[Poj 1182] food chain, poj1182 food chain

[Poj 1182] food chain, poj1182 food chain
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

Eat C, C eat.

There are N animals numbered 1-N. Every animal is A, B, and C, but we do not know

Which one is it.

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.

2 x y indicates that X eats Y.

This person speaks K sentences one by one for N animals in the preceding two statements. Some of these K sentences are true.

Some are false. When one sentence meets the following three conditions, this sentence is a lie, otherwise it is the truth.

• The current statement conflicts with some of the preceding actual statements.

• In the current statement, X or Y is greater than N, which is false.

• 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 and K sentences.

Input/Output Format

Input Format:

The first line has two integers, N, K, indicating N animals and K sentences.

The second line begins with one sentence per line (see the example below as required)

Output Format:

A row, an integer, indicates the total number of false statements.

Input and Output sample input sample #1:

100 71 101 12 1 22 2 32 3 31 1 32 3 11 5 5

Output sample #1:

3

Description

1 ≤ N ≤ 5 ≤ 10 ^ 4

1 ≤ K ≤ 10 ^ 5

Question

Create three elements I-A, I-B, and I-C for each animal I, and create and query sets with these 3 x N elements. This query sets the following information for maintenance:

~ I-x indicates "I belongs to Type x"

~ Check each group in the set to indicate that all elements in the group occur simultaneously or not.

For example, if I-A and j-B are in the same group, it indicates that if I belongs to type A, j must belong to Type B. If j belongs to type

B, then I must belong to category. Therefore, you only need to perform the following operations on each piece of information.

~ First, x and y belong to the same type... merge x-A and y-A, x-B and y-B, x-C, and y-C.

~ In the second type, x eats y ........................... merge x-A, y-B, x-B, y-C, x-C, and y-.

However, before merging, you must first determine whether the merger will produce contradictions. For example, in the case of the first type of information,

Check whether x-A, y-B, or y-C is in the same group.

 

1 # include <cstdio> 2 # include <iostream> 3 using namespace std; 4 const int maxn = 50005; 5 const int maxm = 100005; 6 int n, m; int par [maxn * 3], deep [maxn * 3]; // It must be multiplied by 3, where 7 int T [maxm], X [maxm], Y [maxm]; 8 void init (int n) {9 for (int I = 0; I <n; I ++) {10 par [I] = I; 11 deep [I] = 0; 12} 13} 14 int find (int x) {15 if (par [x] = x) return x; 16 else return par [x] = find (par [x]); 17} 18 void unite (int x, int y) {19 x = find (x ), y = find (y); 20 if (x = y) return; 21 if (deep [x] <deep [y]) par [x] = y; 22 else {23 par [y] = x; 24 if (deep [x] = deep [y]) deep [x] ++; 25} 26} 27 bool same (int x, int y) {28 return find (x) = find (y); 29} 30 int read () {// cin, cout on poj T 31 int x = 0, f = 1; char ch = getchar (); 32 while (ch <'0' | ch> '9 ') {if (ch = '-') f =-1; ch = getchar ();} 33 while (ch> = '0' & ch <= '9 ') {x = x * 10 + ch-'0'; ch = getchar ();} 34 return x * f; 35} 36 int main () {37 n = read (), m = read (); 38 for (int I = 0; I <m; I ++) T [I] = read (), X [I] = read (), Y [I] = read (); 39 init (n * 3); 40 int ans = 0; 41 for (int I = 0; I <m; I ++) {42 // cin> T [I]> X [I]> Y [I]; 43 int t = T [I], x = X [I]-1, y = Y [I]-1; 44 if (x <0 | x> = n | y <0 | y> = n) {ans ++; continue;} 45 if (t = 1) {46 if (same (x, y + n) | same (x, y + 2 * n) ans ++; 47 else {48 unite (x, y ); 49 unite (x + n, y + n); 50 unite (x + 2 * n, y + 2 * n ); 51} 52} 53 else {54 if (same (x, y) | same (x, y + 2 * n) ans ++; 55 else {56 unite (x, y + n); 57 unite (x + n, y + 2 * n); 58 unite (x + 2 * n, y ); 59} 60} 61} 62 printf ("% d \ n", ans );
63 return 0; 64}