题目描述

blablabla

样例

blablabla

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<cstring>
#include<queue>

using namespace std;


const int N = 1010, M = 20010;
int h[N], e[M], ne[M], w[M], idx;
int dist[N];
bool st[N];
int n, m, k;


void add(int a, int b, int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++;
}

bool check(int bound)
{
    memset(dist, 0x3f, sizeof dist);
    memset(st, false, sizeof st);

    deque<int> dq;
    dq.push_back(1);
    dist[1] = 0;

    while(dq.size())
    {
        auto t = dq.front();
        dq.pop_front();

        if (t == n) break;
        if (st[t]) continue;
        st[t] = true;

        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i], c = w[i] > bound;

            if (dist[j] > dist[t] + c)
            {
                dist[j] = dist[t] + c;
                if (c) dq.push_back(j);
                else dq.push_front(j);
            }
        }
    }
    return dist[n] <= k;
}

int main()
{
    cin >> n >> m >> k;

    memset(h, -1, sizeof h);
    while(m --)
    {
        int a, b, c;
        cin >> a >> b >>c;
        add(a, b, c);
        add(b, a, c);
    }

    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 == 1e6 + 1) cout << -1;
    else cout << r;
    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla