HDOJ-1301-POJ-Jungle Roads

Jungle Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5535    Accepted Submission(s): 3998

Problem 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.

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.

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

9A 2 B 12 I 25B 3 C 10 H 40 I 8C 2 D 18 G 55D 1 E 44E 2 F 60 G 38F 0G 1 H 35H 1 I 353A 2 B 10 C 40B 1 C 200

 

Sample Output

21630

 

 

 

最短路问题
题意:输入N代表村子个数,A-Z表示,以后n行:村名(A-Z)--与m个村子相连--相连村子1--路长--相连的村子2--路长......相连的村子m--路长;

保留权值最小的树.

 

思路:可用kruscal及prim均可:

还有个待解决的问题,此题在杭电提交AC,但在poj上以第一种输入通不过(注释部分)

普利姆:

#include<cstdio>#include<cstring>#define INF 0x3f3f3f3fint N,g[30][30];//距离邻接矩阵void prim(){int dis[30],vis[30],k,min;//dis中保存的是距离-整棵树-最短的距离memset(vis,0,sizeof(vis));//初始化岛屿全为未在点集合内for(int i=1;i<=N;i++){dis[i]=g[1][i];/记录i点到集合1的距离}dis[1]=0;//本身到本身的距离为0vis[1]=1;//标记1已在集合内for(int v=1;v<N;v++){//还要纳入n-1个点k=1;min=INF;for(int i=1;i<=N;i++){if(!vis[i]&&dis[i]<min){//如果i点未在集合内  并且小于最小值k=i;//记录最小值min=dis[i];//记录编号}}vis[k]=1;//标记为已在集合内for(int i=1;i<=N;i++){//更新其余点到集合1的距离if(!vis[i] && dis[i]>g[k][i])//如果i未在集合1内并且 当前保存的i到集合的距离大于新更新的值dis[i]=g[k][i];//更新最小值}}int sum=0;for(int i=2;i<=N;i++){sum+=dis[i];}printf("%d\n",sum);return;}int main(){while(scanf("%d",&N),N){char ch1,ch2;int in1,in2;memset(g,0x3f,sizeof(g));//getchar();for(int n=0;n<N-1;n++){//输入1：//HDOJ 1,2均通过 //scanf("%c",&ch1);//getchar();//scanf("%d",&in1);//for(int i=0;i<in1;i++){//getchar();//scanf("%c",&ch2);//getchar();//scanf("%d",&in2);// 输入2： //POJ 仅通过2. scanf("\n%c %d",&ch1,&in1);for(int i=0;i<in1;i++){scanf(" %c %d",&ch2,&in2);g[ch1-'A'+1][ch2-'A'+1]=g[ch2-'A'+1][ch1-'A'+1]=in2;//普利姆此处要对称输入!!!!!!!!!!!!!! }//getchar();}prim();}return 0;}

克鲁斯卡尔:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int per[110000],N,M;struct str{int a,b,x;//起点,终点,权值}edge[100000];int cmp(str a,str b){return a.x<b.x;}void init(){//初始化所有节点for(int i=0;i<=110000;i++)per[i]=i;}int find(int x){查询根节点if(x==per[x])return x;return per[x]=find(per[x]);}bool join(int x,int y){判断是否满足添加条件int i=find(x);int j=find(y);if(i==j)return false;per[i]=j;return true;}int main(){while(scanf("%d",&N),N){char ch1,ch2;int in1,in2;int k=1;getchar();for(int n=0;n<N-1;n++){//输入1：//HDOJ 1,2均通过 //scanf("%c",&ch1);//getchar();//scanf("%d",&in1);//for(int i=0;i<in1;i++){//getchar();//scanf("%c",&ch2);//getchar();//scanf("%d",&in2);// 输入2： //POJ 仅通过2. scanf("\n%c %d",&ch1,&in1);for(int i=0;i<in1;i++){scanf(" %c %d",&ch2,&in2);//edge[k].a=ch1-'A';edge[k].b=ch2-'A';edge[k].x=in2;k++;}getchar();}//printf("--%d %d %d %d\n",edge[k].a,edge[k].b,edge[k].x,k);M=k;sort(edge+1,edge+M,cmp);//按权值排序//for(int m=1;m<M;m++){//printf("--%d %d %d\n",edge[m].a,edge[m].b,edge[m].x);//}init();int sum=0;for(int m=1;m<M;m++){if(join(edge[m].a,edge[m].b))//满足条件添加并加上其权值sum+=edge[m].x;}printf("%d\n",sum);}}