最小生成树 Kruscal与Prime HDU 1233 还是畅通工程 HDU 1301 Jungle Roads

        最小生成树的两种算法，Kruscal的思路是将所有的边由小到大排序，然后依次选出符合条件的最短边，最终让整个图联通，其中用结构体数组储存测试数据，用并查集检查图中 点是否连接；Prime则是对点的研究，任意选取一个点，然后选择与之连接的最短边，在此，使用二维数组存储测试数据，并且用book数组避免形成回路。

        两种算法都能达到相同的目的，对于一般的最小生成树问题，Kruscal算法速度更快一些，而且相对容易理解，但是如果遇到边比较多的问题，Kruscal算法就不是很划算了。

还是畅通工程

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

某省调查乡村交通状况，得到的统计表中列出了任意两村庄间的距离。省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可），并要求铺设的公路总长度为最小。请计算最小的公路总长度。

 

Input

测试输入包含若干测试用例。每个测试用例的第1行给出村庄数目N ( < 100 )；随后的N(N-1)/2行对应村庄间的距离，每行给出一对正整数，分别是两个村庄的编号，以及此两村庄间的距离。为简单起见，村庄从1到N编号。
当N为0时，输入结束，该用例不被处理。

 

Output

对每个测试用例，在1行里输出最小的公路总长度。

 

Sample Input

3
1 2 1
1 3 2
2 3 4
4
1 2 1
1 3 4
1 4 1
2 3 3
2 4 2
3 4 5
0

 

Sample Output

3
5

Hint
Hint
 
Huge input, scanf is recommended.

Kruscal代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int MIXN = 105;

int n;
int pre[MIXN];

struct side
{
    int start;
    int end;
    int length;
}s[MIXN*(MIXN-1)/2];

int cmp (side a, side b)
{
    return a.length < b.length;
}

int find (int v)
{
    if (pre[v] == v)
    {
        return v;
    }
    else
    {
        pre[v] = find (pre[v]);
        return pre[v];
    }
}

int mix (int x, int y)
{
    int t1, t2;
    t1 = find (x);
    t2 = find (y);
    if (t1 != t2)
    {
        pre[t2] = t1;
        return 1;
    }
    return 0;
}

int main ()
{
    int cnt, sum;
    while (scanf ("%d", &n) != EOF)
    {
        if (n == 0)
            break;
        cnt = 0;
        sum = 0;
        memset (s, 0, sizeof (s));
        memset (pre, 0, sizeof (pre));
        for (int i=1; i<=n*(n-1)/2; i++)
        {
            scanf ("%d %d %d",&s[i].start,&s[i].end,&s[i].length);
        }
        sort (s+1, s+n*(n-1)/2+1, cmp);
        for (int i=1; i<=n; i++)
        {
            pre[i] = i;
        }
        for (int i=1; i<=n*(n-1)/2; i++)
        {
            if (mix (s[i].start, s[i].end))
            {
                cnt ++;
                sum += s[i].length;
            }
            if (cnt == n-1)
                break;
        }
        cout << sum << endl;
    }
    return 0;
}

Jungle Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem 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.

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.

 

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

Prime代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int inf = 0x7fffffff;

int main ()
{
    int n, k;
    char a, b;
    int c;
    int e[30][30], dis[30], book[30];
    char s[100];
    int cnt, sum;
    int minx, mini;
    while (scanf ("%d", &n) != EOF)
    {
        if (n == 0)
            break;
        memset (e, 0, sizeof (e));
        memset (dis, 0, sizeof (dis));
        memset (book, 0, sizeof (book));
        sum = 0;
        cnt = 0;
        for (int i=1; i<=n; i++)
        {
            for (int j=1; j<=n; j++)
            {
                if (i == j)
                    e[i][j] = 0;
                else
                    e[i][j] = inf;
            }
        }
        for (int i=1; i<n; i++)
        {
            getchar ();
            scanf ("%c %d", &a,&k);
            for (int j=1; j<=k; j++)
            {
                scanf (" %c %d", &b,&c);
                e[i][b-64] = c;
                e[b-64][i] = c;
            }
        }
        for (int i=1; i<=n; i++)
        {
            dis[i] = e[1][i];
        }
        book[1] = 1;
        cnt ++;
        while (cnt < n)
        {
            minx = inf;
            for (int i=1; i<=n; i++)
            {
                if (book[i]==0 && dis[i]<minx)
                {
                    minx = dis[i];
                    mini = i;
                }
            }
            book[mini] = 1;
            cnt ++;
            sum += dis[mini];
            for (int i=1; i<=n; i++)        //i从2开始也可
            {
                if (book[i]==0 && dis[i]>e[mini][i])        //book[] == 0 即下一站没有连接，避免形成回路
                    dis[i] = e[mini][i];        //如果mini点到未被访问的点i的距离 比1点到i点的距离小，则用e[mini][i]代替dis[i]，即e[mini][i]为连接到i点的最短距离
            }
        }
        cout << sum << endl;
    }
    return 0;
}