http://poj.org/problem?id=3662
Telephone Lines
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions:9310 | Accepted: 3374 |
Description
Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of the cables required to connect his farm to the phone system.
There are N (1 ≤ N ≤ 1,000) forlorn telephone poles conveniently numbered 1..N that are scattered around Farmer John‘s property; no cables connect any them. A total of P (1 ≤ P ≤ 10,000) pairs of poles can be connected by a cable; the rest are too far apart.
The i-th cable can connect the two distinct poles Ai and Bi, with length Li (1 ≤ Li ≤ 1,000,000) units if used. The input data set never names any {Ai, Bi} pair more than once. Pole 1 is already connected to the phone system, and pole N is at the farm. Poles 1 and N need to be connected by a path of cables; the rest of the poles might be used or might not be used.
As it turns out, the phone company is willing to provide Farmer John with K (0 ≤ K < N) lengths of cable for free. Beyond that he will have to pay a price equal to the length of the longest remaining cable he requires (each pair of poles is connected with a separate cable), or 0 if he does not need any additional cables.
Determine the minimum amount that Farmer John must pay.
Input
* Line 1: Three space-separated integers: N, P, and K
* Lines 2..P+1: Line i+1 contains the three space-separated integers: Ai, Bi, and Li
Output
* Line 1: A single integer, the minimum amount Farmer John can pay. If it is impossible to connect the farm to the phone company, print -1.
Sample Input
5 7 1 1 2 5 3 1 4 2 4 8 3 2 3 5 2 9 3 4 7 4 5 6
Sample Output
4
Source
题意：
【感觉题意描述很不清啊，一会长度一会段的】
有$n$个点，$p$条边，每条边有一个权值（花费）。将第$1$个点和第$n$个点连通，并且可以有$k$条边是免费的。剩下的不免费的边的最大值作为最终的花费。求最终花费的最小值。
思路：
刚开始题意理解错了。以为是有$k$长度的是免费的，边的那个权值是长度。想了半天搞不懂。
后来发现其实就是求一个使路径上第$k+1$大的边权尽量小的路径。【虽然还是不会】
因为当我们支付的钱变多时，合法的路径一定包含了花费更少的路径。答案具有单调性。
所以我们可以二分答案。【注意想题目答案的单调性尝试二分】
这时候问题就变成了，把价格超过$mid$的边花费看成是$1$，不超过的边花费看成是$0$，然后求$1~N$的最短路是否不超过$k$就可以了。
要注意考虑$1$和$N$不连通的情况，输出是$-1$
虐狗宝典上还有一个思路是用dp，但是我不是很会写。
用$D[x,p]$表示从$1$号节点到基站$x$，途中已经指定了$p$条电缆免费时，经过的路径上最贵的电缆的花费最小是多少（选择一条$1$到$x$的路径，使路径上第$p+1$大的边权尽量小）。若有一条从$x$到$y$长度是$z$的无向边，用$max(D[x,p],z)$更新$D[y,p]$的最小值，用$D[x,p]$更新$D[y, p+1]$的最小值。前者表示不在电缆$(x,y,z)$上使用免费升级服务，后者表示使用。用迭代的思想，借助SPFA进行动态规划，直至所有状态收敛。
还可以把图中节点拓展到二维，用二元组$(x,p)$表示一个节点，从$(x,p)$到$(y,p)$有长度为$z$的边，从$(x,p)$到$(y,p+1)$有长度为0的边。问题就变成了$N*K$个点，$P*K$条边的广义单源最短路问题。
1 #include<iostream>
2 //#include<bits/stdc++.h>
3 #include<cstdio>
4 #include<cmath>
5 #include<cstdlib>
6 #include<cstring>
7 #include<algorithm>
8 #include<queue>
9 #include<vector>
10 #include<set>
11 #include<climits>
12 using namespace std;
13 typedef long long LL;
14 #define N 100010
15 #define pi 3.1415926535
16
17 const int maxn = 1005;
18 const int maxp = 10005;
19
20 int n, p, k;
21 struct node{
22 int v, w, nxt;
23 }e[maxp * 2];
24 int tot = 0, head[maxn];
25 LL dis[maxn];
26 bool vis[maxn];
27
28 void addedge(int u, int v, int w)
29 {
30 e[tot].v = v;
31 e[tot].w = w;
32 e[tot].nxt = head[u];
33 head[u] = tot++;
34 e[tot].v = u;
35 e[tot].w = w;
36 e[tot].nxt = head[v];
37 head[v] = tot++;
38 }
39
40 int dijkstra(int mid)
41 {
42 memset(dis, 0x3f, sizeof(dis));
43 memset(vis, 0, sizeof(vis));
44 dis[1] = 0;
45 priority_queue<pair<LL, int> >que;
46 que.push(make_pair(0, 1));
47 while(que.size()){
48 int x = que.top().second;que.pop();
49 if(vis[x])continue;
50 vis[x] = true;
51 for(int i = head[x]; i != -1; i = e[i].nxt){
52 int y = e[i].v, z = e[i].w;
53 if(z > mid)z = 1;
54 else z = 0;
55 if(dis[y] > dis[x] + z){
56 dis[y] = dis[x] + z;
57 que.push(make_pair(-dis[y], y));
58 }
59 }
60 }
61 return dis[n];
62 }
63
64 int main()
65 {
66 while(scanf("%d%d%d", &n, &p, &k) != EOF){
67 for(int i = 0; i < n; i++){
68 head[i] = -1;
69 }
70 tot = 0;
71
72 int ed = -1;
73 for(int i = 0; i < p; i++){
74 int u, v, w;
75 scanf("%d%d%d", &u, &v, &w);
76 ed = max(ed, w);
77 addedge(u, v, w);
78 }
79
80 //printf("%d\n", dijkstra(0));
81 if(dijkstra(0) == 1061109567)printf("-1\n");
82 else{
83 int st = 0, ans;
84 while(st <= ed){
85 int mid = (st + ed) / 2;
86 if(dijkstra(mid) <= k){
87 ans = mid;
88 ed = mid - 1;
89 }
90 else{
91 st = mid + 1;
92 }
93 }
94 printf("%d\n", ans);
95 }
96
97
98 }
99 return 0;
100 }
原文地址：https://www.cnblogs.com/wyboooo/p/9966149.html