# POJ 1251 Jungle Roads C++版 Kruskal

POJ题解 2483浏览

Kruskal算法的应用

Description

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

Slyar:题意很简单，就是求最小生成树的路径和，这次依旧使用Kruskal算法解决。多组数据，以0结束，每组数据给出节点个数m，以下m-1行每行开头为起始节点和与它相连的节点数目n，随后n组数据表示和它相连的节点以及边权值。输出最小生成树的路径和。

```#include
#include

using namespace std;

#define MAX 26

/* 定义边(x,y)，权为w */
typedef struct
{
int x, y;
int w;
}edge;

edge e[MAX * MAX];
/* rank[x]表示x的秩 */
int rank[MAX];
/* father[x]表示x的父节点 */
int father[MAX];
int sum;

/* 比较函数，按权值非降序排序 */
bool cmp(const edge a, const edge b)
{
return a.w < b.w;
}

/* 初始化集合 */
void Make_Set(int x)
{
father[x] = x;
rank[x] = 0;
}

/* 查找x元素所在的集合,回溯时压缩路径 */
int Find_Set(int x)
{
if (x != father[x])
{
father[x] = Find_Set(father[x]);
}
return father[x];
}

/* 合并x,y所在的集合 */
void Union(int x, int y, int w)
{

if (x == y) return;
if (rank[x] > rank[y])
{
father[y] = x;
}
else
{
if (rank[x] == rank[y])
{
rank[y]++;
}
father[x] = y;
}
sum += w;
}

int main()
{
int i, j, k, m, n, t;
char ch;
while(cin >> m && m != 0)
{
k = 0;
for (i = 0; i < m; i++) Make_Set(i);
for (i = 0; i < m - 1; i++)
{
cin >> ch >> n;
for (j = 0; j < n; j++)
{
cin >> ch >> e[k].w;
e[k].x = i;
e[k].y = ch - 'A';
k++;
}
}

sort(e, e + k, cmp);

sum = 0;

for (i = 0; i < k; i++)
{
Union(Find_Set(e[i].x), Find_Set(e[i].y), e[i].w);
}

cout << sum << endl;
}
return 0;
}
```

### 网友最新评论 (1)

1. 很好理解，谢谢！
yuhe9年前 (2011-08-25)回复