通信线路

题目描述：

n个点，m条双向边，求1到n的路程中价格第k+1大的边的权值最小是多少，如果路径数量小于k+1，则输出0

思路1:分层图最短路

求第k+1大的边，则可以转换成让k条边的权值是0，再求权值最大值的最小值，考虑分层图最短路，同一层内普通建图，但是层与层之间对应的点建权值为0的边，此外还需要给n * i到n * (i + 1) 之间建权值为0的边，来解决路径数量不足k+1的情况，然后跑最短路，输出dis[n * (k + 1)]的值即可

#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 1000000007
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

typedef long long ll;
typedef pair <int,int> pii;

#define MAX 2000000 + 50
int n, m, k;

int tot;
int head[MAX];
struct ran{
    int to, nex, val;
}tr[MAX];
inline void add(int u, int v, int c){
    tr[++tot].to = v;
    tr[tot].val = c;
    tr[tot].nex = head[u];
    head[u] = tot;
}

int dis[MAX];
bool vis[MAX];
void dijkstra(int s, int t){
    priority_queue<pii, vector<pii>, greater<pii>>q;
    mem(dis, inf);
    q.push(m_p(0, s));dis[s] = 0;
    while (!q.empty()) {
        auto [d, u] = q.top();q.pop();
        if(vis[u])continue;
        vis[u] = 1;
        for(int i = head[u]; i; i = tr[i].nex){
            int v = tr[i].to;
            if(dis[v] > max(tr[i].val, dis[u])){
                dis[v] = max(tr[i].val, dis[u]);
                q.push(m_p(dis[v], v));
            }
        }
    }
    if(dis[t] == inf)cout << -1 << endl;
    else cout << dis[t] << endl;
}


void work(){
    cin >> n >> m >> k;
    for(int i = 1, a, b, c; i <= m; ++i){
        cin >> a >> b >> c;
        add(a, b, c);add(b, a, c);
        for(int j = 1; j <= k; ++j){
            add(a + (j - 1) * n, b + j * n, 0);
            add(b + (j - 1) * n, a + j * n, 0);
            add(a + j * n, b + j * n, c);
            add(b + j * n, a + j * n, c);
        }
    }
    for(int i = 1; i <= k; ++i)add(i * n, (i + 1) * n, 0);
    dijkstra(1, n * k + n);
}


int main(){
    io;
    work();
    return 0;
}

思路2:二分答案 + 最短路

二分权值p，巧妙的建图：权值大于p的边的权值为1，权值小于等于p的权值为0，这样跑最短路得到的就是最少需要几条大于p的边，拿它和k进行比较来作为二分的check函数，非常妙

#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 1000000007
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

typedef long long ll;
typedef pair <int,int> pii;

#define MAX 300000 + 50
int n, m, k;
int a, b, c;

int tot;
int head[MAX];
struct ran{
    int to, nex, val;
}tr[MAX];
inline void add(int u, int v, int c){
    tr[++tot].to = v;
    tr[tot].val = c;
    tr[tot].nex = head[u];
    head[u] = tot;
}

bool vis[1005];
int dis[1005];
int dijkstra(int s, int mid){
    priority_queue<pii, vector<pii>, greater<pii> >q;
    mem(dis, inf);mem(vis, 0);
    q.push(m_p(0, s));dis[s] = 0;
    while (!q.empty()) {
        auto [d, u] = q.top();q.pop();
        if(vis[u])continue;
        vis[u] = 1;
        for(int i = head[u]; i; i = tr[i].nex){
            int v = tr[i].to;
            int p = 0;
            if(tr[i].val > mid)p = 1;
            if(dis[v] > dis[u] + p){
                dis[v] = dis[u] + p;
                q.push(m_p(dis[v], v));
            }
        }
    }
    return dis[n];
}


bool judge(int mid){
    if(dijkstra(1, mid) <= k)return true;
    return false;
}

void work(){
    cin >> n >> m >> k;
    for(int i = 1; i <= m; ++i){
        cin >> a >> b >> c;
        add(a, b, c);
        add(b, a, c);
    }
    
    int l = 0, r = 1000001;
    while (l <= r) {
        int mid = (l + r) / 2;
        if(judge(mid))r = mid - 1;
        else l = mid + 1;
    }
    if(dis[n] == inf)cout << -1 << endl;
    else cout << l << endl;
}


int main(){
    io;
    work();
    return 0;
}