牛客寒假算法基础集训营4 - F Applese 的QQ群 二分+拓扑排序

题目链接

题意：一共 $n$ 个人， $m$ 个操作，每个操作 $x，y$ 表示第 $x$ 个人有一条有向边指向第 $y$ 个人，每次操作后询问当前整个关系网是否存在环。

思路：题意可知，答案的输出必然先全是"Yes"，之后若出现"No"，则之后全是"No"，所以我们可以二分最后一次"Yes"出现的位置，每次判断是否有环用拓扑排序即可。

#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
#define N 100005
int n, m;
queue<int> q;
int head[N] = {}, du[N];
int edge_num = 0;
struct edge
{
    int to, sign, next;
} e[N << 1];
inline bool ToPo(const int &tt)
{
    while(!q.empty())
        q.pop();
    memset(du, 0, sizeof(int) * (n + 1));
    for (int i = 1; i <= n; ++i)
        for (int j = head[i]; j != 0; j = e[j].next)
            if (e[j].sign <= tt)
                ++du[e[j].to];
    for (int i = 1; i <= n; ++i)
        if (du[i] == 0)
            q.push(i);
    int x;
    while(!q.empty())
    {
        x = q.front(), q.pop();
        for (int i = head[x]; i != 0; i = e[i].next)
        {
            if (e[i].sign > tt)
                continue;
            --du[e[i].to];
            if (du[e[i].to] == 0)
                q.push(e[i].to);
        }
    }
    for (int i = 1; i <= n; ++i)
        if (du[i] > 0)
            return false;
    return true;
}
int main()
{
    cin >> n >> m;
    int x, y;
    for (int i = 1; i <= m; ++i)
    {
        cin >> x >> y;
        e[++edge_num].to = y, e[edge_num].sign = i, e[edge_num].next = head[x], head[x] = edge_num;
    }
    int l = 1, r = m;
    int mid;
    while (l < r)
    {
        mid = (l + r) >> 1;
        if (ToPo(mid + 1))
            l = mid + 1;
        else
            r = mid;
    }
    for (int i = 1; i <= l; ++i)
        puts("Yes");
    for (int i = l + 1; i <= m; ++i)
        puts("No");
    return 0;
}