改一处，各要重做几个

（本条路线统一用 gcc -std=c11 -O0 -Wall 编译）把一份最小的依赖表写成数组，顺着它数一数：

#include <stdio.h>
#include <string.h>

/* 一份最小的 Makefile 依赖表：谁依赖谁 */
#define N 5
static const char *name[N] = {"util.h", "util.c", "main.c", "util.o", "app"};
/* dep[i][j] = 1 表示 i 一变，j 就得重做 */
static const int dep[N][N] = {
    /*        util.h util.c main.c util.o app */
    /*util.h*/ {0,     0,     0,     1,     0},
    /*util.c*/ {0,     0,     0,     1,     0},
    /*main.c*/ {0,     0,     0,     0,     1},
    /*util.o*/ {0,     0,     0,     0,     1},
    /*app   */ {0,     0,     0,     0,     0},
};

/* 从 start 出发，顺着依赖一路标记，数出要重做几个 */
static int touched(int start) {
    int mark[N] = {0};
    mark[start] = 1;
    for (int round = 0; round < N; round++)
        for (int i = 0; i < N; i++)
            if (mark[i])
                for (int j = 0; j < N; j++)
                    if (dep[i][j]) mark[j] = 1;
    int n = 0;
    for (int i = 0; i < N; i++) if (mark[i] && i != start) n++;
    return n;
}

int main(void) {
    /* 改 util.h、改 main.c，各要重做几个 */
    printf("%d/%d\n", touched(0), touched(2));
    return 0;
}