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 02:47:51
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: 全椒县| 会理县| 突泉县| 墨脱县| 莎车县| 腾冲县| 贞丰县| 安康市| 汶上县| 旬邑县| 屏东县| 灌阳县| 扬州市| 昭通市| 澄城县| 文化| 兰考县| 肥城市| 常熟市| 平阳县| 隆林| 耒阳市| 嘉鱼县| 永丰县| 伊金霍洛旗| 青州市| 政和县| 巴林右旗| 长武县| 常山县| 屯昌县| 翼城县| 全州县| 临洮县| 城口县| 乌兰浩特市| 和林格尔县| 大城县| 平利县| 平阳县| 安吉县|