C. Timofey and a tree 观察题 ＋ ｄｆｓ模拟

http://codeforces.com/contest/764/problem/C

题意：在n个顶点中随便删除一个，然后分成若干个连通子图，要求这若干个连通子图的颜色都只有一种。

记得边是双向的，wa15的可能是不知道边是双向的吧。

一个观察：如果某条边连接的两个顶点的颜色不同，那么可以看看删除这两个顶点，成立就成立，不成立就不成立。

因为必定要把这两个顶点分开。

然后就是暴力dfs了。

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <assert.h>

#define IOS ios::sync_with_stdio(false)

using namespace std;

#define inf (0x3f3f3f3f)

typedef long long int LL;

#include <iostream>

#include <sstream>

#include <vector>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <bitset>

const int maxn = 2e6 + ;

struct node {

    int u, v, tonext;

}e[maxn];

int num;

int first[maxn];

void add(int u, int v) {

    num++;

    e[num].u = u;

    e[num].v = v;

    e[num].tonext = first[u];

    first[u] = num;

}

int c[maxn], in[maxn];

int u[maxn], v[maxn];

set<int>ss;

int vis[maxn];

int DFN;

int dfs(int cur, int no) {

    ss.insert(c[cur]);

    if (ss.size() >= ) return false;

    bool flag = true;

    for (int i = first[cur]; i; i = e[i].tonext) {

        if (e[i].v == no) continue;

        if (vis[e[i].v] == DFN) continue;

        vis[e[i].v] = DFN;

        flag = flag && dfs(e[i].v, no);

    }

    return flag;

}

bool del(int cur) {

    for (int i = first[cur]; i; i = e[i].tonext) {

        ss.clear();

        DFN++;

        vis[cur] = DFN;

        vis[e[i].v] = DFN;

        if (dfs(e[i].v, inf) == false) return false;

    }

    return true;

}

void work() {

    int n;

    cin >> n;

    for (int i = ; i <= n - ; ++i) {

        cin >> u[i] >> v[i];

    }

    for (int i = ; i <= n; ++i) {

        cin >> c[i];

    }

    int which = inf;

    for (int i = ; i <= n - ; ++i) {

        if (c[u[i]] != c[v[i]] && which == inf) which = i;

        add(u[i], v[i]);

        add(v[i], u[i]);

    }

    if (which == inf) {

        cout << "YES" << endl;

        cout <<  << endl;

        return;

    }

//    cout << which << "  " << root << endl;

    if (del(u[which])) {

        cout << "YES" << endl;

        cout << u[which] << endl;

        return;

    }

    if (del(v[which])) {

        cout << "YES" << endl;

        cout << v[which] << endl;

        return;

    }

    cout << "NO" << endl;

}

int main() {

#ifdef local

    freopen("data.txt", "r", stdin);

//    freopen("data.txt", "w", stdout);

#endif

    work();

    return ;

}