POJ 1182 食物链——并查集

挑战程序设计88页原题

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

using namespace std;

const int maxn = 150000 + 10;

int par[maxn], ran[maxn];

void init(int n) {
    for (int i = 0; i <= n; i++) {
        par[i] = i, ran[i] = 0;
    }
}

int seek(int x) {
    if (par[x] == x) return x;
    else return par[x] = seek(par[x]);
}

void unite(int x, int y) {
    x = seek(x), y = seek(y);
    if (x == y) return;
    if (ran[x] == ran[y]) par[x] = y;
    else {
        par[y] = x;
        if (ran[x] == ran[y]) ran[x]++;
    }
}

bool same(int x, int y) {
    if (seek(x) == seek(y)) return true;
    else return false;
}

int N, K;
int T[maxn], X[maxn], Y[maxn];

void solve() {
    init(N * 3);

    int ans = 0;
    for (int i = 0; i < K; i++) {
        int t = T[i];
        int x = X[i] - 1, y = Y[i] - 1;

        if (x < 0 || N <= x || y < 0 || N <= y) {
            ans++;
            continue;
        }

        if (t == 1) {
            if (same(x, y + N) || same(x, y + 2 * N)) {
                ans++;
            }
            else {
                unite(x, y);
                unite(x + N, y + N);
                unite(x + 2 * N, y + 2 * N);
            }
        }

        else {
            if (same(x, y) || same(x, y + 2 * N)) {
                ans++;
            }
            else {
                unite(x, y + N);
                unite(x + N, y + 2 * N);
                unite(x + 2 * N, y);
            }
        }
    }
    printf("%d\n", ans);
}

int main()
{
    scanf("%d %d", &N, &K);
    for (int i = 0; i < K; i++) {
        scanf("%d %d %d", &T[i], &X[i], &Y[i]);
    }
    solve();
    return 0;
}