A - Black and White Tree

There is a tree with NN vertices numbered 11 through NN. The ii-th of the N1N−1 edges connects vertices aiai and bibi.

Initially, each vertex is uncolored.

Takahashi and Aoki is playing a game by painting the vertices. In this game, they alternately perform the following operation, starting from Takahashi:

  • Select a vertex that is not painted yet.
  • If it is Takahashi who is performing this operation, paint the vertex white; paint it black if it is Aoki.

Then, after all the vertices are colored, the following procedure takes place:

  • Repaint every white vertex that is adjacent to a black vertex, in black.

Note that all such white vertices are repainted simultaneously, not one at a time.

If there are still one or more white vertices remaining, Takahashi wins; if all the vertices are now black, Aoki wins. Determine the winner of the game, assuming that both persons play optimally.

Constraints

 

  • 2N1052≤N≤105
  • 1ai,biN1≤ai,bi≤N
  • aibiai≠bi
  • The input graph is a tree.

Input

 

Input is given from Standard Input in the following format:

NN
a1a1 b1b1
:
aN1aN−1 bN1bN−1

Output

 

Print First if Takahashi wins; print Second if Aoki wins.

Sample Input 1

 

3
1 2
2 3

Sample Output 1

 

First

Below is a possible progress of the game:

  • First, Takahashi paint vertex 22 white.
  • Then, Aoki paint vertex 11 black.
  • Lastly, Takahashi paint vertex 33 white.

In this case, the colors of vertices 1122 and 33 after the final procedure are black, black and white, resulting in Takahashi's victory.

Sample Input 2

 

4
1 2
2 3
2 4

Sample Output 2

 

First

Sample Input 3

 

6
1 2
2 3
3 4
2 5
5 6

Sample Output 3

 

Second
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#include <cassert>
//#include <unordered_set>
//#include <unordered_map>
#define ll              long long
#define pii             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define forn(i, n)      for(int i = 0; i < int(n); i++)
using namespace std;
int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-6;
const ll mod = 998244353;

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}

ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}

/**********************************************************/
const int N = 1e5 + 5;

vector<int> g[N];
bool leaf[N];


bool dfs(int x, int fa)
{
    if (g[x].size() == 1 && g[x][0] == fa)
    {
        leaf[x] = 1;
        return 0;
    }
    int cnt0 = 0;
    for (auto to : g[x])
    {
        if (to != fa)
        {
            if (dfs(to, x))
                return 1;
            if(leaf[to])
                cnt0++;
        }
    }
    if (cnt0 > 1)
        return 1;
    if (fa == -1 && cnt0 == 0)
        return 1;
    if (cnt0)
        leaf[x] = 0;
    else
        leaf[x] = 1;
    return 0;
}

int main()
{
#ifdef _DEBUG
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    int n;
    cin >> n;
    rep(i, 1, n-1)
    {
        int u, v;
        scanf("%d%d",&u,&v);
        g[u].push_back(v);
        g[v].push_back(u);
    }
    int fg = 0;
    rep(i, 1, n)
    {
        if (g[i].size() > 1)
        {
            fg = dfs(i, -1);
            break;
        }
    }
    if (fg)
        puts("First");
    else
        puts("Second");
    return 0;
}

 

posted @ 2020-10-08 12:19  DeaL57  阅读(154)  评论(0编辑  收藏  举报