>Link
ybtoj道路与航线
>Description
>解题思路
直接二分答案，BFS判断是否能通过在这个时间 mid 之前建的边 从a到b / 从b到a
>代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define N 6000010
using namespace std;
queue<int> Q;
struct edge
{
int to, nxt;
} e[N * 2];
struct node
{
int u, v, k;
} p[N];
int n, m, a, b, cnt, h[N], ans;
bool vis[N];
int read ()
{
int ret = 0;
char c = getchar();
while (c < '0' || c > '9') c = getchar();
while (c >= '0' && c <= '9')
{
ret = ret * 10 + c - '0';
c = getchar();
}
return ret;
}
void write (int x)
{
if (x < 10) {
putchar (x + '0'); return;}
write (x / 10);
putchar (x % 10 + '0');
}
bool check (int x, int S, int T)
{
for (int i = 1; i <= n; i++) vis[i] = 0, h[i] = 0;
cnt = 0;
for (int i = 1; i <= x; i++)
if (!p[i].k)
{
e[++cnt] = (edge){
p[i].v, h[p[i].u]};
h[p[i].u] = cnt;
e[++cnt] = (edge){
p[i].u, h[p[i].v]};
h[p[i].v] = cnt;
}
else
{
e[++cnt] = (edge){
p[i].v, h[p[i].u]};
h[p[i].u] = cnt;
}
while (!Q.empty()) Q.pop();
Q.push (S), vis[S] = 1;
int u, v;
while (!Q.empty())
{
u = Q.front();
Q.pop();
for (int i = h[u]; i; i = e[i].nxt)
{
v = e[i].to;
if (vis[v]) continue;
if (v == T) return 1;
vis[v] = 1;
Q.push (v);
}
}
return 0;
}
int main()
{
freopen ("road.in", "r", stdin);
freopen ("road.out", "w", stdout);
n = read(), m = read(), a = read(), b = read();
for (int i = 1; i <= m; i++)
p[i].u = read(), p[i].v = read(), p[i].k = read();
int l = 1, r = m, mid;
ans = m;
while (l <= r)
{
mid = (l + r) / 2;
if (check (mid, a, b)) ans = min (ans, mid), r = mid - 1;
else l = mid + 1;
}
write (ans), putchar (' ');
l = 1, r = m;
ans = m;
while (l <= r)
{
mid = (l + r) / 2;
if (check (mid, b, a)) ans = min (ans, mid), r = mid - 1;
else l = mid + 1;
}
write (ans);
return 0;
}