제출 #342961

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
342961		sealnot123	Star Trek (CEOI20_startrek)	C++14	43 / 100	1096 ms	492 KiB

startrek

/*
	Author: AquaBlaze
	Time: 2021-01-03 13:25:24
	Generated by powerful Codeforces Tool :^)
 	You can download the binary file in here https://github.com/xalanq/cf-tool (Windows, macOS, Linux)	
    Keqing best girl :)
    Nephren will always be in my heart
*/
#include<bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define eb emplace_back
#define all(a) (a).begin(),(a).end()
#define SZ(a) (int)(a).size()
#define FOR(i, a, b) for(int i=(a); i<=(b); ++i)
#define ROF(i, a, b) for(int i=(a); i>=(b); --i)
#define make_unique(a) sort(all((a))), (a).resize(unique(all((a)))-(a).begin())
#define pc(x) putchar(x)
#define MP make_pair
#define MT make_tuple

using namespace std;

typedef long long i64;
typedef tuple<int,int,int> iii;
typedef pair<int,int> pii;
typedef pair<i64,i64> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;

const int N = 1005;
const int mod = 1000000007;

int add(int a, int b){ return ((a+=b)>=mod)?a-mod:a; }
void adding(int &a, int b){ a = add(a, b); }
int mul(int a, int b){ return a*1ll*b%mod; }

pii operator + (const pii& A, const pii& B){
    return pii(add(A.x,B.x), add(A.y,B.y));
}
pii operator - (const pii& A, const pii& B){
    return pii(add(A.x,mod-B.x), add(A.y,mod-B.y));
}

vector<int> g[N];
int result[N];
pii dp[N][2];

void add_node(int u, int v){
    result[u] -= !result[v];
    if(result[u]){
        dp[u][1] = dp[u][1]+dp[v][0]+dp[v][1];
    }else{
        dp[u][0] = dp[u][0]+dp[v][1];
        dp[u][1] = dp[u][1]+dp[v][0];
    }
    result[u] += !result[v];
}

void adjust(int u){
    if(!result[u]){
        dp[u][0] = dp[u][0]+pii(1, 0);
        dp[u][1] = dp[u][1]+pii(0, 1);
    }else dp[u][1] = dp[u][1]+pii(1, 1);
}

void dfs(int u, int p){
    for(const int &e : g[u]){
        if(e == p) continue;
        dfs(e, u);
        result[u] += (!result[e]);
    }
    adjust(u);
    for(const int &e : g[u]){
        if(e == p) continue;
        add_node(u, e);
    }
}

void solve(){
    int n;
    i64 d;
    cin >> n >> d;
    FOR(i, 2, n){
        int a, b;
        cin >> a >> b;
        g[a].eb(b);
        g[b].eb(a);
    }
    vvi m(2, vi(2,0));
    vi res(2, 0);
    ROF(i, n, 1){
        memset(dp, 0, sizeof dp);
        memset(result, 0, sizeof result);
        dfs(i, -1);
        int r = !(!result[i]);
        res[1-r]++;
        FOR(j, 0, 1){
            adding(m[j^1][0], dp[i][j].x);
            adding(m[j^1][1], dp[i][j].y);
        }
    }
    FOR(t, 2, d){
        vi tmp(2, 0);
        FOR(i, 0, 1){
            FOR(j, 0, 1){
                adding(tmp[i], mul(res[j], m[i][j]));
            }
        }
        res = tmp;
    }
    int ans = 0;
    adding(ans, mul(dp[1][1].x, res[0]));
    adding(ans, mul(dp[1][1].y, res[1]));
    printf("%d",ans);
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    solve();
	return 0;
}
/*
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 */

• Subtask 1: 0 / 0
• Subtask 2: 0 / 7
• Subtask 3: 8 / 8
• Subtask 4: 15 / 15
• Subtask 5: 0 / 15
• Subtask 6: 20 / 20
• Subtask 7: 0 / 20
• Subtask 8: 0 / 15