POJ 1182 食物链——并查集

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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;}

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