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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 2010, M = 6010;
// SG函数模板题
int n, m, k;
int f[N];
int h[N], e[M], ne[M], idx;
void add(int a, int b) {
e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
int sg(int u) {
//记忆化搜索
if (~f[u]) return f[u];
//找出当前结点u的所有出边看看哪个sg值没有使用过
set<int> S;
for (int i = h[u]; ~i; i = ne[i])
S.insert(sg(e[i]));
//找到第一个没有出现的过的自然数, 0,1,2,3,4,...
for (int i = 0;; i++)
if (S.count(i) == 0) {
f[u] = i;
break;
}
return f[u];
}
int main() {
memset(h, -1, sizeof h);
cin >> n >> m >> k;
while (m--) {
int a, b;
cin >> a >> b;
add(a, b);
}
memset(f, -1, sizeof f); //初始化sg函数的结果表
int res = 0;
while (k--) {
int u;
cin >> u;
res ^= sg(u); //计算每个出发点的sg(u),然后异或在一起
}
if (res) //所有出发点的异或和不等于0,先手必胜
puts("win");
else //所有出发点的异或和等于0先手必败
puts("lose");
return 0;
}