Bitwise Exclusive-OR Sequence

题意：

有n个点， m个关系每个关系 a b c 表示点a的值异或点b的值，为c

分析

首先可能是多个图，对于每个图中的每一位，取0 或者取1 都是可以确定图上的其他数字取0 或者取1 。所以我们可以去枚举取0 或者取1 ，来取个min 得到结果。
所以这里我们用并查集，最大数为 2的30次，所以我们对于每一位都做一个并查集，就是30个并查集。 f[a] 表示当前位置取1 f[a+n] 表示当前位取0
这样就可以枚举区间了。

代码

#include <bits/stdc++.h>using namespace std;
typedef long long LL;const int N = 2e5+10;LL n, m, fa[N], sz[N];
LL a[N];struct Node
{LL a, b;LL val;
}num[N];LL find(LL x)
{if(fa[x] != x) fa[x] = find(fa[x]);return fa[x];
}void init()
{for(LL i = 1; i <= n; i ++){fa[i] = i, sz[i] = 1;fa[i+n] = i+n, sz[i+n] = 0;}
}void merge(LL a, LL b)
{sz[a] += sz[b];fa[b] = a;
}int main()
{cin >> n >> m;for(int i = 1; i <= m; i ++){scanf("%lld%lld%lld", &num[i].a, &num[i].b, &num[i].val);}LL res = 0;for(int i = 0; i < 30; i ++){init(); // 每一次做并查集都要初始化！for(int j = 1; j <= m; j ++){int aa = find(num[j].a);int bb = find(num[j].b);int aaa = find(num[j].a + n);int bbb = find(num[j].b + n);if((num[j].val >> i) & 1 ) // 不同为1 ，一个取0一个取1{if(aa == bb){cout << -1 << endl;return 0;}if(aa == bbb) continue;merge(aa, bbb);merge(bb, aaa);}else // 相同为0， 都取1 或者都取0{if(aa == bbb ){cout << -1 << endl;return 0;}if(aa == bb) continue;merge(bb, aa);merge(bbb, aaa);}}for(int j = 1; j <= n; j ++) // 对于每个数字计算结果{res += (LL)min(sz[find(j)], sz[find(j+n)]) * (1 << i);sz[find(j)] = 0;sz[find(j+n)] = 0;}}cout << res << endl;return 0;
}
/*
3 3
1 2 1
2 3 1
1 3 1 */

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Bitwise Exclusive-OR Sequence

Bitwise Exclusive-OR Sequence

题意：

分析

代码

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