Uva12232-exclusive-or (with permission and query set)

Question: uva12316-exclusive-or (with permission and query set)

I P V indicates that the value of XP is v. Or I p q V represents x P ^ x I + 1 ^ x I + 2... ^ x ^ q = V; then, Q k P1 P2 p3... PK asks the difference or value of these numbers.

Solution: first, we need to clarify x ^ y = V, x ^ z = W, then y ^ z = V ^ W. Therefore, we can use and query the set to solve this problem, because as long as two numbers are different from the same number or, they belong to the same set, even if they are unknown to the value of the same number, however, the differences or values of these two numbers are unknown. The differences or values of multiple numbers are the same. as long as these two numbers are different from one of the same values, we can know that, if there is a mismatch, we cannot find it, note that a value here is unknown. If it is known, the other value can also be obtained. Then, in the case of I p v, we may wish to give it a virtual node so that it can be the same as I P Q v. Then the difference between all numbers and 0 is itself, so the initial value is 0. This requires that if there are two facts that are distinct, then the subsequent steps should not be done.

Code:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 2e4 + 5;const int maxn = 4e4 + 5;int n, Q;int p[N];int v[N];struct Query {char type;int a, b, val;int k, l[100];} q[maxn];void init () {for (int i = 0; i <= n; i++) {p[i] = i;v[i] = 0;}}int getParent (const int a) {if (a == p[a])return a;int t = p[a];p[a] = getParent (p[a]);v[a] ^= v[t];return p[a];}void solve () {int cnt = 0;for (int i = 1; i <= Q; i++) {if (q[i].type == 'I') {cnt++;int p1 = getParent (q[i].a);int p2 = getParent (q[i].b);if (p1 == p2) {if ((v[p1] ^ v[q[i].a] ^ v[q[i].b]) != q[i].val) {printf ("The first %d facts are conflicting.\n", cnt);return;}} else {if (p1 == n)swap (p1, p2);p[p1] = p2;v[p1] = (v[q[i].a] ^ v[q[i].b] ^ q[i].val);}} else {int ans = 0;int parent[N];bool flag = 1;memset (parent, 0, sizeof (parent));for (int j = 0; j < q[i].k; j++) {getParent (q[i].l[j]);ans ^= v[q[i].l[j]];if (p[q[i].l[j]] != n)parent[p[q[i].l[j]]] ^= 1;}for (int j = 0; j <= n; j++) {if (parent[j]) {flag = 0;break;}}if (flag)printf ("%d\n", ans);elseprintf ("I don't know.\n");}}}int main () {char str[105];int a, b, val;int cas = 0;while (scanf ("%d%d", &n, &Q) && (n || Q)) {init ();printf ("Case %d:\n", ++cas); for (int i = 1; i <= Q; i++) {scanf ("%s", str);q[i].type = str[0];if (q[i].type == 'I') {gets(str);if (sscanf (str, "%d%d%d", &q[i].a, &q[i].b, &q[i].val) == 2) {q[i].val = q[i].b;q[i].b = n;}} else {scanf ("%d", &q[i].k);for (int j = 0; j < q[i].k; j++)scanf ("%d", &q[i].l[j]);}}solve();printf ("\n");}return 0;}

Uva12232-exclusive-or (with permission and query set)