docs: 添加仓库地址到README；chore: 整理实验分析；feat: Kruskal最小生成树与图存储结构

2025-12-14 14:06:31 +08:00
parent be03307655
commit f2b8d1fe4c
52 changed files with 4600 additions and 291 deletions
@@ -0,0 +1,119 @@
+#include <stdio.h>
+#include <stdlib.h>
+#ifdef _WIN32
+#include <windows.h>
+#endif
+
+// 图的边的表示
+typedef struct
+{
+    int u, v;
+    int w;
+} Edge;
+
+// 边的比较函数，用于排序
+static int cmp_edge(const void *a, const void *b)
+{
+    int wa = ((const Edge *)a)->w;
+    int wb = ((const Edge *)b)->w;
+    return (wa > wb) - (wa < wb);
+}
+
+// 并查集的查找函数
+static int find(int *parent, int x)
+{
+    if (parent[x] == x)
+    {
+        return x;
+    }
+    return parent[x] = find(parent, parent[x]);
+}
+
+// 并查集的合并函数
+static void unite(int *parent, int *sz, int a, int b)
+{
+    if (sz[a] < sz[b])
+    {
+        int t = a;
+        a = b;
+        b = t;
+    }
+    parent[b] = a;
+    sz[a] += sz[b];
+}
+
+int main(void)
+{
+#ifdef _WIN32
+    system("chcp 65001 > nul");
+    SetConsoleOutputCP(65001);
+    SetConsoleCP(65001);
+#endif
+
+    // 读取图的边数和节点数
+    int n, m;
+    if (scanf("%d %d", &n, &m) != 2)
+    {
+        printf("输入错误\n");
+        return 0;
+    }
+
+    // 读取图的边
+    Edge *e = (Edge *)malloc((size_t)m * sizeof(Edge));
+    for (int i = 0; i < m; ++i)
+    {
+        int u, v, w;
+        scanf("%d %d %d", &u, &v, &w);
+        e[i].u = u;
+        e[i].v = v;
+        e[i].w = w;
+    }
+
+    // 初始化并查集
+    int *parent = (int *)malloc((size_t)(n + 1) * sizeof(int));
+    int *sz = (int *)malloc((size_t)(n + 1) * sizeof(int));
+    for (int i = 1; i <= n; ++i)
+    {
+        parent[i] = i;
+        sz[i] = 1;
+    }
+
+    // 对边按权重排序
+    qsort(e, (size_t)m, sizeof(Edge), cmp_edge);
+
+    // 最小生成树的计算
+    long long ans = 0;
+    int cnt = 0;
+    for (int i = 0; i < m && cnt < n - 1; ++i)
+    {
+        // 跳过无效的边
+        int u = e[i].u, v = e[i].v;
+        if (u < 1 || u > n || v < 1 || v > n)
+        {
+            continue;
+        }
+        // 检查是否形成环
+        int ru = find(parent, u), rv = find(parent, v);
+        if (ru != rv)
+        {
+            ans += e[i].w;
+            unite(parent, sz, ru, rv);
+            ++cnt;
+        }
+    }
+
+    // 检查是否所有节点都在最小生成树中
+    if (cnt == n - 1)
+    {
+        printf("%lld\n", ans);
+    }
+    else
+    {
+        printf("impossible\n");
+    }
+
+    free(sz);
+    free(parent);
+    free(e);
+    return 0;
+}