「USACO08JAN」电话线Telephone Lines

传送门
Luogu

解题思路

考虑二分，每次把大于二分值的边的权设为1，小于等于的设为0，如果最短路<=k则可行，记得判无解

细节注意事项

• 咕咕咕

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while (!isdigit(c)) f |= (c == '-'), c = getchar();
    while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
    s = f ? -s : s;
}

const int _ = 1002;
const int __ = 10002;

int tot, head[_], nxt[__ << 1], ver[__ << 1], w[__ << 1];
inline void Add_edge(int u, int v, int d)
{ nxt[++tot] = head[u], head[u] = tot, ver[tot] = v, w[tot] = d; }

int n, m, k, exi[_], dis[_];

inline bool check(int mid) {
    static queue < int > Q;
    memset(exi + 1, 0, sizeof (int) * n);
    memset(dis + 1, 0x3f, sizeof (int) * n);
    dis[1] = 0, exi[1] = 1, Q.push(1);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop(), exi[u] = 0;
        for (rg int i = head[u]; i; i = nxt[i]) {
            int v = ver[i], d = (w[i] > mid);
            if (dis[v] > dis[u] + d) {
                dis[v] = dis[u] + d;
                if (!exi[v]) Q.push(v);
            }
        }
    }
    return dis[n] <= k;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.in", "r", stdin);
#endif
    read(n), read(m), read(k);
    for (rg int u, v, d; m--; )
        read(u), read(v), read(d), Add_edge(u, v, d), Add_edge(v, u, d);
    int l = 0, r = 1e6 + 1;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (check(mid)) r = mid;
        else l = mid + 1;
    }
    if (r == (int) 1e6 + 1) puts("-1");
    else printf("%d\n", r);
    return 0;
}

完结撒花 \(qwq\)