POJ 3662 Telephone Lines （dijstra+二分）

题意：有N个独立点，其中有P对可用电缆相连的点，要使点1与点N连通，在K条电缆免费的情况下，问剩下的电缆中，长度最大的电缆可能的最小值为多少。

分析：

1、二分临界线（符合的情况的点在右边），找可能的最小值，假设为mid。

2、将大于mid的边变为1，小于等于mid的边变为0（表示这些边由自己承包），由此算出1~N的最短路长度为x。x即为所用的大于mid的电缆个数。

3、若x<=K,则符合情况，但是想让所用的免费电缆条数x更多，所以让mid更小一些，这样自己承包的边也减少，x将更大，即r = mid;

4、若x>K，则所用免费电缆条数x超过了K条，不再符合题意，自己承包的边过少了，所以l = mid + 1;

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int N, P, K;
struct Edge{
    int from, to, dist;
    Edge(int f, int t, int d):from(f), to(t), dist(d){}
};
struct HeapNode{
    int d, u;
    HeapNode(int dd, int uu):d(dd), u(uu){}
    bool operator < (const HeapNode& rhs)const{
        return d > rhs.d;
    }
};
struct Dijkstra{
    int n, m;
    vector<Edge> edges;
    vector<int> G[MAXN];
    bool done[MAXN];
    int d[MAXN];
    int p[MAXN];
    void init(int n){
        this -> n = n;
        for(int i = 0; i < n; ++i) G[i].clear();
        edges.clear();
    }
    void AddEdge(int from, int to, int dist){
        edges.push_back(Edge(from, to, dist));
        m = edges.size();
        G[from].push_back(m - 1);
    }
    void dijkstra(int s, int t){
        priority_queue<HeapNode> Q;
        for(int i = 0; i < N; ++i) d[i] = INT_INF;
        d[s] = 0;
        memset(done, 0, sizeof done);
        Q.push(HeapNode(0, s));
        while(!Q.empty()){
            HeapNode x = Q.top();
            Q.pop();
            int u = x.u;
            if(done[u]) continue;
            done[u] = true;
            for(int i = 0; i < G[u].size(); ++i){
                Edge& e = edges[G[u][i]];
                int tmp = e.dist > t ? 1 : 0;
                if(d[e.to] > d[u] + tmp){
                    d[e.to] = d[u] + tmp;
                    p[e.to] = G[u][i];
                    Q.push(HeapNode(d[e.to], e.to));
                }
            }
        }
    }
}dij;
bool judge(int x){
    dij.dijkstra(0, x);
    if(dij.d[N - 1] <= K) return true;
    return false;
}
int solve(){
    int l = 0, r = 1e6;
    while(l < r){
        int mid = l + (r - l) / 2;
        if(judge(mid)) r = mid;
        else l = mid + 1;
    }
    if(judge(r)) return r;
    return -1;
}
int main(){
    dij.init(N);
    scanf("%d%d%d", &N, &P, &K);
    for(int i = 0; i < P; ++i){
        int a, b, l;
        scanf("%d%d%d", &a, &b, &l);
        dij.AddEdge(a - 1, b - 1, l);
        dij.AddEdge(b - 1, a - 1, l);
    }
    printf("%d\n", solve());
    return 0;
}