add

public/c/ch6/tsp/fenzhi/tsp2.c

@@ -0,0 +1,78 @@
+#include <stdio.h>
+#include <limits.h>
+#define MAX_CITIES 4  // 设定城市数量为4
+#define INF INT_MAX 
+int n = MAX_CITIES;  // 城市数量
+int dist[MAX_CITIES][MAX_CITIES] = {  // 距离矩阵
+    {0, 30, 6, 4},  // 城市 0 到其他城市的距离
+    {30, 0, 5, 10},   // 城市 1 到其他城市的距离
+    {6, 5, 0, 20},   // 城市 2 到其他城市的距离
+    {4, 10, 20, 0}    // 城市 3 到其他城市的距离
+};
+int bestPath[MAX_CITIES];  // 最优路径
+int bestCost = INF;  // 最优路径的总成本
+// 计算从节点current到下一个节点的界限
+int calculateLowerBound(int path[], int current) {
+    int lowerBound = 0;
+    int visited[MAX_CITIES] = {0}; 
+    // 计算已访问的路径的距离
+    for (int i = 0; i < current; i++) {
+        lowerBound += dist[path[i]][path[i + 1]];
+        visited[path[i]] = 1;
+    }
+    // 计算当前节点到其他未访问节点的最短距离
+    for (int i = 0; i < n; i++) {
+        if (!visited[i]) {
+            int minDist = INF;
+            for (int j = 0; j < n; j++) {
+                if (!visited[j] && dist[i][j] < minDist) {
+                    minDist = dist[i][j];
+                }
+            }
+            lowerBound += minDist;
+        }
+    } 
+    return lowerBound;
+} 
+// 分支限界法求解TSP
+void branchAndBound(int path[], int current, int cost) {
+    if (current == n) {
+        // 所有城市都已访问，检查是否是最优解
+        cost += dist[path[current - 1]][path[0]];  // 加上返回起点的距离
+        if (cost < bestCost) {
+            bestCost = cost;
+            for (int i = 0; i < n; i++) {
+                bestPath[i] = path[i];
+            }
+        }
+        return;
+    } 
+    // 计算当前路径的界限
+    int lowerBound = cost + calculateLowerBound(path, current); 
+    // 如果当前解已经不可能优于最优解，剪枝
+    if (lowerBound >= bestCost)  return;  
+    // 继续扩展路径
+    for (int i = 0; i < n; i++) {
+        int flag = 0;
+        for (int j = 0; j < current; j++) {
+            if (path[j] == i) {
+                flag = 1;  // 检查当前城市是否已访问
+                break;
+            }
+        }
+        if (flag == 0) {
+            path[current] = i;
+            branchAndBound(path, current + 1, cost + dist[path[current - 1]][i]);
+        }
+    }
+} 
+int main() {
+    int path[MAX_CITIES];
+    path[0] = 0;  // 从第一个城市出发
+    branchAndBound(path, 1, 0); 
+    printf("\n最优路径是: ");
+    for (int i = 0; i < n; i++) 
+        printf("%d ", bestPath[i]); 
+    printf("\n总路程: %d\n", bestCost); 
+    return 0;
+}

public/c/ch6/tsp/huisu/travelroute.c

@@ -0,0 +1,43 @@
+#include <stdio.h>
+#include <limits.h>
+#define MAX_CITIES 4  // 城市数量
+#define INF INT_MAX
+int n = MAX_CITIES;  // 城市数量
+int dist[MAX_CITIES][MAX_CITIES] = {
+    {0, 30, 6, 4},   {30, 0, 5, 10},  
+    {6, 5, 0, 20},  {4, 10, 20, 0} 
+};
+int visited[MAX_CITIES];  // 标记城市是否访问
+int bestPath[MAX_CITIES];  // 最优路径
+int bestCost = INF;  // 最优路径的总成本
+// 回溯法求解TSP
+void backtrack(int currentCity, int count, int cost, int path[]) {
+    if (count == n) { // 所有城市都已访问，检查是否是最优解 
+        cost += dist[currentCity][0];  // 加上返回起点的距离
+        if (cost < bestCost) {
+            bestCost = cost;
+            for (int i = 0; i < n; i++) 
+                bestPath[i] = path[i]; 
+        }
+        return;
+    } 
+    for (int i = 0; i < n; i++) {// 尝试所有未访问的城市
+        if (!visited[i]) { // 访问城市i 
+            visited[i] = 1; path[count] = i; 
+            // 递归访问下一个城市
+            backtrack(i, count + 1, cost + dist[currentCity][i], path); 
+            visited[i] = 0; // 回溯
+        }
+    }
+}
+int main() {
+    int path[MAX_CITIES];
+    visited[0] = 1;  // 从第一个城市出发
+    path[0] = 0;     // 起点是城市0
+    backtrack(0, 1, 0, path);  // 从城市0开始回溯  
+    printf("\n最优路径是: ");
+    for (int i = 0; i < n; i++) 
+        printf("%d ", bestPath[i]); 
+    printf("\n最小路程: %d\n", bestCost); 
+    return 0;
+}