POJ 1182 食物链

题目链接：POJ 1182 食物链

题目大意：

题解：

并查集+思维

开了三倍大小的标记数组来表示三个物种，\(1\)到\(n\)为\(A\)物种，\(n+1\)到\(2 \times n\)为\(B\)物种，\(2 \times n + 1\)到\(3 \times n\)为\(C\)物种。
如果\(u\)吃\(v\)，则相对的\(u+n\)与\(v\)为一个物种，\(u+2\times n\)与\(v + n\)为一个物种，\(u\)与\(v + 2\times n\)为一个物种。
通过并查集关联同一物种。

#include <iostream>
#include <cstdio>
#include <ctype.h>
using namespace std;
#define MAXN 100010
#define INF 100000000
#define ll long long

ll read() {     // fast read
    ll ans = 0; int f = 1;  char x = getchar();
    while (!isdigit(x)) {
        if (x == '-')
            f = -1;
        x = getchar();
    }
    while (isdigit(x)) {
        ans = (ans << 3) + (ans << 1) + (x ^ 48);
        x = getchar();
    }
    return ans * f;
}

int fa[150050], n, k, act, u, v, ans;

int find(int x) {   // find father
    return fa[x] == x ? x : fa[x] = find(fa[x]);
}

int main() {
    n = read(), k = read();
    for (int i = 1; i <= n; ++i) {
        fa[i] = i;
        fa[i + n] = i + n;
        fa[i + n * 2] = i + n * 2;
    }
    while (k--) {
        act = read(), u = read(), v = read();
        if (u > n || v > n) {
            ans++;
            continue;
        }
        else if (act == 1) {
            if (find(u) == find(v + n) || find(v) == find(u + n)) {
                ans++;
            }
            else {  // same father
                fa[find(u)] = find(v);
                fa[find(u + n)] = find(v + n);
                fa[find(u + n * 2)] = find(v + n * 2);
            }
        }
        else {
            if (find(u) == find(v + n) || find(u) == find(v)) {
                ans++;
            }
            else {  // u -> v
                fa[find(u + n)] = find(v);
                fa[find(u + n * 2)] = find(v + n);
                fa[find(u)] = find(v + n * 2);
            }
        }
    }
    cout << ans;
    return 0;
}

---

带权并查集

// 带权并查集
#include <cstdio>
#include <iostream>
using namespace std;

int f[50005], d[50005], n, k, d1, x, y, ans;

int find(int x) {
    if (x != f[x]) {
        int xx = f[x];
        f[x] = find(f[x]);
        d[x] = (d[x] + d[xx]) % 3;
    }
    return f[x];
}

int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) {
        f[i] = i;
        d[i] = 0;
    }
    for (int i = 1; i <= k; i++) {
        scanf("%d%d%d", &d1, &x, &y);
        if ((d1 == 2 && x == y) || (x > n || y > n)) {
            ans++;
            continue;
        }
        if (d1 == 1) {
            if (find(x) == find(y)) {
                if (d[x] != d[y]) ans++;
            } else {
                d[f[x]] = (d[y] - d[x] + 3) % 3;
                f[f[x]] = f[y];
            }
        }
        if (d1 == 2) {
            if (find(x) == find(y)) {
                if (d[x] != (d[y] + 1) % 3) ans++;
            } else {
                d[f[x]] = (d[y] - d[x] + 4) % 3;
                f[f[x]] = f[y];
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}