[备战NOI同步赛]Kruskal最小生成树模板

/*
Kruscal最小生成树模板 
By:qpswwww(ZYK)
调用接口:并查集模板 、最短路模板(结构体Line) 
*/
int Kruscal()
{
    int i,j,tot=0;
    sort(edges+1,edges+n,cmp);
    for(i=0;i<MAXN;i++) makeSet(i);
    for(i=1;i<n;i++)
    {
        if(findSet(edges[i].from)!=findSet(edges[i].to)) 
		{
			Union(findSet(edges[i].from),findSet(edges[i].to)); 
			tot+=edges[i].w; //加边
		}
    }
	return tot;
}

例题:POJ 1251 Jungle Roads

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

Mid-Central USA 2002

 

此题就是个裸最小生成树，连预处理都不需要，根据输入(每一行第一个字母为起点,第一个数字为与起点连接的边数，后面的每个字母、数字为终点、边权)直接建立单向边即可，但是我还是在Kruskal上写错被坑了一把。。。

/*
Kruscal最小生成树模板 
By:qpswwww(ZYK)
调用接口:并查集模板 、最短路模板(结构体Line) 
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>

#define INF 0x3f
#define MAXN 1000
#define MAXM 10000 

using namespace std;

int f[MAXN],depth[MAXN],num[MAXN]; //f[i]=点i的父节点,depth[i]=点i的深度,num[i]=i结点对应下面的元素个数

struct Line
{
	int from,to,w,next; //起点,终点,权值,以该边起点为起点的下一条边在邻接表中对应编号
}edges[MAXN]; //静态邻接表，SPFA用

int head[MAXM],n=1; //head[i]=以i为起点的第一条边在edges数组中的位置,n=边的个数 

int cmp(Line a,Line b)
{
    return a.w<b.w;
}

void makePic(int u,int v,int w) //SPFA用,建立u->v,边权为w的有向边,输入边前一定要注意把head数组全部置为-1,非常重要,否则SPFA会死循环!!!!
{
	edges[n].from=u;
	edges[n].to=v;
	edges[n].w=w;
	edges[n].next=head[u]; //记录上一条边的编号
	head[u]=n++; //更新第一条(最后一条)边的编号
}

void makeSet(int i) //并查集初始化
{
	f[i]=i;
	num[i]=1;
	depth[i]=1;
}

int findSet(int x) //带路径压缩的并查集查找
{
	if(f[x]==x) return x;
	return f[x]=findSet(f[x]);
}

void Union(int a,int b) //将a、b合并,有深度优化
{
	if(a==b) return;
	if(depth[a]>depth[b]) {f[b]=a; num[a]+=num[b];}
	else if(depth[a]<depth[b]) {f[a]=b; num[b]+=num[a];}
	else {f[a]=b; depth[b]=depth[b]+1; num[b]+=num[a];}
}

int Kruscal()
{
    int i,j,tot=0;
    sort(edges+1,edges+n,cmp);
    for(i=0;i<MAXN;i++) makeSet(i);
    for(i=1;i<n;i++)
    {
        if(findSet(edges[i].from)!=findSet(edges[i].to)) 
		{
			Union(findSet(edges[i].from),findSet(edges[i].to)); 
			tot+=edges[i].w; //加边
		}
    }
	return tot;
}

int main()
{
    int i,j,num,k,s;
    while(1)
    {
		n=1;
        memset(head,-1,sizeof(head));
        scanf("%d",&num);
        if(num==0) return 0;
        for(i=1;i<num;i++)
        {
            char u[3],v[3];
            scanf("%s%d",u,&k);
            for(j=0;j<k;j++)
            {
                scanf("%s%d",v,&s);
                makePic(u[0]-'A'+1,v[0]-'A'+1,s); //建边 
            }
        }
        printf("%d\n",Kruscal()); 
    }
}