Codeforces 776D The Door Problem 【并查集】

题目链接：http://codeforces.com/contest/776/problem/D

题意：

有一些锁，初始状态为ri，其中ri为1表示unlock，为0表示lock，给出锁与锁之间的关系，即拉一下第i个锁，与第i个锁相连的其它锁都要改变一下状态，问有没有可能让所有的锁都变成unlock状态（即都为1）

题解：

并查集维护即可，使x表示执行了第x号命令，x+m表示不执行第x号命令，维护即可。

P.S：这题评测时间真心长，请大家耐心等待。

代码：

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

using namespace std;

typedef long long LL;

const int inf = 1 << 26;

vector<int> sav[200005];
int fath[400005], a[200005];

int find(int x) {
if(fath[x] == x) return fath[x];
return fath[x] = find(fath[x]);
}

void merge(int x, int y) {
int fx = find(x), fy = find(y);
if(fx == fy) return ;
fath[fy] = fx;
}

int main(){
int n, m;
scanf("%d %d", &n, &m);
for ( int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
for ( int i = 1; i <= m; i ++ ) {
int x;
scanf("%d", &x);
for ( int j = 0; j < x; j ++ ) {
int y;
scanf("%d", &y);
sav[y].push_back(i);
}
}

for ( int i = 0; i <= 400000; i ++ ) fath[i] = i;
for ( int i = 1; i <= n; i ++ ) {
int x = sav[i][0], y = sav[i][1];
if(a[i]) {
merge(x, y);
merge(x+m, y+m);
} else {
merge(x+m, y);
merge(x, y+m);
}
}

for ( int i = 1; i <= m; i ++ ) {
if(find(i) == find(i+m)) {
puts("NO");
return 0;
}
}
puts("YES");

return 0;
}