Poj 1182 food chain (type and query set)

Food Chain Time Limit:1000 MS Memory Limit:10000 K Total Submissions:41805 Accepted:12160 Description The animal kingdom contains three types of animals A, B, and C. The food chains of these three types constitute an interesting ring. A eats B, B eats C, and C eats. There are N animals numbered 1-N. Every animal is one of A, B, and C, but we don't know which one it is. There are two ways to describe the relationship between the food chains of the N animals: The first statement is "1 x y", indicating that X and Y are similar. The second statement is "2 x y", which indicates that X eats Y. This person speaks K sentences one by one for N animals in the preceding two statements. These K sentences are true or false. When one sentence meets the following three conditions, this sentence is a lie, otherwise it is the truth. 1) The current statement conflicts with some of the preceding actual statements; 2) In the current statement, X or Y is greater than N, which is false; 3) The current statement indicates that X eats X, which is a lie. Your task outputs the total number of false statements based on the given N (1 <= N <= 50,000) and K statements (0 <= K <= 100,000. Input The first line is two integers N and K, separated by a space. Each row in the following K rows contains three positive integers, D, X, and Y, which are separated by a space. D indicates the type of the statement. If D = 1, X and Y are of the same type. If D = 2, X eats Y. Output Only one integer indicates the number of false statements. Sample Input 100 71 101 1 2 1 22 2 3 2 3 3 1 1 3 2 3 1 1 5 5 Sample Output 3 Source Noi 01

The idea comes from and detailed explanations can be found in the drifting mavericks

Ideas:

Check the set and maintain an array of pre-root nodes and an array of links (offsets) from rel-root nodes to this point. Rel has only three values: 0, 1, and 2. 0 indicates the same relationship, 1 indicates that the root node eats the node, and 2 indicates that the root node is eaten by the node.

When op, x, and y are input, fx = Find (x), fy = Find (y); If fx! = Fy, merge the fx set and fy set. pre [fy] = fx; fx-> fy = fx-> x + x-> y + y-> fy; x-y is the op-1, that is, rel [fy] = (rel [x] + op-1-rel [y] + 3) % 3; If fx = fy, check whether there is a conflict, x-> y = x-> fx + fx-> y, that is, test (-rel [x] + rel [y] + 3) Whether % 3 is equal to the op-1.

Update the rel array of each vertex when searching. Update the rel Array Based on the previous rel and the current rel of its father. rel [x] = (rel [px] + rel [x]) % 3;

Ps: This question is very difficult. It only requires a single set of data. WA is required for multiple groups.

Code:

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            # Pragma comment (linker, "/STACK: 102400000,102400000") # define maxn 50005 # define MAXN 100005 # define OO (1 <31) -1 # define mod 90003 # define INF 0x3f3f3f3f # define pi acos (-1.0) # define eps 1e-6typedef long ll; using namespace std; int n, m, ans, tot, flag, cnt; int pre [maxn], rel [maxn]; int Find (int x) {if (x = pre [x]) return x; int tmp = pre [x]; pre [x] = Find (pre [x]); rel [x] = (rel [tmp] + rel [x]) % 3; return pre [x];} bool Merge (int op, int x, int y) {int fx = Find (x), fy = Find (y); if (fx = fy) // determine whether a conflict exists in the same set {if (-rel [x] + rel [y] + 3) % 3! = Op-1) return true;} else // merge {pre [fy] = fx; rel [fy] = (rel [x] + op-1-rel [y] + 3) % 3;} return false;} int main () {int I, j, t, op, x, y; scanf ("% d", & n, & m); for (I = 1; I <= n; I ++) pre [I] = I, rel [I] = 0; ans = 0; for (I = 1; I <= m; I ++) {scanf ("% d", & op, & x, & y ); if (x> n | y> n | op = 2 & x = y) {ans ++; continue;} if (Merge (op, x, y) ans ++;} printf ("% d \ n", ans); return 0 ;}