poj-1634

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//952K3094MSC++#include <cstdio>#include <cstring>#include <cstdlib>using namespace std;const int MAX = 30010;struct Employee {int ID;int height;int salary;int subordinatesNum;int parentPos;int nextBroPos;};typedef struct Employee Employee;int cmp(const void * a, const void * b) {return (*((Employee *)b)).salary - (*((Employee *)a)).salary;}Employee employees[MAX];int childListHead[MAX];int caseNum;int employeedNum;int queryNum;int IDMap[999999];void addInTree(int parentPos, Employee & curEmpolyee, int curEmpolyeePos) {employees[parentPos].subordinatesNum++;int curChildPos = childListHead[parentPos];// add as child's childwhile(curChildPos) {const Employee & curChildEmplyee = employees[curChildPos];if (curChildEmplyee.salary > curEmpolyee.salary &&curChildEmplyee.height >= curEmpolyee.height) {addInTree(curChildPos, curEmpolyee, curEmpolyeePos);return;}curChildPos = employees[curChildPos].nextBroPos;}// add as parent's childcurEmpolyee.parentPos = parentPos;curEmpolyee.nextBroPos = childListHead[parentPos];childListHead[parentPos] = curEmpolyeePos;}int main() {scanf("%d", &caseNum);for (int i = 1; i <= caseNum; i++) {scanf("%d %d", &employeedNum, &queryNum);memset(childListHead, 0, sizeof(childListHead));memset(employees, 0, sizeof(employees));memset(IDMap, 0xff, sizeof(IDMap));for (int j = 0; j < employeedNum; j++) {scanf("%d %d %d",&(employees[j].ID),&(employees[j].salary),&(employees[j].height));}qsort(employees, employeedNum, sizeof(Employee), cmp);for (int j = 0; j < employeedNum; j++) {IDMap[employees[j].ID] = j;}int rootPos = 0;employees[rootPos].parentPos = -1;for (int j = 1; j < employeedNum; j++) {addInTree(rootPos, employees[j], j);}for (int j = 1; j <= queryNum; j++) {int queryId;scanf("%d", &queryId);// for (int k = 0; k < employeedNum; k++) {// if (employees[k].ID == queryId) {// int parentId = 0;// if (employees[k].parentPos != -1) {// parentId = employees[employees[k].parentPos].ID;// }// printf("%d %d\n", parentId, employees[k].subordinatesNum);// }// }int parentId = 0;if (employees[IDMap[queryId]].parentPos != -1) {parentId = employees[employees[IDMap[queryId]].parentPos].ID;}printf("%d %d\n", parentId, employees[IDMap[queryId]].subordinatesNum);}}}

考察的应该是建造一颗多叉树，不过我用这种思路做，竟然3000+ms, 而另外一种类似于枚举的竟然只有94ms......

http://blog.csdn.net/neofung/article/details/7654076

后来搜了一下，貌似只要用多叉树做的，不进行特别优化的话，好像至少都是1000+向上。

用多叉树的思想比较简单：

首先在输入完以后，按照salary进行从大到小排序，那么第一个一定是boss，

然后以其为根，将后面的employee插入到这棵树中，

如果当前的employee的salary （小于）和 height （小于等于）当前根的某个child，那么再递归的将该employee插入到该child，

直到到达某个节点，没有child，或者所有child不满足上面的关系，那么将此employee作为此节点的child插入，更新其parentId，

同时在插入的过程中，更新每个父节点的childNum。

最后查询时，遍历所有的employee，找到和queryID相同employee，输出其parentId和 childNum（本来以为遍历费劲，后来专门搞了一个map来映射ID到数组位置，还是3000+ms....）

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