59 lines
1.5 KiB
C++
59 lines
1.5 KiB
C++
#include <iostream>
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#include <cstdio>
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#include <cstring>
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using namespace std;
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const int N = 1e5 + 7;
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int n, m;
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int a[N];
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int d[N];
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("HDU5883.in", "r", stdin);
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#endif
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int T;
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scanf("%d", &T);
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while (T--) {
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memset(d, 0, sizeof d);
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scanf("%d%d", &n, &m);
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for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
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while (m--) {
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int a, b;
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scanf("%d%d", &a, &b);
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d[a]++, d[b]++;
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}
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// 奇数度点的个数
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int cnt = 0;
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for (int i = 1; i <= n; i++)
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if (d[i] & 1) cnt++;
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// 不是0，而且个数不是2，那么肯定没有欧拉通路，也没有欧拉回路
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if (cnt && cnt != 2) {
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puts("Impossible");
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continue;
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}
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int ans = 0;
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for (int i = 1; i <= n; i++) {
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int x = (d[i] + 1) >> 1;
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if (x & 1) ans ^= a[i];
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}
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// 如果没有度数为奇数的点，也就是欧拉回路,出发点终点是同一个点，这样会造成起点的权值被异或干掉
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// 因为欧拉回路可以以任意一个点做为起点和终点，所以，需要枚举一遍，分别尝试以i这起点，找出最大异或值
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int res = ans;
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if (cnt == 0)
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for (int i = 1; i <= n; i++) res = max(res, ans ^ a[i]);
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// 输出异或和
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printf("%d\n", res);
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}
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return 0;
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} |