题意:一共 个人, 个操作,每个操作 表示第 个人有一条有向边指向第 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;
}