分析
按照逻辑关系建图,跑tarjan,如果上下点在一个环中,说明不可能,不然就可能。
代码
#include <bits/stdc++.h>
#define ll long long
#define ms(a, b) memset(a, b, sizeof(a))
#define inf 0x3f3f3f3f
using namespace std;
template <typename T>
inline void read(T &x) {
x = 0; T fl = 1;
char ch = 0;
while (ch < '0' || ch > '9') {
if (ch == '-') fl = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 1) + (x << 3) + (ch ^ 48);
ch = getchar();
}
x *= fl;
}
#define N 2000005
struct edge {
int to, nt;
}E[N << 1];
int dfn[N], low[N], H[N], S[N], belong[N];
bool vis[N];
int cnt, tot, top, scc, n, m;
void add_edge(int u, int v) {
E[++ cnt] = (edge){v, H[u]};
H[u] = cnt;
}
void tarjan(int u) {
dfn[u] = low[u] = ++ tot;
vis[u] = 1;
S[top ++] = u;
for (int e = H[u]; e; e = E[e].nt) {
int v = E[e].to;
if (!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
}
else if (vis[v]) {
low[u] = min(low[u], low[v]);
}
}
int j;
if (dfn[u] == low[u]) {
scc ++;
do {
j = S[-- top];
belong[j] = scc;
vis[j] = 0;
}
while (u != j);
}
}
int main() {
read(n); read(m);
for (int i = 1; i <= m; i ++) {
int u, v, a, b;
read(u); read(a); read(v); read(b);
add_edge(u + (a ^ 1) * n, v + b * n);
add_edge(v + (b ^ 1) * n, u + a * n);
}
for (int i = 1; i <= 2 * n; i ++) {
if (!dfn[i]) tarjan(i);
}
for (int i = 1; i <= n; i ++) {
if (belong[i] == belong[i + n]) {
printf("IMPOSSIBLE\n");
return 0;
}
}
printf("POSSIBLE\n");
for (int i = 1; i <= n; i ++) {
printf("%d ", belong[i] > belong[i + n]);
}
return 0;
}