算法：最小生成树-模板+例题+（prim代码模板）
例题：
There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.
有N个村庄，编号从1到N，你应该建造一些道路，使每两个村庄可以相互连接。我们说两个村A和B是相连的，当且仅当A和B之间有一条道路，或者存在一个村C以便在A和C之间有一条道路，并且C和B相连。
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
我们知道一些村庄之间已经有一些道路，你的工作就是修建一些道路，使所有村庄都连通起来，所有道路的长度都是最小的。我们知道一些村庄之间已经有一些道路，你的工作就是修建一些道路，使所有村庄都连通起来，所有道路的长度都是最小的。
Input
The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.
输入
第一行是整数N（3 <= N <= 100），这是村庄的数量。然后是N行，其中第i个包含N个整数，这些N个整数中的第j个是村庄i和村庄j之间的距离（距离应该是[1,1000]内的整数）。
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
然后是整数Q（0 <= Q <= N *（N + 1）/ 2）。然后是Q行，每行包含两个整数a和b（1 <= a
Output
You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.
输出
您应该输出一个包含整数的行，该整数是要构建的所有道路的长度，以便连接所有村庄，并且此值最小。
Sample Input
3
0 990 692
990 0 179
692 179 0
1
1 2
Sample Output
179
ac代码：
#include
#include
#include
#define MAX 0x3f3f3f3f
using namespace std;
int logo[1010];//用来标记0和1 表示这个点是否被选择过
int map1[1010][1010];//邻接矩阵用来存储图的信息
int dis[1010];//记录任意一点到这个点的最近距离
int n;//点个数
void prim()
{
int i, j, now;
int sum = 0;
/*初始化*/
for (i = 1; i <= n; i++)
{
dis[i] = MAX;
logo[i] = 0;
}
/*选定1为起始点，初始化*/
for (i = 1; i <= n; i++)
{
dis[i] = map1[1][i];
}
dis[1] = 0;
logo[1] = 1;
/*循环找最小边，循环n-1次*/
for (i = 1; i < n; i++)
{
now = MAX;
int min1 = MAX;
for (j = 1; j <= n; j++)
{
if (logo[j] == 0 && dis[j] < min1)
{
now = j;
min1 = dis[j];
}
}
if (now == MAX)
break;//防止不成图
logo[now] = 1;
sum += min1;
for (j = 1; j <= n; j++)//添入新点后更新最小距离
{
if (logo[j] == 0 && dis[j] > map1[now][j])
dis[j] = map1[now][j];
}
}
if (i < n)
printf("?\n");
else
printf("%d\n", sum);
}
int main()
{
int m;
int s, e;
while (~scanf("%d", &n))
{
memset(map1, 0, sizeof(map1)); //清空vis背包
for (int i = 1; i <= n; i++) //输入格式处理
{
for (int j = 1; j <= n; j++)
{
scanf("%d", &map1[i][j]);
}
}
scanf("%d", &m);
while (m--)
{
scanf("%d %d", &s, &e); //每输入一条两个村庄之间已经有的道路，也就是道路距离为0
map1[s][e] = 0;
map1[e][s] = 0;
}
prim(); //开始寻找权重总和
}
return 0;
}
模板代码；
#include
#include
#include
#define MAX 0x3f3f3f3f
using namespace std;
int logo[1010];//用来标记0和1 表示这个点是否被选择过
int map[1010][1010];//邻接矩阵用来存储图的信息
int dis[1010];//记录任意一点到这个点的最近距离
int n;//点个数
int prim()
{
int i, j, now;
int sum = 0;
/*初始化*/
for (i = 1; i <= n; i++)
{
dis[i] = MAX;
logo[i] = 0;
}
/*选定1为起始点，初始化*/
for (i = 1; i <= n; i++)
{
dis[i] = map[1][i];
}
dis[1] = 0;
logo[1] = 1;
/*循环找最小边，循环n-1次*/
for (i = 1; i < n; i++)
{
now = MAX;
int min1 = MAX;
for (j = 1; j <= n; j++)
{
if (logo[j] == 0 && dis[j] < min1)
{
now = j;
min1 = dis[j];
}
}
if (now == MAX)
break;//防止不成图
logo[now] = 1;
sum += min1;
for (j = 1; j <= n; j++)//添入新点后更新最小距离
{
if (logo[j] == 0 && dis[j] > map[now][j])
dis[j] = map[now][j];
}
}
if (i < n)
printf("?\n");
else
printf("%d\n", sum);
}
int main()
{
while (scanf("%d", &n), n)//n是点数
{
int m = n * (n - 1) / 2;//m是边数
memset(map, 0x3f3f3f3f, sizeof(map));//map是邻接矩阵存储图的信息
for (int i = 0; i < m; i++)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
if (c < map[a][b])//防止重边
map[a][b] = map[b][a] = c;
}
prim();
}
}
这个算法模板的使用，基本就是输入格式不一样而已，
原模板输入格式是第一行输入一个数n ，然后接下来的n行分别输入三个数，第一个数表示某个点，第二个数表示另一个点，第三个数表示这两个点之间的距离
总的来说这个算法里面的函数，把对应点和对应边输入处理好后直接就有最小权重和。