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

exercise/32.c

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+// 计算含运算符 '&'（取两数较小值）与 '|'（取两数较大值）的表达式
+// 两者同优先级，按从左到右结合；遇到括号优先计算括号内表达式。
+// 输入：一行字符串，仅由数字、&、|、(、) 组成
+// 输出：一个整数，表示表达式的值
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <ctype.h>
+
+static inline int min_i(int a, int b) { return a < b ? a : b; }
+static inline int max_i(int a, int b) { return a > b ? a : b; }
+
+static inline void apply_top(int *vals, int *vtop, char *ops, int *otop) {
+    char op = ops[--(*otop)];
+    int b = vals[--(*vtop)];
+    int a = vals[--(*vtop)];
+    int r = (op == '&') ? min_i(a, b) : max_i(a, b);
+    vals[(*vtop)++] = r;
+}
+
+int main(void) {
+    // 读取整行输入
+    char *line = NULL;
+    size_t cap = 0;
+    {
+        // getline 若不可用则用 fgets 简易读取
+        // Windows 平台无 getline，这里使用 fgets
+        size_t bufcap = 16384; // 大于题目约束的 10000
+        line = (char*)malloc(bufcap);
+        if (!line) return 1;
+        if (!fgets(line, (int)bufcap, stdin)) return 0;
+        size_t len = strlen(line);
+        // 去掉行尾 CR/LF
+        while (len && (line[len-1] == '\n' || line[len-1] == '\r')) line[--len] = '\0';
+        cap = len + 1;
+    }
+
+    size_t L = strlen(line);
+    int *vals = (int*)malloc((L + 1) * sizeof(int));
+    char *ops = (char*)malloc(L + 1);
+    if (!vals || !ops) return 1;
+    int vtop = 0, otop = 0;
+
+    for (size_t i = 0; i < L;) {
+        char c = line[i];
+        if (isspace((unsigned char)c)) { ++i; continue; } // 忽略空白
+        if (c >= '0' && c <= '9') {
+            int x = 0;
+            while (i < L && line[i] >= '0' && line[i] <= '9') {
+                x = x * 10 + (line[i] - '0');
+                ++i;
+            }
+            vals[vtop++] = x;
+        } else if (c == '&' || c == '|') {
+            // 同优先级：在遇到 '(' 之前，应用所有已入栈的运算符
+            while (otop > 0 && ops[otop - 1] != '(') {
+                apply_top(vals, &vtop, ops, &otop);
+            }
+            ops[otop++] = c;
+            ++i;
+        } else if (c == '(') {
+            ops[otop++] = c;
+            ++i;
+        } else if (c == ')') {
+            while (otop > 0 && ops[otop - 1] != '(') {
+                apply_top(vals, &vtop, ops, &otop);
+            }
+            if (otop > 0 && ops[otop - 1] == '(') --otop; // 弹出左括号
+            ++i;
+        } else {
+            // 非法字符：跳过以增强鲁棒性
+            ++i;
+        }
+    }
+
+    while (otop > 0) {
+        apply_top(vals, &vtop, ops, &otop);
+    }
+
+    if (vtop > 0) printf("%d\n", vals[vtop - 1]);
+
+    free(vals);
+    free(ops);
+    free(line);
+    return 0;
+}