HDU - 3715 Go Deeper (二分 ＋ 2-SAT)

题目大意：给出一个递归函数，问这个递归函数最多能递归几层

解题思路：二分枚举递归层数，然后依此建边

如果给出的c[i]为0时，那么x[a[i]]和x[b[i]]中的其中一个要为真，连边即为 !a[i]->b[i], !b[i] -> a[i]

如果给出的c[i]为1时，那么x[a[i]]和x[b[i]]两个要么都为真，要么都为假，连边即为a[i]->b[i], b[i]->a[i], !a[i]->!b[i], !b[i]->!a[i]

如果给出的c[i]为2时，那么x[a[i]]和x[b[i]]两个中的一个要为假
，连边即为a[i]->!b[i], b[i]->!a[i]

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

#define N 10010
#define M 410

struct Node {
    int a, b, c;
}node[N];

vector<int> G[M];
bool mark[M];
int S[M];
int n, m, top;

void init() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i++) {
        scanf("%d%d%d", &node[i].a, &node[i].b, &node[i].c);
    }
}

bool dfs(int u) {
    if (mark[u ^ 1])
        return false;
    if (mark[u])
        return true;
    mark[u] = true;
    S[++top] = u;
    for (int i = 0; i < G[u].size(); i++)
        if (!dfs(G[u][i]))
            return false;
    return true;
}

void AddEdge(int x, int valx, int y, int valy) {
    G[2 * x + valx].push_back(y * 2 + valy);
}

bool judge(int mid) {
    for (int i = 0; i < 2 * n; i++)
        G[i].clear();

    for (int i = 1; i <= mid; i++) {
        if (node[i].c == 0) {
            AddEdge(node[i].a, 0, node[i].b, 1);
            AddEdge(node[i].b, 0, node[i].a, 1);
        }
        else if (node[i].c == 1) {
            AddEdge(node[i].a, 0, node[i].b, 0);
            AddEdge(node[i].b, 0, node[i].a, 0);
            AddEdge(node[i].a, 1, node[i].b, 1);
            AddEdge(node[i].b, 1, node[i].a, 1);
        }
        else if (node[i].c == 2) {
            AddEdge(node[i].a, 1, node[i].b, 0);
            AddEdge(node[i].b, 1, node[i].a, 0);
        }
    }

    memset(mark, 0, sizeof(mark));
    for (int i = 0; i < 2 * n; i += 2) {
        if (!mark[i] && !mark[i ^ 1]) {
            top = 0;
            if (!dfs(i)) {
                while (top) mark[S[top--]] = false;
                if (!dfs(i ^ 1))
                    return false;
            }
        }
    }
    return true;
}

void solve() {
    int l = 1, r = m, mid;
    while (l <= r) {
        mid = (l + r) / 2;
        if (judge(mid))
            l = mid + 1;
        else
            r = mid - 1;
    }
    printf("%d\n", l - 1);
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}