POJ Butterfly

使用队列，核心思想还是找到矛盾的地方，用一个一维数组保存种类，一个二维数组保存关系，如果遍历之后，两种蝴蝶种类的异或与关系矩阵中的内容不符合，那么就是有矛盾，不可能实现。

#include <cstring>
#include <iostream>
#include <queue>
#include <math.h>
#include <vector>
#include <algorithm>
#include <map>

#define MAX_N 1010
#define MAX_M 100010
using namespace std;

int n, m;
bool visited[MAX_N];
bool typee[MAX_N];
int relations [MAX_N][MAX_N];

int a, b, c;
int main()
{
    while (scanf(\"%d%d\", &n, &m)!=EOF) {
        memset(visited, false, sizeof(visited));
        memset(typee, true, sizeof(typee));
        memset(relations, -1, sizeof(relations));
        for (int i = 0; i < m; i++) {
            scanf(\"%d%d%d\", &a, &b, &c);
            relations[a][b] = relations[b][a] = c;
        }
        int visnum = 0, cur;
        queue<int> q;
        while (visnum != n) {
            for (int i = 0; i < n; i++) {
                if (!visited[i]) {
                    q.push(i);
                    visnum++;
                    visited[i] = true;
                    break;
                }
            }
            if (visnum == n) {
                break;
            }
            while (!q.empty()) {
                cur = q.front();
                q.pop();
                for (int i = 0; i < n; i++) {
                    if (relations[cur][i] != -1) {
                        if (!visited[i]) {
                            if (relations[cur][i] == 0) {
                                typee[i] = typee[cur];
                            }else{
                                typee[i] = !typee[cur];
                            }
                            q.push(i);
                            visnum++;
                            visited[i] = true;
                        }
                    }
                }
            }
        }
        bool flag = true;
        for (int i = 0; flag && i < n; i++) {
            for (int j = 0; flag && j < n; j++) {
                if (relations[i][j] == -1) {
                    continue;
                }
                if (relations[i][j] != (typee[i] ^ typee[j])) {
                    flag = false;
                }
            }
        }
        if (flag) {
            cout << \"YES\\n\";
        }else{
            cout << \"NO\\n\";
        }
    }
    
    return 0;
}