POJ - 1251- Jungle Roads
题目链接：https://cn.vjudge.net/contest/238509#problem/J
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
这个题目是个很明显的求最小生成树的题目，这种题目一般就是克鲁斯科尔算法，把所有边都记录下来，然后排序，从小到大开始取边并且不能构成环，因为是树，所以取到n-1条边时就已经取完了。
下面这个代码我第一次交的时候AC了，但是后来又做到交的时候，一样的代码却超时了，我也不知道怎么回事
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 30;
struct edge
{
int u, v;
int w;
bool operator < (const edge& obj) const
{
return w < obj.w;
}
}e[1000];
int f[maxn];
int Find(int x)
{
return x == f[x] ? x : f[x] = Find(f[x]);
}
int join(int a, int b)
{
int x = Find(a), y = Find(b);
if(x == y) return 0;
f[x] = y;
return 1;
}
int main()
{
int n;
while(~scanf("%d", &n), n)
{
memset(e, 0, sizeof(e));
for(int i = 0; i <= n; i++) f[i] = i;
int cnt = 0, k, a;
for(int i = 1; i < n; i++)
{
getchar();
a = getchar()-'A';
scanf("%d", &k);
for(int j = 0; j < k; j++)
{
getchar();
e[cnt].v = getchar()-'A';
scanf("%d", &e[cnt].w);
e[cnt++].u = a;
}
}
sort(e, e+cnt);
int ans = 0, sum = 0;
for(int i = 0; i < cnt && sum < n-1; i++)
{
int u = e[i].u, v = e[i].v, w = e[i].w;
if(join(u, v))
{
ans += w;
sum++;
}
}
printf("%d\n", ans);
}
return 0;
}
上面那个我不确定他到底是不是真的能过
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 27;
struct Edge
{
int v, u;
int cost;
}e[N*(N - 1) / 2];
int set[N];
int Find(int x)
{
if(x == set[x]) return x;
return Find(set[x]);
}
bool operator < (const Edge& a, const Edge& b)
{
return a.cost < b.cost;
}
int main()
{
int n, k, w, minCost;
char c1, c2;
while (cin >> n && n)
{
minCost = 0;
int j = 0; //j为边数
for (int i = 1; i < N; i++) set[i] = i;
for (int i = 1; i < n; i++)
{
cin >> c1 >> k;
for (; k--; j++)
{
cin >> c2 >> w;
e[j].v = c1 - 'A';
e[j].u = c2 - 'A';
e[j].cost = w;
}
}
sort(e, e + j);
//for(int i = 0; i < j; ++i)
//cout << e[i].v << " " << e[i].u << " " << e[i].cost << endl;
for (int i = 0; i < j; ++i)
{
int set1 = Find(e[i].v);
int set2 = Find(e[i].u);
if (set1 != set2)
{
minCost += e[i].cost;
if(set1 < set2)
set[set2] = set1;
else if(set1 > set2)
set[set1] = set2;
}
}
cout << minCost << endl;
}
}这个代码是一位大佬写的，是我很久以前看到的了，但我忘记是谁的了，还是贴一下吧，总得给一个正解吧，不过声明：这确实是我搜来的代码，原创大佬请见谅。