这题挺考验阅读理解能力的。。
如果能读懂的话,不难发现这就是在统计有多少入度为\(0\)的点
缩点后判断一下即可
当然有一种例外情况是\(1 -> 3, 2 -> 3\),也就是存在一个孤立点,判掉即可
判断的时候应当满足三个条件:所在联通块大小为\(2\),入度为0,所有指向的点入度均大于\(2\)
另外就是题目中没有说有没有重边,我没判过了,但是最好还是判一下。
/*
*/
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define rg register
#define pt(x) printf("%d ", x);
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7, B = 1;
const double eps = 1e-9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, a[MAXN], b[MAXN], f[MAXN];
void solve(int val) {
f[N] = val;
for(int i = N - 1; i >= 1; i--) {
bool flag = 0;
for(int j = 0; j <= 3; j++) {
int nx = f[i + 1]; f[i] = j;
if((a[i] == (f[i] | f[i + 1])) && (b[i] == (f[i] & f[i + 1]))) {flag = 1; break;}
}
if(!flag) return ;
}
puts("YES");
for(int i = 1; i <= N; i++) printf("%d ", f[i]);
exit(0);
}
main() {
//Fin(a);
N = read();
for(int i = 1; i <= N - 1; i++) a[i] = read();
for(int i = 1; i <= N - 1; i++) b[i] = read();
solve(0); solve(1); solve(2); solve(3);
puts("NO");
return 0;
}
/*
*/