POJ - 1182 食物链

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 50003;

int n, k, ans, d, x, y;
int par[MAXN], relation[MAXN];//par[x]表示x与par[x]的关系能够确定，relation[x]表示x与par[x]的关系：0表示同类，1表示par[x]吃x，2表示x吃par[x]

int getPar(int a) {
    if(par[a] != a) {
        int pa = getPar(par[a]);
        relation[a] = (relation[a] + relation[par[a]]) % 3;
        par[a] = pa;
    }
    return par[a];
}

void merg(int d, int a, int b) {
    int pa = getPar(a), pb = getPar(b);
    if(pa == pb) {
        if((relation[b] - relation[a] + 3) % 3 != d) {
            ++ans;
        }
        return ;
    }
    par[pb] = pa;
    relation[pb] = (relation[x] - relation[y] + d + 3) % 3;
}

int main() {
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; ++i) {
        par[i] = i;
        relation[i] = 0;
    }
    ans = 0;
    while(k-- > 0) {
        scanf("%d%d%d", &d, &x, &y);
        if(x > n || y > n || (d == 2 && x == y)) {
            ++ans;
            continue;
        }
        merg(d - 1, x, y);
    }
    printf("%d\n", ans);
    return 0;
}