国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

POJ1251- Jungle Roads(Kruskal)

2019-11-11 02:58:34
字體:
來源:轉載
供稿:網友

Jungle Roads Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 24836 Accepted: 11691 Description

The Head Elder of the tropical island of Lagrishan has a PRoblem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems.

Input

The input consists of one to 100 data sets, followed by a final line containing only 0. Each data set starts with a line containing only a number n, which is the number of villages, 1 < n < 27, and the villages are labeled with the first n letters of the alphabet, capitalized. Each data set is completed with n-1 lines that start with village labels in alphabetical order. There is no line for the last village. Each line for a village starts with the village label followed by a number, k, of roads from this village to villages with labels later in the alphabet. If k is greater than 0, the line continues with data for each of the k roads. The data for each road is the village label for the other end of the road followed by the monthly maintenance cost in aacms for the road. Maintenance costs will be positive integers less than 100. All data fields in the row are separated by single blanks. The road network will always allow travel between all the villages. The network will never have more than 75 roads. No village will have more than 15 roads going to other villages (before or after in the alphabet). In the sample input below, the first data set goes with the map above. Output

The output is one integer per line for each data set: the minimum cost in aacms per month to maintain a road system that connect all the villages. Caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit. Sample Input

9 A 2 B 12 I 25 B 3 C 10 H 40 I 8 C 2 D 18 G 55 D 1 E 44 E 2 F 60 G 38 F 0 G 1 H 35 H 1 I 35 3 A 2 B 10 C 40 B 1 C 20 0 Sample Output

216 30

又一道樣例給錯的題…

#include<iostream>#include<string.h>#include<algorithm>using namespace std;const int maxn =30;int fa[maxn];void init(){ for(int i=0;i<maxn;i++) fa[i]=i;}int Find(int x){ if(fa[x] == x) return fa[x]; else fa[x]= Find(fa[x]);}void Union(int x,int y){ int fx=Find(x),fy=Find(y); if(fx != fy) fa[fx] = fy;}typedef struct node{ int st,ed,cost;}Edge;Edge edge[10000];int cmp(Edge a,Edge b){ return a.cost <b.cost;}int main(){ int n; while(cin>>n){ if(!n) break; char start,end; int num,cost,tot_cost=0,m=0; for(int i=0;i < n-1;i++){ cin>>start>>num; for(int j=0 ;j<num;j++){ cin>>end>>cost; edge[m].st = start-'A'; edge[m].ed = end-'A'; edge[m].cost = cost; m++; } } init(); sort(edge,edge+m,cmp); int rst=n; for(int i=0;i< m && rst > 1 ;i++){ if(Find(edge[i].st) != Find(edge[i].ed)){ Union(edge[i].st,edge[i].ed); tot_cost += edge[i].cost; rst--; } } // cout<<"rst = "<<rst<<endl; cout<<tot_cost<<endl; } return 0;}
發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 牡丹江市| 荔浦县| 平潭县| 奉贤区| 桦南县| 阿坝县| 万盛区| 乌苏市| 逊克县| 明溪县| 黄浦区| 固镇县| 凤台县| 宁晋县| 广宗县| 盐边县| 岑巩县| 楚雄市| 龙江县| 天长市| 筠连县| 大丰市| 高安市| 夏津县| 富宁县| 运城市| 双城市| 平昌县| 东明县| 光山县| 义马市| 灵山县| 吴川市| 威海市| 友谊县| 淮滨县| 隆尧县| 班戈县| 革吉县| 安国市| 都兰县|