DFS nbut1218 You are my brother

传送门：点击打开链接

题意：告诉你n个关系，关系(u,v)表示v是u的父亲，现在A是节点1，B是节点2，问A和B的辈分关系

思路：这肯定是树型结构，先建立(u,v)的有向边，找到A的祖先root，然后再从root出发看A和B的深度是多少

然后比较A和B的深度。

trick：A和B可能并不是一个祖先，也就是说A和B可能没有血缘关系，给出的数据并不是一棵树，而是森林，此时应该输出You are my brother

#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define FIN freopen("input.txt","r",stdin)

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

const int MX = 6e3 + 5;
const int INF = 0x3f3f3f3f;

int Head[MX], Next[MX], rear;

struct Edge {
    int u, v, cost;
} E[MX], S[MX];

void edge_init() {
    rear = 0;
    memset(Head, -1, sizeof(Head));
}

void edge_add(int u, int v, int cost = 0) {
    E[rear].u = u;
    E[rear].v = v;
    E[rear].cost = cost;
    Next[rear] = Head[u];
    Head[u] = rear++;
}

int DFS(int u) {
    for(int i = Head[u]; ~i; i = Next[i]) {
        int v = E[i].v;
        return DFS(v);
    }
    return u;
}

void solve(int u, int d, int &a, int &b) {
    if(u == 1) a = d;
    if(u == 2) b = d;

    for(int i = Head[u]; ~i; i = Next[i]) {
        int v = E[i].v;
        solve(v, d + 1, a, b);
    }
}

int main() {
    int n, u, v; //FIN;
    while(~scanf("%d", &n)) {
        edge_init();
        for(int i = 1; i <= n; i++) {
            scanf("%d%d", &u, &v);
            S[i].u = u; S[i].v = v;
            edge_add(u, v);
        }
        int root = DFS(1);

        edge_init();
        for(int i = 1; i <= n; i++) {
            edge_add(S[i].v, S[i].u);
        }

        int a = -1, b = -1;
        solve(root, 0, a, b);

        if(a == -1 || b == -1) {
            printf("You are my brother\n");
            continue;
        }

        if(a == b) {
            printf("You are my brother\n");
        } else if(a > b) {
            printf("You are my elder\n");
        } else {
            printf("You are my younger\n");
        }
    }
    return 0;
}