[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"$fB7VUMtuFU_F6aLqDtnLhBNG3PudfP6x6orOd3wrVMcE":3},{"answer":4,"createTime":5,"id":6,"options":7,"origin":12,"question":18,"related":19,"source":27,"type":32},[],"2023-06-21 23:15:03",897740645,[8,9,10,11],"回溯","分支限界","回溯和分支限界","回溯求解子集树问题",{"courseId":13,"courseImg":14,"courseName":15,"workId":16,"workName":17},"1000008932","https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002F565a18baa1564c24b33cd6f84945033a.jpg","算法分析与设计","40286319","第九章单元测试","在对问题的解空间树进行搜索的方法中,一个活结点最多有一次机会成为活结点的是( )",[20,29,33,43,52,61,70,75,83,92],{"answer":21,"createTime":5,"id":22,"options":23,"question":26,"source":27,"type":28},[],897740633,[24,25],"对","错","使用限界函数作优先级, 第一个扩展的叶子就是最优解","v2",3,{"answer":30,"createTime":5,"id":6,"options":31,"question":18,"source":27,"type":32},[],[8,9,10,11],0,{"answer":34,"createTime":5,"id":35,"options":36,"question":41,"source":27,"type":42},[],897740765,[37,38,39,40],"求解目标不同","搜索方式不同","对扩展结点的扩展方式不同","存储空间的要求不同","分支限界法与回溯法的不同点是什么",1,{"answer":44,"createTime":5,"id":45,"options":46,"question":51,"source":27,"type":32},[],897740767,[47,48,49,50],"使用限界函数作优先级, 第一个加入队列的叶子就是最优解","用约束函数在扩展结点处剪去不满足约束的子树","用限界函数剪去得不到最优解的子树","回溯和分支限界都是动态生成解空间树","下面说法不正确的是()",{"answer":53,"createTime":5,"id":54,"options":55,"question":60,"source":27,"type":42},[],897740772,[56,57,58,59],"针对所给问题,定义问题的解空间(对解进行编码)","确定易于搜索的解空间结构(按树或图组织解)","定义最优子结构","以广度优先或以最小耗费(最大收益)优先的方式搜索解空间,并在搜索过程中用剪枝函数避免无效搜索","用分支限界法设计算法的步骤是",{"answer":62,"createTime":5,"id":63,"options":64,"question":69,"source":27,"type":32},[],897740775,[65,66,67,68],"先进先出","后进先出","结点的优先级","随机","优先队列式分支限界法选取扩展结点的原则是",{"answer":71,"createTime":5,"id":72,"options":73,"question":74,"source":27,"type":28},[],897740783,[24,25],"分支限界法不能解决0\u002F1背包问题",{"answer":76,"createTime":5,"id":77,"options":78,"question":82,"source":27,"type":32},[],897740857,[79,9,80,81],"回溯算法","动态规划","贪心算法","FIFO是( )的搜索方式",{"answer":84,"createTime":5,"id":85,"options":86,"question":91,"source":27,"type":32},[],897740883,[87,88,89,90],"子集树","排列树","深度优先生成树","广度优先生成树","分支限界法解旅行商问题时的解空间树是",{"answer":93,"createTime":5,"id":94,"options":95,"question":96,"source":27,"type":28},[],897740893,[24,25],"优先队列式分支限界法按照队列先进先出的原则,选取下一个节点为扩展结点"]