poj 1251 最小生成树 prim算法
来源:互联网 发布:芜湖网络推广公司 编辑:程序博客网 时间:2024/05/29 18:35
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.
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
21630
这个题目就是一道最小生成树的模板题,第一行输入一个数字,接下来的n-1行就是有几个节点,拿第一组数据来说,A是一个节点,2是它所连接的节点数,B和I都是A连接的节点,而12和25分别是它们之间的权值,把这些节点连接起来的最小权值,就是一个最小生成树模板。
话不多说,直接上代码:
#include <iostream>#include <stdio.h>#include <string.h>const int INF = 0x3f3f3f3f;using namespace std;int map[1010][1010];int dist[1010];int vis[1010];int n,x,y,z,m;void Init(){ for(int i = 0; i <= n; i ++) for(int j = 0; j <= n; j ++) map[i][j] = map[j][i] = INF;}int prim(int v0){ int i,j,sum = 0; for(int i = 1; i <= n; i ++) { dist[i] = map[v0][i]; vis[i] = v0; } vis[v0] = -1; for(int i = 1; i < n; i ++) { int min = INF,v = -1; for(int j = 1; j <= n; j ++) { if(vis[j] != -1 && dist[j] < min) { v = j; min = dist[j]; } } if(v != -1) { vis[v] = -1; sum += dist[v]; for(int j = 1; j <= n; j ++) { if(vis[j] != -1 && map[v][j] < dist[j]) { dist[j] = map[v][j]; vis[j] = v; } } } } return sum;}int main(){ while(~scanf("%d",&n)) { if(n==0) break; char p[5]; char mp[5]; Init(); for(int i = 0; i < n-1; i ++) { scanf("%s%d",&p,&m); x=p[0]-'A'+1; for(int j=0; j<m; j++) { scanf("%s%d",&mp,&z); y=mp[0]-'A'+1; if(map[x][y] > z) map[x][y] = map[y][x] = z; } } printf("%d\n",prim(1)); } return 0;}
阅读全文
0 0
- POJ 1251 最小生成树prim算法
- poj 1251 最小生成树 prim算法
- POJ-1258 最小生成树 prim算法
- POJ 1258 -- 最小生成树(prim算法)
- poj 2485 prim算法最小生成树
- 最小生成树Prim算法 Highways POJ
- POJ 1251 Jungle Roads (最小生成树 Prim普里姆算法)
- POJ-1251 Jungle Roads 最小生成树(prim算法)
- POJ 1258-Agri Net 最小生成树Prim算法
- POJ 2485-Highways 最小生成树Prim算法
- POJ 1789 Truck History 图论 prim算法 最小生成树
- POJ 2485 Highways 图论 prim算法 最小生成树
- POJ 1258 Agri-Net 图论 prim算法 最小生成树
- poj 1789 prim 图最小生成树算法
- POJ 1258 Agri-Net(最小生成树prim算法)
- 【最小生成树之prim算法】POJ 1789---Truck History
- 【最小生成树之prim算法】POJ-1258---Agri-Net
- 最小生成树 ———prim算法 poj 1258
- Struts2学习笔记(七)——Action处理请求参数
- lua随机数生成问题
- dos遍历文件夹,把文件写入同一文件中
- eclipse安装js提示插件
- Farm Tour POJ
- poj 1251 最小生成树 prim算法
- 百度笔试题:malloc/free与new/delete的区别
- 斐波纳契数列
- Linux的常用命令
- 2017多校-3
- fastJson的使用
- B/S学习之路—DOM(1)
- UE4 C++ 碰撞检测(Overlap)
- NTT(快速数论变换)用到的各种素数及原根