Sleepy Game CodeForces - 936B

大意: 给定有向图, 初始点S, 两个人轮流移动, 谁不能移动则输, 但后手睡着了, 先手可以控制后手操作, 求最后先手结果.

 

刚开始看错了, 还以为后手也是最优策略.... 实际上判断是否有偶数个节点的路径即可, 每个点拆成奇数跟偶数, 这样图就是一个DAG, 可以DP求出路径, 若不存在的话找一下是否有环, 有环则平局, 无环则输.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, m, ok, dp[2][N], fa[2][N], inq[N];
vector<int> g[N];

int dfs(int tp, int x) {
	if (ok) return 0;
	if (g[x].empty()&&tp==0) return ok=1;
	if (dp[tp][x]) return 0;
	dp[tp][x] = 1;
	for (int y:g[x]) if (dfs(tp^1,y)) return fa[tp][x]=y;
	return 0;
}
int dfs(int x) {
	inq[x] = 1;
	for (int y:g[x]) if (inq[y]||dfs(y)) return inq[x]=0,1;
	return inq[x] = 0;
}

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,n) {
		int k, t;
		scanf("%d", &k);
		REP(j,1,k) scanf("%d", &t),g[i].pb(t);
	}
	int s;
	scanf("%d", &s);
	if (dfs(1,s)) {
		puts("Win");
		int now = 1;
		while (s) printf("%d ", s), s=fa[now][s],now^=1;
		return hr,0;
	}
	puts(dfs(s)?"Draw":"Lose");
}