DFS nbut1218 You are my brother

传送门：点击打开链接

告诉你一些a和b，b是a的父亲，问节点1和2的辈分关系。

思路：很明显这是一棵树形结构。先从节点连向根建图，然后从节点1出发DFS，这样能找到根节点

然后再从根节点往节点建图，找节点1和节点2的深度，再比较深度

容易错的地方：节点1和节点2可能并不在同一颗树中，这时要输出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;
}