HDU 3715 Go Deeper 二分 + 2-sat

题目：

http://acm.hdu.edu.cn/showproblem.php?pid=3715

题意：

给定一段递归伪代码，问执行这段伪代码递归的最深层数

思路：

二分枚举答案用2-sat判定是否可行。具体建图如下：如果c[i] == 0，那么a[i] OR b[i]，如果c[i] == 1，那么（a[i] AND b[i]） OR （~a[i] AND ~b[i]），如果c[i] == 2，那么NOT（a[i] AND b[i]），然后强连通缩点判断i和~i是否在同一个环内

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;

const int N = 410;
const double eps = 1e-8;
struct edge
{
    int to, next;
} g[N*N*2];
int cnt, head[N], cnt1, head1[N];
int dfn[N], low[N], scc[N], st[N], top, num, idx;
int a[N*N], b[N*N], c[N*N];
bool vis[N];
int n, m;
void add_edge(int v, int u)
{
    g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;
}
void init()
{
    memset(head, -1, sizeof head);
    memset(dfn, -1, sizeof dfn);
    memset(vis, 0, sizeof vis);
    top = num = idx = cnt = 0;
}
void tarjan(int v)
{
    dfn[v] = low[v] = ++idx;
    vis[v] = true, st[top++] = v;
    int u;
    for(int i = head[v]; i != -1; i = g[i].next)
    {
        u = g[i].to;
        if(dfn[u] == -1)
        {
            tarjan(u);
            low[v] = min(low[v], low[u]);
        }
        else if(vis[u]) low[v] = min(low[v], dfn[u]);
    }
    if(dfn[v] == low[v])
    {
        num++;
        do
        {
            u = st[--top];
            vis[u] = false;
            scc[u] = num;
        }
        while(u != v);
    }
}
bool work(int mid)
{
    init();
    for(int i = 0; i < mid; i++)
    {
        if(c[i] == 0)
        {
            add_edge(a[i] + n, b[i]), add_edge(b[i] + n, a[i]);
        }
        else if(c[i] == 1)
        {
            add_edge(a[i], b[i]), add_edge(b[i], a[i]);
            add_edge(a[i] + n, b[i] + n), add_edge(b[i] + n, a[i] + n);
        }
        else if(c[i] == 2)
        {
            add_edge(a[i], b[i] + n), add_edge(b[i], a[i] + n);
        }
    }
    for(int i = 0; i < 2*n; i++)
        if(dfn[i] == -1) tarjan(i);
    for(int i = 0; i < n; i++)
        if(scc[i] == scc[i+n]) return false;
    return true;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        for(int i = 0; i < m; i++) scanf("%d%d%d", &a[i], &b[i], &c[i]);
        int l = 0, r = m, res;
        while(l <= r)
        {
            int mid = (l + r) / 2;
            if(work(mid)) l = mid + 1, res = mid;
            else r = mid - 1;
        }
        printf("%d\n", res);
    }
    return 0;
}