SOJ-1008（将LCA问题转化为RMQ）

/****************************************************************************************************** ** Copyright (C) 2011.05.01 - 2013.07.01 ** Author: famousDT <[e-mail]> ** Edit date: 2011-04-23******************************************************************************************************/#include <stdio.h>#include <stdlib.h>//abs,atof(string to float),atoi,atol,atoll#include <math.h>//atan,acos,asin,atan2(a,b)(a/b atan),ceil,floor,cos,exp(x)(e^x),fabs,log(for E),log10#include <vector>#include <queue>#include <map>#include <set>#include <string>#include <iostream>#include <string.h>//memcpy(to,from,count#include <ctype.h>//character process:isalpha,isdigit,islower,tolower,isblank,iscntrl,isprint#include <algorithm>using namespace std;//typedef long long int ll;#define PI acos(-1)#define MAX(a, b) ((a) > (b) ? (a) : (b))#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MALLOC(n, type) ((type *)malloc((n) * sizeof(type)))#define FABS(a) ((a) >= 0 ? (a) : (-(a)))/***********************************************解题思路*****************************************将LCA问题转化为RMQ**http://www.topcoder.com/tc?module=Static&d1=tutorials&d2=lowestCommonAncestor**********************************************************************************************/struct node{    int parent;      //父亲    int position;     //RMQ表位置    vector<int> child;};vector<node> tree;//分别用于存储RQM的遍例次序和深度vector<int> order, depth;//由树建立RQM表，递归方式void build_RMQ_table(int nod, int dep){    //进入当前节点，存储其编号和深度    order.push_back(nod);    depth.push_back(dep);    node &n = tree[nod];    //第一次遍例该节点，记录节点在表中的位置    n.position = n.position == -1 ? (order.size() - 1) : n.position;    //多叉树的标准深度遍例方式，依次访问所有子节点    vector<int>::iterator iter;    for (iter = n.child.begin(); iter != n.child.end();) {        build_RMQ_table(*iter++, dep + 1);        order.push_back(nod);        depth.push_back(dep);    }}int main(){    map<string, int> nametable;    nametable.clear();    //新节点的初始值，父为-1，RMQ表位置为-1    node newnode;    newnode.parent = newnode.position = -1;    string c, p;    string s1, s2;    vector<node>::iterator iter;    nametable.clear();    tree.clear();    while (cin>>c>>p && c != "no.child") {        if (nametable.end() == nametable.find(p)) {            nametable[p] = tree.size();            tree.push_back(newnode);        }        if (nametable.end() == nametable.find(c)) {            nametable[c] = tree.size();            tree.push_back(newnode);        }        //建立父子关系        tree[nametable[p]].child.push_back(nametable[c]);        tree[nametable[c]].parent = nametable[p];    }    //如果存在森林，简历虚拟的总跟，放在列表最后    tree.push_back(newnode);    //遍历所有节点    for (iter = tree.begin(); iter != tree.end() - 1; ++iter) {        if (iter->parent == -1) {//没有父亲的节点放到总节点儿子列表            iter->parent = tree.size() - 1;            tree.back().child.push_back(iter - tree.begin());        }    }    //从总跟开始地鬼简历RMQ表    build_RMQ_table(tree.size() - 1, 0);    while (cin>>s1>>s2) {        map<string, int>::iterator i1 = nametable.find(s1);        map<string, int>::iterator i2 = nametable.find(s2);        if (i1 == nametable.end() || i2 == nametable.end()) {            cout << "no relation" << endl;            continue;        }        int n1 = tree[i1->second].position;        int n2 = tree[i2->second].position;        if (depth[n1] > depth[n2]) {            swap(n1, n2);        }        //RMQ查询        vector<int>::iterator itertmp = min_element(depth.begin() + MIN(n1, n2),                                           depth.begin() + MAX(n1, n2) + 1);        //如果最小共同祖先(LCS)为总根，认为无关        if (tree[order[itertmp - depth.begin()]].parent == -1) {            cout << "no relation" << endl;             continue;        }        int removed = depth[n2] - depth[n1];        int cousin = depth[n1] - *itertmp;        //二者有一个人是LCS        if (cousin == 0) {            for (; removed > 2; --removed) {                cout << "great ";              }            if (removed > 1)                cout << "grand ";            cout << (tree[i1->second].position == n1 ? "parent" : "child") << endl;         }        else if (cousin == 1 && removed == 0)            cout << "sibling" << endl;          else {            cout << cousin - 1 << " cousin";            if (removed > 0) {                  cout << " removed " << removed;            }            cout << endl;         }    }    return 0;}