"Original" POJ-----1182 food chain Problem Solving report

Title Address:

http://poj.org/problem?id=1182

Topic content:

Food chains

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48791		Accepted: 14222

Description

There are three types of animal a,b,c in the animal kingdom, and the food chain of these three animals is an interesting ring. A eat B, B eat c,c eat a.
Existing n animals, numbered with 1-n. Every animal is one of the a,b,c, but we don't know what it is.
Some people describe the food chain relationship between these n animals in two ways:
The first argument is "1 x y", which means that x and Y are homogeneous.
The second argument is "2 x y", which means x eats y.
This person to n animals, with the above two statements, a sentence after sentence to say K sentence, this k sentence some is true, some false. When one sentence satisfies one of the following three, the sentence is a lie, otherwise it is the truth.
1) The current words conflict with some of the preceding words, which is false;
2) The current word in x or y is greater than N, is a lie;
3) The current words say x eats x, is a lie.
Your task is to output the total number of falsehoods according to the given N (1 <= n <= 50,000) and the K-sentence (0 <= K <= 100,000).

Input

The first line is two integers n and K, separated by a single space.
The following k lines each line is three positive integer d,x,y, and two numbers are separated by a space, where D denotes the kind of claim.
If d=1, it means that x and Y are homogeneous.
If d=2, it means x eats y.

Output

There is only one integer that represents the number of false lies.

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 Problem-solving ideas: A well-liked food chain, is worthy of Noi's topic, indeed very difficult. The general direction is to use the right and check set to do, or see a merge one, see two merge a pair, and then update the node weights. If two nodes are already in the collection, and the accumulated node weights and the given relationship contradiction, then even a mistake. Of course, out of scope is also wrong. Well, let's define the right to take it. Define array Num[n], where num[i] (i < N) means the first animal relationship to its root node。 If it is 0, then I and the root node are the same species, if it is 1, then I am eaten by the root node, if it is 2, then I eat root node. Strictly speaking, at any one time, Num[i] can only guarantee that the stored value is the relationship between I and the previous node, and does not necessarily preserve the relationship with the root node. But because before querying num[i], we would call the Find_root method, which would direct I to the root node and maintain the value of num[i], so we can assume that Num[i] maintains the relationship between I and the root node. The next question is how we update the value of Num[i] in Find_root. First, we have to solve a problem. What are the relationships between A and b known and B and C, and the relationship between A and C? That is, for example: a----> B----> C (meaning A is eaten by B, B is eaten by C) 1 1 easy to see, a----> C is also equal to 2, that is, A is eaten by B, B is eaten by C, a can eat C sign convention: may be false We have 3 animals. A, B, c,a, and B are related to x, whereas the relationship between B and C is Y, then the relationship between A and C is [X][y]. There are: [0][1] = 1[0][0] = 0[0][2] = 2 [1][0] = 1[1][1] = 2[1][2] = 0 [2][0] = 2[2][1] = 0[2][2] = 1 It can be seen that the relationship between a and C is actually equal to (x + y)% 3. Therefore, in the recursion of Find_root, we have completed an update to the node on the current node, we can use this relationship, the previous node as a bridge, to update the current node to the root node after the weight. There are several formulas that need to be used: if the value of a----> B is known to be x, then the value of B----> A is (3-x)% 3. If a----> b----> C----> The relationships of neighboring nodes in D are known, then we can ask A to D based on a----> b----> C, then the problem is converted to a----> C----> D, You can use that formula to find the relationship between A and d. All code:

#include <stdio.h>intanimal[50001 ];intrelation[50001 ];intn,k;//1 is eaten by the root node, 0 is the same as the root node, 2 means eating root nodeintFind_root (intChild ) {  if(animal[child] = =0 )    returnChild ; intTMP =animal[Child]; animal[Child]=Find_root (animal[child]); relation[Child]= (relation[child] + relation[TMP])%3; returnanimal[Child];}intUnion_set_and_judge (intOneintBoth,intrel) {  if(rel = =1) rel=0; intFAT1 =Find_root (one); intFAT2 =Find_root (both); if(FAT1! =fat2) {animal[FAT1]=Fat2; intFat2one = (3-relation[one])%3; intFat2two = (fat2one + rel)%3; intFat2fat = (fat2two + relation[))%3; relation[FAT1]=Fat2fat;    return 0; }  intFat2two = (3-relation[])%3; intOne2two = (relation[one] + fat2two)%3;  returnOne2two = = rel?1:2;}intMainvoid) {scanf ("%d%d", &n, &k); intCount =0;  for(inti =0; I < K; i + + ) {    intD,x,y; scanf ("%d%d%d", &d, &x, &y); if(X > N | | y >N) {count++; Continue; }    intFlag =Union_set_and_judge (x, y, D); if(Flag = =2) {Count++; }} printf ("%d\n", Count); return 0;}

"Original" POJ-----1182 food chain Problem Solving report