poj 1251 Jungle Roads （最小生成树）
Link： http://poj.org/problem?id=1251
Jungle Roads
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Time Limit: 1000MS |
Memory Limit: 10000K |
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Total Submissions: 23507 |
Accepted: 11012 |
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
Source
题意：
给定多个点，和点与点之间的距离， 求最小让其连接起来的线路的总和。
经典的最小生成树问题。 使用 prime算法。
prime算法：
隔离出已经访问的点和未访问的点，寻找到访问过的点的树中到未访问点的最小距离，收之，
生成一个树， 继续。
// 1251
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
const int maxn = 30; int n, mp[maxn][maxn]; int Prime(int cur){
int i,j, minlen, sum, pt = cur, dist[maxn];
bool visited[maxn];
memset(visited, false, sizeof(visited));
visited[cur] = true;
sum = 0;
for(i=0; i<n; i++){
dist[i] = mp[cur][i];
}
for(i=1; i<n; i++){
minlen = 0x3f3f3f3f;
for(j=0; j<n; j++){
if(!visited[j] && minlen > dist[j]){
minlen = dist[j];
pt = j;
}
}
visited[pt] = true;
sum += minlen;
for(j=0; j<n; j++){
if(!visited[j] && dist[j] > mp[pt][j]){
dist[j] = mp[pt][j];
}
}
}
return sum;
} int main(){
freopen("in.txt", "r", stdin); int i,j, num2, num1, ans;
char ch1, ch2;
while(cin>>n && n){
memset(mp, 0x3f3f3f3f, sizeof(mp));
for(i=1; i<n; i++){
cin>>ch1>>num1;
for(j=1; j<=num1; j++){
cin>>ch2>>num2;
mp[ch1-'A'][ch2-'A'] = mp[ch2-'A'][ch1-'A'] = num2;
}
}
ans = Prime(ch1-'A');
cout<<ans<<endl;
}
return 0;
}