洛谷 [P1948] 电话线

二分答案

首先,最大值最小,就是二分答案

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN = 10005;
int head[MAXN], nume, n, m, k, l, r, mid, dis[MAXN];
bool f[MAXN];
struct edge {
	int to, nxt, dis;
}e[MAXN << 1];
void adde(int from, int to, int dis) {
	e[++nume].to = to;
	e[nume].nxt = head[from];
	e[nume].dis = dis;
	head[from] = nume;
}
queue <int> q;
bool chk(int x) {
	memset(dis, 0x3f, sizeof(dis));
	memset(f, 0, sizeof(f));
	dis[1] = 0;
	q.push(1);
	while(!q.empty()) {
		int u = q.front();q.pop();
		for(int i = head[u]; i; i = e[i].nxt) {
			int v = e[i].to;
			int t = (e[i].dis > x);
			if(dis[v] > dis[u] + t) {
				dis[v] = dis[u] + t;
				q.push(v);
			}
		}
	}
	//cout << dis[n] << endl;
	return (dis[n] <= k);
}
int main() {
	cin >> n >> m >> k;
	for(int i = 1; i <= m; i++ ){
		int u, v, di;
		cin >> u >> v >> di;
		r = max(r, di);
		adde(u, v, di); adde(v, u, di);
	}
	int t = r;
	while(l <= r) {
		mid = (l + r) >> 1;
		//printf("%d %d %d\n", l, r, mid);
		if(chk(mid)) {
			r = mid - 1;
		}else l = mid + 1;
	}
	if(l <= t) cout << l << endl;
	else cout << -1 << endl;
	return 0;
}

还可以用 DP 的思路, 在SPFA中转移

也可以对于每种情况,建点连边,跑最短路