通信线路（最短路+二分）

题意

给定一个无向图，可以选择一条线路，将其$k$条边边权变为$0$，使得剩下边的边权最大值最小。求这个最小值。

做法

一看最大值最小，首先考虑二分，二分答案。二分过程中，如果大于$mid$的边数大于$k$，那么这个最小值一定大于当前二分到的答案。

代码1（Dijkstra）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 1010, M = 20010, inf = 0x3f3f3f3f;

typedef pair<int,int> pii;

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

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

bool check(int limit)
{
    
    
    memset(dist,0x3f,sizeof(dist));
    memset(st,false,sizeof(st));
    dist[1] = 0;
    priority_queue<pii,vector<pii>,greater<pii>> heap;
    heap.push({
    
    0,1});
    while(heap.size()){
    
    
        auto t = heap.top();
        heap.pop();
        int ver = t.second, distance = t.first;
        if(st[ver]) continue;
        st[ver] = true;
        for(int i=h[ver];~i;i=ne[i]){
    
    
            int j = e[i];
            int x = w[i] > limit;
            if(dist[j]>distance+x){
    
    
                dist[j] = distance + x;
                heap.push({
    
    dist[j],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 << endl;
    else cout << r << endl;
    return 0;
}

代码2（spfa）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 1010, M = 20010, inf = 0x3f3f3f3f;

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

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

bool check(int limit)
{
    
    
    memset(dist,0x3f,sizeof(dist));
    memset(st,false,sizeof(st));
    dist[1] = 0;
    queue<int> que;
    que.push(1);
    st[1] = true;
    while(que.size()){
    
    
        int t = que.front();
        que.pop();
        st[t] = false;
        for(int i=h[t];~i;i=ne[i]){
    
    
            int j = e[i];
            int distance = w[i] > limit;
            if(dist[j]>dist[t]+distance){
    
    
                dist[j] = dist[t] + distance;
                if(!st[j]){
    
    
                    st[j] = true;
                    que.push(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 << endl;
    else cout << r << endl;
    return 0;
}