POJ--1251--Jungle Roads



Jungle RoadsCrawling in process...Crawling failedTime Limit:1000MS    Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

SubmitStatus Practice POJ 1251

 

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

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

题意：求丛林中所有点都连通时所需的最小花费，即构造最小生成树。

思路：用 Kruskal 或 prim 算法构造最小生成树（将字符处理成字符串输入）

代码如下：

Kruskal:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>int flag[100], n;struct node{int s, d, c;}road[100];bool cmp(const node& a, const node& b){return a.c<b.c;}void init(){for(int i=0; i<n; i++)//下标从0 开始flag[i]=i;        //初始化根节点为其本身下标}int find(int x)//找该点所属的集合的根节点{return x==flag[x] ? x : flag[x]=find(flag[x]);}using namespace std;int main(){#ifdef OFFLINEfreopen("t.txt","r",stdin);#endifint i, j, k, p, w;char s[3], d[3];while(scanf("%d", &n)){if(n==0) break;j=n-1;     p=0;while(j--){ scanf("%s %d", s, &k);while(k--){scanf("%s %d", d, &w);road[p].s=s[0]-'A';road[p].d=d[0]-'A';road[p++].c=w; }}sort(road, road+p, cmp);init();int num=0, cost=0;for(i=0;i<p;i++){int L=find(road[i].s), R=find(road[i].d);if(L != R){num++;flag[L]=R;cost+=road[i].c;}if(num==n-1)  break;}printf("%d\n", cost);}return 0;}

prim:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define INF 0x3f3f3f3fint nearvex[300], lowest[300], road[300][300], n;int prim(int u){int i, j, k, min, cost=0;for(i=0;i<n;i++){//顶点下标从0开始lowest[i]=road[u][i];nearvex[i]=u;//距离i的前一节点}nearvex[u]=-1;//-1表示已经该点已加入的最小生成树for(i=1;i<n;i++){//再加 (n-1) 个点(每次在未加入最小生成树的点集中找距离树点集的最小值)min=INF, k=-1;for(j=0;j<n;j++){if(nearvex[j] != -1&&lowest[j]<min){min=lowest[j];  k=j;}}if(k != -1){nearvex[k]=-1;cost+=lowest[k];for(j=0;j<n;j++){if(nearvex[j] != -1&&road[k][j]<lowest[j])lowest[j]=road[k][j];//从k出发更新最小值}}}return cost;}using namespace std;int main(){#ifdef OFFLINEfreopen("t.txt","r",stdin);#endifint k, t, w; char u[3], v[3];while(scanf("%d", &n)){if(n==0)  break;memset(road, INF, sizeof(road));k=n-1;while(k--){scanf("%s %d", u, &t);//将字符当成字符串读入while(t--){scanf("%s %d", v, &w);road[u[0]-'A'][v[0]-'A']=w;//无向路(2条)road[v[0]-'A'][u[0]-'A']=w;}}  int ans=prim(0);printf("%d\n", ans);}return 0;}