题目地址:
https://ac.nowcoder.com/acm/contest/6112/B
思路:
对于异或运算,只需要特判两种情况,
1.当mp[x]^mp[y]==k时,一步到达
2.当mp[x]==mp[y]时,如果存在中间值使得mp[x]^mp[i]==k即可两步到达,这题时间卡的紧,使用快读读入,开数组记录读入的数据。
代码:
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=2e6+5;
int mp[maxn], vis[maxn];
inline int read()
{
int x = 0, w = 1;
char ch = 0;
while (ch < '0' || ch > '9')
{
if (ch == '-')
w = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
x = (x << 3) + (x << 1) + ch - '0';
ch = getchar();
}
return x * w;
}
int main()
{
int n, q;
n = read();
q = read();
for (int i = 1; i <= n; i++)
{
mp[i] = read();
vis[mp[i]] = 1;
}
int k, x, y;
for (int i = 0; i < q; i++)
{
k = read();
x = read();
y = read();
if ((mp[x] ^ mp[y]) == k)
printf("1\n");
else if (mp[x] == mp[y] && vis[k ^ mp[x]])
printf("2\n");
else
printf("-1\n");
}
return 0;
}