C语言实现关键路径计算：示例代码及步骤详解

以下是一个使用数据结构来计算关键路径和最迟开始时间的C语言代码示例：\n\nc\n#include <stdio.h>\n#include <stdlib.h>\n\n// 定义活动的结构体\ntypedef struct {\n char name; // 活动名称\n int duration; // 活动所需时间\n int earliest; // 最早开始时间\n int latest; // 最迟开始时间\n} Activity;\n\nint main() {\n // 定义活动数组\n Activity activities[9];\n\n // 初始化活动数组\n activities[0].name = 'A';\n activities[0].duration = 3;\n activities[0].earliest = 0;\n activities[0].latest = 0;\n\n activities[1].name = 'B';\n activities[1].duration = 2;\n activities[1].earliest = -1;\n activities[1].latest = -1;\n\n activities[2].name = 'C';\n activities[2].duration = 4;\n activities[2].earliest = -1;\n activities[2].latest = -1;\n\n activities[3].name = 'D';\n activities[3].duration = 5;\n activities[3].earliest = -1;\n activities[3].latest = -1;\n\n activities[4].name = 'E';\n activities[4].duration = 8;\n activities[4].earliest = -1;\n activities[4].latest = -1;\n\n activities[5].name = 'F';\n activities[5].duration = 3;\n activities[5].earliest = -1;\n activities[5].latest = -1;\n\n activities[6].name = 'G';\n activities[6].duration = 5;\n activities[6].earliest = -1;\n activities[6].latest = -1;\n\n activities[7].name = 'H';\n activities[7].duration = 10;\n activities[7].earliest = -1;\n activities[7].latest = -1;\n\n activities[8].name = 'I';\n activities[8].duration = 2;\n activities[8].earliest = -1;\n activities[8].latest = -1;\n\n // 设置先前工作的依赖关系\n activities[1].earliest = activities[0].duration;\n activities[2].earliest = activities[0].duration;\n activities[3].earliest = activities[1].earliest + activities[1].duration;\n activities[4].earliest = activities[1].earliest + activities[1].duration;\n activities[5].earliest = activities[2].earliest + activities[2].duration;\n activities[6].earliest = activities[5].earliest + activities[5].duration;\n activities[7].earliest = activities[4].earliest + activities[4].duration;\n activities[8].earliest = activities[3].earliest + activities[3].duration + activities[6].earliest + activities[6].duration + activities[7].earliest + activities[7].duration;\n\n // 输出关键路径\n printf("关键路径为：");\n for (int i = 0; i < 9; i++) {\n if (activities[i].earliest == activities[i].latest) {\n printf("%c ", activities[i].name);\n }\n }\n printf("\n");\n\n // 计算最迟开始时间\n activities[8].latest = activities[8].earliest;\n activities[7].latest = activities[8].latest - activities[7].duration;\n activities[6].latest = activities[7].latest - activities[6].duration;\n activities[5].latest = activities[6].latest - activities[5].duration;\n activities[4].latest = activities[7].latest - activities[4].duration;\n activities[3].latest = activities[4].latest - activities[3].duration;\n activities[2].latest = activities[5].latest - activities[2].duration;\n activities[1].latest = activities[3].latest - activities[1].duration;\n activities[0].latest = activities[1].latest - activities[0].duration;\n\n // 输出最迟开始时间\n printf("该项目最迟开始时间为：%d\n", activities[0].latest);\n\n return 0;\n}\n\n\n在这个示例中，我们使用了一个包含9个活动的活动数组来存储每个活动的信息。然后，根据先前工作的依赖关系，我们计算出每个活动的最早开始时间。根据最早开始时间，我们可以确定关键路径，即最早开始时间和最迟开始时间相等的活动。最后，我们计算出最迟开始时间，即从最后一个活动开始倒退计算每个活动的最迟开始时间。最后，我们输出关键路径和该项目的最迟开始时间。\n\n本代码示例展示了如何使用C语言数据结构和算法来计算关键路径和项目最迟开始时间，帮助您更好地理解和应用关键路径方法。\n