对于一个任务，有n个步骤，其中一些步骤要前置步骤完成后才能做，每个步骤要求在1号或者2号实验室做，要求在两个实验室之间往返的次数最少
初始点题目没说，就是1或2都可以，我们要分类讨论，然后拓扑即可，往返移动
#include<iostream>
#include<cstdio>
#include<queue>
//#include<windows.h>
#define rep(i,j,k) for(register int i(j);i<=k;++i)
#define drp(i,j,k) for(register int i(j);i>=k;--i)
#define bug cout<<"~~~~~~~~~~~~~"<<'
';
#define bugout(x) cout<<x<<endl;
typedef long long lxl;
template<typename T>
inline T max(T &a, T &b) {
return a > b ? a : b;
}
template<typename T>
inline T min(T a, T b) {
return a < b ? a : b;
}
inline char gt() {
static char buf[1 << 21], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
template <typename T>
inline void read(T &x) {
register char ch = gt();
x = 0;
int w(0);
while(!(ch >= '0' && ch <= '9'))w |= ch == '-', ch = gt();
while(ch >= '0' && ch <= '9')x = x * 10 + (ch & 15), ch = gt();
w ? x = ~(x - 1) : x;
}
template <typename T>
inline void out(T x, char cc) {
if(x < 0) x = -x, putchar('-');
char ch[20];
int num(0);
while(x || !num) ch[++num] = x % 10 + '0', x /= 10;
while(num) putchar(ch[num--]);
putchar(cc);
}
const int N = 3e5 + 79;
const int M = 2e6 + 79;
int degree[N], n, m, a[N], tot, in[N];
struct graph {
int head[N], tot, next[M], ver[M];
inline void add(int a, int b) {
ver[++tot] = b;
next[tot] = head[a];
head[a] = tot;
}
} G;
int q[N];
int temp[N], cnt;
inline int topsort(int st) {
int step(0);
rep(i, 1, n) in[i] = degree[i];
q[q[0] = 1] = 0;
a[0] = st;
int pos(st);
while(1) {
step++;
rep(i, 1, cnt) {
q[++q[0]] = temp[i];
}
cnt = 0;
rep(kkk, 1, q[0]) {
int x = q[kkk];
for(register int i(G.head[x]); i; i = G.next[i]) {
int y(G.ver[i]);
in[y]--;
if(!in[y] && a[y] == pos) q[++q[0]] = y;
else if(!in[y]) temp[++cnt] = y;
}
}
q[0] = 0;
pos++;
if(pos == 3) pos = 1;
if(!cnt) break;
}
return step - 1;
}
int main() {
freopen("save.in", "r", stdin);
freopen("save.out", "w", stdout);
read(n);
read(m);
rep(i, 1, n) {
read(a[i]);
G.add(0, i);
degree[i]++;
}
int x, y;
rep(i, 1, m) {
read(x);
read(y);
G.add(x, y);
degree[y]++;
}
out(min(topsort(1), topsort(2)), '
');
return 0;
}