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.

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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

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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.

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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

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Sample Output

4

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Source

USACO 2008 January Silver 二分枚舉每條邊的長度l，然后以這條邊的長度l為界限重新建立地圖，大于l的賦值為1，小于的賦值為0，然后跑一個dijkstra求出大于l的有多少個，然后繼續二分枚舉。 注意一開始邊要排序！！！ 注意一開始要先建立個以0為界限的圖，判斷是否可以到達，或者0的特判！！！ 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<cmath> 7 using namespace std; 8 #define inf 1<<30 9 #define N 10006 10 #define M 1006 11 int n,p,k; 12 struct Node{ 13 int u,v,l; 14 }node[N]; 15 int mp[M][M]; 16 void build_map(int length){//建圖 17 for(int i=0;i<=n;i++){ 18 for(int j=0;j<=n;j++){ 19 mp[i][j]=inf; 20 } 21 } 22 for(int i=0;i<p;i++){ 23 int a=node[i].u; 24 int b=node[i].v; 25 int c=node[i].l; 26 if(c>length){ 27 mp[a][b]=mp[b][a]=1; 28 } 29 else{ 30 mp[a][b]=mp[b][a]=0; 31 } 32 } 33 34 } 35 36 int dijkstra(int st){//dijkstra求從1到n的距離 37 int vis[M]; 38 int dis[M]; 39 for(int i=0;i<=n;i++){ 40 vis[i]=0; 41 dis[i]=inf; 42 } 43 vis[st]=1; 44 dis[st]=0; 45 int x=n; 46 while(x--){ 47 for(int i=1;i<=n;i++){ 48 if(dis[st]+mp[st][i]<dis[i]){ 49 dis[i]=dis[st]+mp[st][i]; 50 } 51 } 52 int minn=inf; 53 for(int i=1;i<=n;i++){ 54 if(!vis[i] && dis[i]<minn){ 55 minn=dis[i]; 56 st=i; 57 } 58 } 59 vis[st]=1; 60 } 61 return dis[n]; 62 } 63 bool solve(int mid){//二分搜索的判斷函數 64 build_map(node[mid].l); 65 int ans=dijkstra(1); 66 if(ans<=k) return true; 67 return false; 68 } 69 void go(){//二分搜索 70 int low=0; 71 int high=p; 72 while(low<high){ 73 int mid=(low+high)>>1; 74 if(solve(mid)){ 75 high=mid; 76 } 77 else{ 78 low=mid+1; 79 } 80 } 81 printf("%d\n",node[low].l); 82 } 83 bool cmp(Node a,Node b){//排序函數！！！ 84 return a.l<b.l; 85 } 86 int main() 87 { 88 while(scanf("%d%d%d",&n,&p,&k)==3){ 89 for(int i=0;i<p;i++){ 90 scanf("%d%d%d",&node[i].u,&node[i].v,&node[i].l); 91 } 92 sort(node,node+p,cmp);//二分搜索一定要先排序！！！？？？ 93 build_map(0);//先以0為界限建立圖 94 int ans=dijkstra(1); 95 //printf("%d\n",ans); 96 if(ans==inf){//如果不能到達，輸出-1 97 printf("-1\n"); 98 } 99 else{ 100 if(ans<=k){//如果一開始就不用付出，則輸出0 101 printf("0\n"); 102 } 103 else{ 104 go();//二分搜索 105 } 106 } 107 } 108 return 0; 109 } View Code

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