Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

2-SAT 模板

用于求解逻辑判断的问题。

查看代码
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#include <iostream>
#include <cstdio>
#include <stack>
using namespace std;
const int N = 2e6 + 10, M = 2e6 + 10;
int n, m;
int idx, hd[N], nxt[M], edg[M];
bool vis[N];
int cnt, stmp, id[N], dfn[N], low[N];
stack<int> stk;
void tarjan(int x)
{
dfn[x] = low[x] = ++stmp;
stk.push(x);
vis[x] = true;
for (int i = hd[x]; ~i; i = nxt[i])
if (!dfn[edg[i]])
{
tarjan(edg[i]);
low[x] = min(low[x], low[edg[i]]);
}
else if (vis[edg[i]])
low[x] = min(low[x], dfn[edg[i]]);
if (low[x] == dfn[x])
{
int y;
cnt++;
do
{
y = stk.top();
stk.pop();
vis[y] = false;
id[y] = cnt;
} while (y != x);
}
}
void add(int a, int b)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 0; i < (n << 1); i++)
hd[i] = -1;
while (m--)
{
int i, a, j, b;
scanf("%d%d%d%d", &i, &a, &j, &b);
i--, j--;
add(i << 1 | !a, j << 1 | b);
add(j << 1 | !b, i << 1 | a);
}
for (int i = 0; i < n << 1; i++)
if (!dfn[i])
tarjan(i);
for (int i = 0; i < n; i++)
if (id[i << 1] == id[i << 1 | 1])
{
printf("IMPOSSIBLE\n");
return 0;
}
printf("POSSIBLE\n");
for (int i = 0; i < n; i++)
printf("%d ", id[i << 1] > id[i << 1 | 1]);
return 0;
}