通信线路【二分】【spfa】

>Link

ybtoj通信线路

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>解题思路

求最大值最小，可以考虑二分答案
二分以后如何处理呢？我们将二分出来的mid作为标准值，边权大于mid的为1，小于等于的为0，这样跑一个spfa，如果1到n的最短路超过了k，说明路径上大于mid的数多于k，当前mid作为答案是不行的

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>代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define N 1010
#define M 10010
using namespace std;

struct edge
{
	int to, next, w;
} e[2 * M];
int n, m, k, h[N], cnt, l, r, mid, c[N], ans;
bool y[N];

void add (int u, int v, int w)
{
	e[++cnt] = (edge){v, h[u], w}; h[u] = cnt;
	e[++cnt] = (edge){u, h[v], w}; h[v] = cnt;
}
bool spfa (int x)
{
	memset (c, 0x7f, sizeof (c));
	queue<int> q;
	q.push(1); y[1] = 1;
	c[1] = 0;
	while (!q.empty())
	{
		int u = q.front();
		q.pop();
		y[u] = 0;
		for (int i = h[u]; i; i = e[i].next)
		{
			int v = e[i].to;
			if (c[v] > c[u] + (e[i].w > x))
			{
				c[v] = c[u] + (e[i].w > x);
				if (!y[v])
				{
					q.push(v);
					y[v] = 1;
				}
			}
		}
	}
	if (c[n] == c[0])
	{
		printf ("-1");
		exit (0); //注意要判断是否有通路的情况
	}
	return c[n] <= k;
}

int main()
{
	freopen ("phone.in", "r", stdin);
	freopen ("phone.out", "w", stdout);
	
	int u, v, w;
	scanf ("%d%d%d", &n, &m, &k);
	for (int i = 1; i <= m; i++)
	{
		scanf ("%d%d%d", &u, &v, &w);
		add (u, v, w);
	}
	l = 0, r = 1000000;
	while (l < r)
	{
		mid = (l + r) / 2;
		if (spfa (mid)) r = mid;
		else l = mid + 1;
	}
	printf ("%d", r);
	return 0;
}