POJ-3662 Telephone Lines---二分+最短路+最小化第k+1大

题目链接：

https://cn.vjudge.net/problem/POJ-3662

题目大意：

求一条路径从1到n使第k+1大的边最小。

解题思路：

二分答案mid，当原边权小于等于mid新边权为0，否则新边权为1.

求最短路，若小于等于k说明满足条件

 

注意：最开始的l必须是0，而不是这些边中的最小边，因为最小化第k+1大的话答案可能为0

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define MID(l, r) (l + (r - l) / 2)
#define lson(o) (o * 2)
#define rson(o) (o * 2 + 1)
using namespace std;
typedef long long ll;
const int INF = 1e9 +7;
const int maxn = 1e6 + 10;
int n, m, k;
struct edge{
    int u, v, w;
    edge(int u, int v, int w):u(u), v(v), w(w){}
    edge(){}
};
struct Heapnode
{
    int d, u;
    Heapnode(){}
    Heapnode(int d, int u):d(d), u(u){}
    bool operator <(const Heapnode & a)const
    {
        return d > a.d;
    }
};
bool v[maxn];
int d[maxn];

int dijkstra(int s, int t, vector<edge>edges, vector<int>G[])
{
    priority_queue<Heapnode>q;
    for(int i = 0; i <= n; i++)d[i] = INF;
    d[s] = 0;
    memset(v, 0, sizeof(v));
    q.push(Heapnode(0, s));
    while(!q.empty())
    {
        Heapnode now = q.top();
        q.pop();
        int u = now.u;
        if(v[u])continue;
        v[u] = 1;
        for(int i = 0; i < G[u].size(); i++)
        {
            edge& e = edges[G[u][i]];
            int v = e.v;
            if(d[v] > d[u] + e.w)
            {
                d[v] = d[u] + e.w;
                q.push(Heapnode(d[v], v));
            }
        }
    }
    return d[t];
}

vector<edge>edges;
vector<int>G[maxn];
vector<edge>edges2;
void addedge(int u, int v, int w)
{
    edges.push_back(edge(u, v, w));
    int m = edges.size();
    G[u].push_back(m - 1);
}
void init(int mid)
{
    edges2.clear();
    for(int i = 0; i < edges.size(); i++)
    {
        edge& e = edges[i];
        if(e.w <= mid)
        {
            edges2.push_back(edge(e.u, e.v, 0));
        }
        else
        {
            edges2.push_back(edge(e.u, e.v, 1));
        }
    }
}
int main()
{
    scanf("%d%d%d", &n, &m, &k);
    int l = 0, r = 0, u, v, w;
    while(m--)
    {
        scanf("%d%d%d", &u, &v, &w);
        addedge(u, v, w);
        addedge(v, u, w);
        r = max(r, w);
    }
    int ans = -1;
    while(l <= r)//l最开始要设置成0，不能设置成最小边，因为第k大可能为0
    {
        int mid = (l + r) / 2;
        init(mid);
        int t = dijkstra(1, n, edges2, G);
        //cout<<mid<<" "<<t<<endl;
        if(t <= k)
        {
            ans = mid;
            r = mid - 1;
        }
        else
        {
            l = mid + 1;
        }
    }
    cout<<ans<<endl;
    return 0;
}