; i < odd.size(); i += 2) {
g[odd[i]][odd[i + 1]] = g[odd[i + 1]][odd[i]] = dist(points[odd[i]], points[odd[i + 1]]);
}
vector<int> circuit = euler_circuit();
double sum = 0.0;
for (int i = 0; i < circuit.size() - 1; i++) {
sum += g[circuit[i]][circuit[i + 1]];
}
sum += g[circuit[circuit.size() - 1]][circuit[0]];
return sum;
}
int main() {
cin >> n;
points.resize(n);
for (int i = 0; i < n; i++) {
cin >> points[i].x >> points[i].y;
}
double ans = christofides();
cout << "最小哈密顿回路长度为:" << ans << endl;
return 0;
}
```