模拟 csu1711 Kinfolk

传送门:点击打开链接

题意:一颗完全二叉树,表示辈分关系,节点从0开始。随便告诉你两个节点,问v节点是u节点的什么关系。

思路:顶级模拟题!感谢这道题目,我终于搞懂了各种头衔的含义。。

思路大概是先求出u和v到lca的距离,然后就利用这个距离去讨论。

刚开始可以写一部分出来,发现规律后就可以直接按规律来写了。

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout << "[" << x << "]"
#define FIN freopen("input.txt", "r", stdin)
#define FOUT freopen("output.txt", "w+", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
 
const int MX = 40000;
int vis[MX];
 
int u, v; char s[10];
int h1, h2;
 
void find_h(int id) {
    int t = u;
    while(t) {
        vis[t] = id;
        t /= 2;
    }
    t = v;
    int lca;
    while(t) {
        if(vis[t] == id) {
            lca = t;
            break;
        }
        t /= 2;
    }
    h1 = h2 = 0;
    for(t = u; t != lca; t /= 2) h1++;
    for(t = v; t != lca; t /= 2) h2++;
}
 
string pre[] = {"", "grand", "great-grand", "great-great-grand"};
string num[] = {"", "1st", "2nd", "3rd"};
string last[] = {"", "once", "twice", "thrice"};
 
string DFS(int h1, int h2) {
    if(h1 == 0 && h2 == 0) return "self";
    if(h1 == 1 && h2 == 0) {
        if(s[0] == 'M') return "father";
        else return "mother";
    }
    if(h1 == 0 && h2 == 1) {
        if(s[0] == 'M') return "son";
        else return "daughter";
    }
    if(h1 == 1 && h2 == 1) {
        if(s[0] == 'M') return "brother";
        else return "sister";
    }
    if(h1 == 2 && h2 == 1) {
        if(s[0] == 'M') return "uncle";
        else return "aunt";
    }
    if(h1 == 1 && h2 == 2) {
        if(s[0] == 'M') return "nephew";
        else return "niece";
    }
 
    int d = abs(h1 - h2), Min = min(h1, h2);
    if(Min >= 2) {
        if(Min - 1 <= 3 && d <= 3) {
            if(d == 0) return num[Min-1] + " cousin";
            else return num[Min-1] + " cousin " + last[d] + " removed";
        } else return "kin";
    } else {
        if(d - 1 > 3) return "kin";
        else {
            if(h1 > h2) return pre[d-1] + DFS(h1 - d + 1, h2);
            if(h1 < h2) return pre[d-1] + DFS(h1, h2 - d + 1);
        }
    }
}
 
int main() {
    int sz = 0; //FIN;
    while(~scanf("%d%d%s", &u, &v, s), u >= 0) {
        u++; v++;
        find_h(++sz);
        printf("%s\n", DFS(h1, h2).c_str());
    }
    return 0;
}


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